The Renard 8 DC

[christmas-light]

Hi Every one.

I have designed a new controller to the family (Only the scematick (yet :D )), it is a 8 Channel controller there only can control DC power.

I know that I have a old Thread with a DC 16 controller there is under construction, But I gave up that controller, because the scematick was way too bad :(

So Now I have made a New scematick there is much easier to read, and hope fully, it should also be okay.

It is based on the SSR DC (Wjohn) and the Simpel Renard 8

Why I have made this controller scematick, it is because I am going to replace alot fo my lights with 24 VDC lights, and to controll the new lights I need new controllers :D

Here is the scematick I hope there is some good souls out there, there want to help me check it :)

[christmas-light]

Any One ?

[kychristmas]

I'm not good enough to provide input, but I was curious if you were planning on Home Etching this new design or having them manufactured?

[christmas-light]

Well if it is possible then I want it to be Home Etching but I am not sure that it is enough room on the pcb for that, also, so I think it will be manufactured ;)

[dirknerkle]

A single-sided Renard can be tough to design while a 2-sided board is hard to home-etch. You've got your work cut out for you!

[P. Short]

What is a 788TV?

The opto-isolators are somewhat pointless when you have the same +5V going to both sides of the opto. Perhaps that was just a 'typo'?

In any case, I don't think that it would be impossible to make this a one-sided, home-etch board.

[steve_hirst]

we home etched some simple16s this year and they went great.

[dirknerkle]

Not saying it's not possible, just that you have to have your ducks in a row...

[christmas-light]

I will find out the Etching later, But is my scematick right ?

[christmas-light]

What is a 788TV?

The opto-isolators are somewhat pointless when you have the same +5V going to both sides of the opto. Perhaps that was just a 'typo'?

In any case, I don't think that it would be impossible to make this a one-sided, home-etch board.

The Tow 788TV is tow volt regulators, I has not found out which one I need yet, but I just took the component and added it to the secmatick, just to show that there is tow volt reg.

"The opto-isolators are somewhat pointless when you have the same +5V going to both sides of the opto. Perhaps that was just a 'typo'?"

Can you tell me more about that ?

[Skunberg]

Not saying it's not possible, just that you have to have your ducks in a row...

You mean I can't have my duck's in a basket? VBG

[christmas-light]

Is there anyone there have a idea about my scematick is right or not ?? :confused:

[dirknerkle]

You mean I can't have my duck's in a basket? VBG

Nope. No ducks in baskets. They HATE baskets. :lol:

[Slashdude]

I'm not an expert, but here is a start at what I see: (Experts out there, Please correct me if I'm wrong :))
On the RJ-45's, I believe you need pins 1 and 2 connected to ground.

R3 & JP1 should be on the other side of R1 and R2

I don't know if it matters, but R4 & the 2 diodes are connected to the RJ45 pin 4 on the REN ss

Why is Pin 4 on ST485-IC2 connected to pin1 on ST485-IC3? (This shows NC - not connected)

Hope that helps

[ErnieHorning]

"The opto-isolators are somewhat pointless when you have the same +5V going to both sides of the opto. Perhaps that was just a 'typo'?"

Can you tell me more about that ?Google doesn't return anything on a regulator called 788TV. Usually if Google doesn't know about something, it doesn't exist. Maybe it's not the full part number?:confused:

[christmas-light]

Google doesn't return anything on a regulator called 788TV. Usually if Google doesn't know about something, it doesn't exist. Maybe it's not the full part number?:confused:

It is because it is a eagle part nr. not a real volt reg part nr. ;)


I'm not an expert, but here is a start at what I see: (Experts out there, Please correct me if I'm wrong :))
On the RJ-45's, I believe you need pins 1 and 2 connected to ground.

R3 & JP1 should be on the other side of R1 and R2

I don't know if it matters, but R4 & the 2 diodes are connected to the RJ45 pin 4 on the REN ss

Why is Pin 4 on ST485-IC2 connected to pin1 on ST485-IC3? (This shows NC - not connected)

Hope that helps

Thanks

I will say that most of it sounds right, but is there some one there can confirm this ? :)

[budude]

OK - a things... Why are there two voltage regulators? The one should be able to handle the load for just 8 channels? Also - I would simply copy the ENTIRE front-end of the RenSS series as I did with the Ren48LSD - this is to allow for ZC, power passage to downstream devices if necessary. It's easy to add it now with a few traces even if you don't need it now.

[christmas-light]

The reason why there is two voltage regulators is because if I want to control eg 48 VDC then I will be needing the two voltage regulators because I don´t think that one voltage regulator can handel a drop down from 48 VDC to 5 VDC ?

Can you explain deeper the thing you is saying about the ZC ? Because I just removed it and added a 10 K ohm resistor instead ?

I will take a deeper look at the renard 8 SS scematick.

[christmas-light]

Okay Now I have updated the secmatick, but I know that there is at leat one problem, on IC2 and IC3 I don´t know where pin 4 and pin 1 shall go to ?

[budude]

Well - none of the "common" regulators (LM78xx for example) go up to 48vdc for an input - it's only like 32 or so (going from memory) and even that can lead to high power dissipation for it. There are some DC-to-DC converters that might work - they're used a ton for POE stuff in some form. Provide a link to the regulators you plan to use to confirm.

As for ZC, etc - what I mean is that the "standard" Renards pass signals/power from the input to the output RJ45. Pin 3 for example can be used to provide ZC from whatever source, Pins 7 & 8 can pass power (don't recall specs) - this is mainly for the older Renards like the Ren16, etc. You probably won't need or miss it but since it costs nothing to add, I would just toss in a few traces to be the same as the RenSS line - the schematics for them are in the Wiki - just copy that - again - just a suggestion...


The reason why there is two voltage regulators is because if I want to control eg 48 VDC then I will be needing the two voltage regulators because I don´t think that one voltage regulator can handel a drop down from 48 VDC to 5 VDC ?

Can you explain deeper the thing you is saying about the ZC ? Because I just removed it and added a 10 K ohm resistor instead ?

I will take a deeper look at the renard 8 SS scematick.

[christmas-light]

Well - none of the "common" regulators (LM78xx for example) go up to 48vdc for an input - it's only like 32 or so (going from memory) and even that can lead to high power dissipation for it. There are some DC-to-DC converters that might work - they're used a ton for POE stuff in some form. Provide a link to the regulators you plan to use to confirm.

As for ZC, etc - what I mean is that the "standard" Renards pass signals/power from the input to the output RJ45. Pin 3 for example can be used to provide ZC from whatever source, Pins 7 & 8 can pass power (don't recall specs) - this is mainly for the older Renards like the Ren16, etc. You probably won't need or miss it but since it costs nothing to add, I would just toss in a few traces to be the same as the RenSS line - the schematics for them are in the Wiki - just copy that - again - just a suggestion...

Thanks for your suggestion budude.

I have not found out which volt reg I need yet, but I am going to find out soon.

Have you seen my update on the secmatick ?

[ctmal]

To solve the regulator issue you could always bring the dc for the pic side of the board in separately from the power for the outputs. You could bring it in on the RJ45 port like the XMUS board does or a separate jack. This would also give you options down the road if you just wanted to run 24v, 12v, etc. items on the output.

[aussiephil]

Re Voltage Reg

For Linear regulators there are only a couple i know that will handle input >36V and to drop down to 5V would require some LARGE heatsinks even at 100mA.

The only effective way is to use a switching regulator or provide a seperate board power feed.

Phil

[ErnieHorning]

You can use a zener on the input to drop some of the voltage. It has to be able to pass enough current to charge your input cap the first time though and whatever your max run current is.

You can use a common base transistor to drop the voltage. It's still linear so watts still comes out.

Depending on what your run current is, you can use a series resistor to limit the input current and thus the voltage. Again there's heat and wasted power.

A switching supply takes space and more parts but doesn't waste power into heat.

Externally regulating the voltage down would make your controllers easier to design though.

[christmas-light]

To solve the regulator issue you could always bring the dc for the pic side of the board in separately from the power for the outputs. You could bring it in on the RJ45 port like the XMUS board does or a separate jack. This would also give you options down the road if you just wanted to run 24v, 12v, etc. items on the output.

I have selected to do what ctmal says because it will be much easier.

I have updated the secmatick again, and this is what I have updated:

I have added / removed some parts, so basically it is the power modul from the renard 16 xmus

I have added a capacitor on the pic 16F688

I have added 1 large capacitor at the power in, just to keep noise away and also, if the board meed "alot" of mA then it can get it from that capacitor, (it will be the same cab that there is used on the Renard 64 XC)

I still don´t know where pin 4 and pin 1 on IC2 and IC3 shall go to ?

And I also think that there is a problem at the opto / trace ?
:p

Best regards

[ErnieHorning]

I still don´t know where pin 4 and pin 1 on IC2 and IC3 shall go to ?

Pin 4 of IC2 should be grounded because it's an input and you don't want to leave these floating. Pin 1 can be left open; It's an output that's not being used.

And I also think that there is a problem at the opto / trace ?Well if both sides are connected to +5 then there is really no point in having it there. To make isolation effective, you should connect the output + to the supply that you're using for the FET's. Or are you supplying +5 from two seperate supplies that are't connected in any way?

[P. Short]

A couple of comments on 'Renard DC8 v2.0.pdf'

1) The LEDs on page 2 appear to be oriented in the wrong direction. Based on connecting +5V to the resistor network, I'm assuming that the outputs of the PIC are supposed to be active low. This means that you have the cathodes and anodes of the status LEDs reversed.

2) As I've said before (but didn't really explain) the opto-isolators are somewhat useless as long as you have the same +5v on both sides of the opto. In the DC SSR boards (such as those from wjohn) the input and output sides of the SSR have complete galvanic isolation from each other. This means that there isn't any path where current can flow from the input side of the board to the output side (or vice-versa). In particular, those boards have two separate 'grounds' and two separate DC power sources (without any connection between them), and the opto-isolator makes sure that there isn't any direct connection through the control paths.

This separation is a carry-over from the AC SSRs, where the separation is absolutely mandatory for safety reasons (so that you don't get shocks, and to protect the computers from AC power-line voltages). This separation was carried over into the DC SSR designs, as an added safety precaution.

In your schematic, you have the same +5V power signal on both sides of the opto-isolator. This provides a DC path between the input and output sides of the SSR section, completely defeating any measures at galvanic isolation. For your design you must decide if you want this galvanic isolation (it would be a good idea, as it adds a degree of flexibility in using the board, and makes it easier to prevent ground loops). In order to do this, you need to have two separate ground circuits (one for the RS485 jacks/RS485 chips/PICs) and one for the MOSFETS (and possibly the load that they are driving). You also need two sources of DC power (again, one for the RS485 chips/PICs/LED resistors and one for the MOSFET gate circuitry), isolated from each other.

[christmas-light]

Okay I see now, thanks ;)

Well the part have to be connected that way, that means that I will be needing a volt regulator that can handle up to 50 VDC and dorp it down to 5 VDC?

That is hard to find, but I have been looking at this one http://dk.mouser.com/ProductDetail/Fairchild-Semiconductor/LM317AHVT/?qs=sGAEpiMZZMtUqDgmOWBjgKu8a7X7FsEiay9rqSFuX5c%3d but I don´t know if that will works ?

Also I have updatede the secmatick:

I have updated:
1 the led´s
2 IC2 on ping 4
3 The opto / Traice problem, but not added the volt regulator, Have to finde one there can be used to this controller and can handle up to 50 VDC

Best regards

[P. Short]

For the power supply on the output side of the opto's I would tend to use just a resistor and a zener diode, such as the 1N5993 (selected for the relatively low VZT). The gate drive for the MOSFETs shouldn't take that much current.

[ErnieHorning]

BTW, the LED in the ILQ1 requires 20mA. The PIC spec only allows 95mA total out the ground pin. The ILQ2 only requires 5mA drive which keeps the PIC in spec.

Oh and speaking of the PIC, power and ground are backward.:roll:

[christmas-light]

BTW, the LED in the ILQ1 requires 20mA. The PIC spec only allows 95mA total out the ground pin. The ILQ2 only requires 5mA drive which keeps the PIC in spec.

Oh and speaking of the PIC, power and ground are backward.:roll:

Sorry, I have just edited that, Thanks ;)


For the power supply on the output side of the opto's I would tend to use just a resistor and a zener diode, such as the 1N5993 (selected for the relatively low VZT). The gate drive for the MOSFETs shouldn't take that much current.

I am sorry that I am asking all these dum questions :roll:

But can you explain that more deeper, what you mean ?

Do you suggest that I am adding a resistor and a diode on each indput/output on the MOSFET ?

If so then how many ohm should the resistor be on ? :smile:

I have made 4 examples, of what I think that you is saying.

They can bee seen here:

http://pcjulelys.dk/suggestions/1.png
http://pcjulelys.dk/suggestions/2.png
http://pcjulelys.dk/suggestions/3.png
http://pcjulelys.dk/suggestions/4.png

I have also updated the secmatick:

Best regards

[ErnieHorning]

I have made 4 examples, of what I think that you is saying.

Either 3 or 4, the order doesn't matter except the symbol should look more like this with the little flags on the side.
9478

[christmas-light]

Once again THANKS :smile:

Here is the updated secmatick:

am I right that the resistor valu changes ~ How many V there is going in to the controller (just like the AC SSR ) ?

Best regards

[ErnieHorning]

In this case the zener drops the voltage by the value of the specific part. The resistor limits the current in the zener. The value of the resistor depends on the supply voltage, the voltage value of the zener and how much current is needed for the zener to operate.

Do a Google search for "how zerners work" for further explanation.

[christmas-light]

Okay Thanks.

I have found this calculator:

http://www.reuk.co.uk/Zener-Diode-Voltage-Regulator.htm

I am wondering if it is use full ?

I am just wondering how I calculate it ?

Also it the rest of the secmatick looking okay ? :)

Best regards

[Matt_Edwards]

I don't see this being addressed in the thread, but is there a reason for the OPTOs? Generally if the SSR portion of the cct is on the same board as the PIC and you PIC power supply is generated from the same as your light source you can get away without the OPTO. It may same you some space.

[P. Short]

I don't see this being addressed in the thread, but is there a reason for the OPTOs? Generally if the SSR portion of the cct is on the same board as the PIC and you PIC power supply is generated from the same as your light source you can get away without the OPTO. It may same you some space.

The issue isn't so much the power supply or PIC as isolation between the RJ45 connectors and the SSR outputs (assuming that you want isolation). However, the usefulness of isolation is a bit questionable IMO for DC SSRs, since it is usually not an issue to connect all of the grounds together.

[Matt_Edwards]

So this is not an issue with Renards as such, it is the fact that most operate using RS323. If RS485 was used and the screen on the comms cable is not used, it shouldn't be a problem.

[P. Short]

I'm not sure what you're saying...

[Matt_Edwards]

There is quite a lot of DMX DC kit already in use that don't use Optos nor have any issues. I was wondering if this was a renard only issue due to RS232 using the Ground / return signal.

[P. Short]

For any sort of controller for AC, you need opto-isolators (or something else that provides substantially the same degree of galvanic isolation.

As far as the Renard DC controllers are concerned, I personally do not see the need for galvanic isolation in the majority of cases, and would not have included it myself if I were designing the boards. There may be some circumstances where it would be useful, perhaps to avoid ground loops when using multiple power supplies. I do not know how often people would actually need this isolation.

However, the people who designed most of the DC SSRs felt that it would be a good idea to have isolation, and that has been carried forward from those initial designs. SInce the DC SSRs were not my project, and I didn't think that it hurt anything to have isolation, I decided not to enter into any real discussion on this.

[christmas-light]

The reason why I have used the DC SSR Design is because I think I am going to use diffrent power supply´s on the same board and also because of the saftey and also I know that the DC SSR can handle alot of Volt and Amp so that is fine with me :)

can some one explain how to calculate the resistor value for the Zener Diode ?

also how is the rest of the secmatick looking ?

[Matt_Edwards]

For any sort of controller for AC, you need opto-isolators (or something else that provides substantially the same degree of galvanic isolation.

As far as the Renard DC controllers are concerned, I personally do not see the need for galvanic isolation in the majority of cases, and would not have included it myself if I were designing the boards. There may be some circumstances where it would be useful, perhaps to avoid ground loops when using multiple power supplies. I do not know how often people would actually need this isolation.
i would have thought for Remote SSR's, it would be worthwhile in some circumstances.


However, the people who designed most of the DC SSRs felt that it would be a good idea to have isolation, and that has been carried forward from those initial designs. SInce the DC SSRs were not my project, and I didn't think that it hurt anything to have isolation, I decided not to enter into any real discussion on this.
No worries. I am all for people having a go at doing things themselves. I won't cloud the discussion here.

[christmas-light]

I am all for people having a go at doing things themselves.

I don´t know if I shall say me too :lol: ;)

[ErnieHorning]

Anything that isn’t powered directly from line voltage, I would expect to be isolated anyway. The transformers in the supplies would provide the isolation. At work, we rely on the transformers for isolation but if we’re connecting to equipment that runs directly off of line voltage, we’ll isolate it with a 1:1 transformer.

That said, you can’t guarantee that unknown low voltage power supplies will always be 100% isolated. We have some equipment at work that uses a 120 to 24 AC transformer. You would think that it’s isolated from the line. It turns out that one side of the 24 AC output is connected to earth ground. I wanted to monitor a microcontroller’s UART port, so I connected an RS232 level translator board to the board on the equipment. .I connected a Notebook computer’s (Compaq I think) RS232 port to a level translator and promptly blew out the communications port. The Notebook’s power supply was external with only a two prong polarized plug. That particular model was the only one that had this problem. I measured something like 50 volts of leakage. It was an older computer at the time, so it wasn’t a huge loss, just disappointing.

Something else to ponder: If everything in your display is low voltage DC then I doubt that you will have a problem. If you’re also running high voltage AC, it’s certainly possible to return a substantial amount of voltage back on the ground or control line if you don’t provide some type of filtering. Sometimes it’s virtually impossible to keep the wires separated enough. You could be hanging several feet of wire outside and depending on the number of DC channels; this would have an additive effect. In this case the isolation would be a benefit.

Maybe this is all hog wash but I like to think of the “what if’s” before instead of the “aw crap” later.

[dirknerkle]

maybe this is all hog wash but i like to think of the “what if’s” before instead of the “aw crap” later.



roflmao!!

[christmas-light]

Anything that isn’t powered directly from line voltage, I would expect to be isolated anyway. The transformers in the supplies would provide the isolation. At work, we rely on the transformers for isolation but if we’re connecting to equipment that runs directly off of line voltage, we’ll isolate it with a 1:1 transformer.

That said, you can’t guarantee that unknown low voltage power supplies will always be 100% isolated. We have some equipment at work that uses a 120 to 24 AC transformer. You would think that it’s isolated from the line. It turns out that one side of the 24 AC output is connected to earth ground. I wanted to monitor a microcontroller’s UART port, so I connected an RS232 level translator board to the board on the equipment. .I connected a Notebook computer’s (Compaq I think) RS232 port to a level translator and promptly blew out the communications port. The Notebook’s power supply was external with only a two prong polarized plug. That particular model was the only one that had this problem. I measured something like 50 volts of leakage. It was an older computer at the time, so it wasn’t a huge loss, just disappointing.

Something else to ponder: If everything in your display is low voltage DC then I doubt that you will have a problem. If you’re also running high voltage AC, it’s certainly possible to return a substantial amount of voltage back on the ground or control line if you don’t provide some type of filtering. Sometimes it’s virtually impossible to keep the wires separated enough. You could be hanging several feet of wire outside and depending on the number of DC channels; this would have an additive effect. In this case the isolation would be a benefit.

Maybe this is all hog wash but I like to think of the “what if’s” before instead of the “aw crap” later.


Yes your right and that is why I have designed the controller with the design of the DC SSR.

Is there any one there pls can help me / explain how I calculate the resistor value to the Zener Diode

and it there some one there pls can check my secmatick a last time just to make sure that there is no mistakes

please :oops:

[ErnieHorning]

Is there any one there pls can help me / explain how I calculate the resistor value to the Zener Diode
1. Determine how much current you need to flow thru the Zener to achieve the desired operating point. In general, the test current from the data sheet is a good value to use. Lets assume its Iz = 10 mA.

2. Determine the minimum unregulated voltage. Let's assume its Vinmin=35V.

3. Determine the maximum Zener voltage at the test current. Let's assume its Vxmax = 12.2V.

4. Calculate the resistor value that will provide the required Zener current under worst case (minimum) input voltage and worst case (maximum) Zener voltage.

5. R = (Vinmin-Vzmax)/ Iz = (35-12.2)/0.01A = 2280 Ohms.

6. Power = i^2R = (.01^2) X 2280 = .228 W.

A 1/4W resistor would barely squeak by but would run hot, so I’d step up to a 1/2 watt.

[christmas-light]

Okay thanks.

Here is my try, just to see if I have understood it right

My power supply is 48 VDC

The Zener diode is a 1N5993

The Zener Max. DC Current is 98 VDC

Normal Zener Volt is 5.1 V, Test Current is 5.0 mA

R = (48-98 ) / 0.005A = ?

My resistor value is then ?

Some thing is not correct in my calculation :roll:

What did I do wrong?

I have uploaded the datasheet on the 1N5993

Best regards

[ErnieHorning]

In line 3 above, it was supposed to read: Vzmax = 12.2V. This is the zener voltage at whatever the test current was, which could increase the voltage slightly.

So, R = (Vinmin-Vzmax)/ Iz = (48-5.1)/0.005A = 8580 Ohms.

Power = i^2R = (.005^2) X 8580 = .2145 W.

A 1/4W resistor would still run too hot. Never go more than 80% on a part which would be .200 watts on a .25 watt part. I like to design for no more than 50% (which is an industry recommendation) so I’d still step up to a 1/2 watt.

8580 Ohms would be a precision part, which you don’t need for this application. The next closest common one down would be 8200 or 8.2K. This would increase the zener current to about.0012 and the watts would go to .224.

At 98 volts the resistor would dissipate 1.17 watts, plus an additional 20% is 1.404 watts. So if you want to allow 98 volts, you’ll need at least a 1.5 watt resistor.

[christmas-light]

Thanks ErnieHorning

I have finally understood it :D

I have also made a "calculator" for it

Can I get you to do one thing more, just check my secmatick a last time, just to be sure that it is okay ?

The calculator can be found here:

www.pcjulelys.dk/Zener%20Resistor%20Value.xlsx

Best regards

[P. Short]

If 48V is the maximum input voltage that you would be using, what is the minimum? For calculating the value of the resistor you need to use the minimum input voltage, although you need to use the maximum input voltage for calculating the power rating on the resistor (while also ensuring that you don't exceed the power rating of the zener). I think that Ernie was making an assumption for the minimum voltage...

[P. Short]

And if you are offended by the relatively huge amount of power dissipated by the resistor in the zener circuit, there are some fairly inexpensive shunt regulators that work on the same principle as the zener but will regulate at much lower current levels. They are the LM431, TL431 and the TLV 431. The first two parts will generally maintain regulation at 1 mA, the the third part (a bit more expensive) will go down to 0.1 mA.

[christmas-light]

well I think that the minimum input voltage is also eg 48 V because if I want to get the controller to run on 24 V the I have to change the resistors ?

[P. Short]

well I think that the minimum input voltage is also eg 48 V because if I want to get the controller to run on 24 V the I have to change the resistors ?

Not necessarily. If you used a resistor with 1/2 the value given above but twice the power rating, you could run with either voltage. You should run the calculations, of course.

[christmas-light]

Not necessarily. If you used a resistor with 1/2 the value given above but twice the power rating, you could run with either voltage. You should run the calculations, of course.
I am a bit confused, lost and much dum, can you explain some, my english is not the best, if you can cut it in to peaces

[P. Short]

OK...

To determine the resistor value, use the minimum input voltage. Assume that it is 24VDC, and use the numbers from your example. The resistor value should then be (24V - 5.1V)/.005A = 3780 Ω.

For determining the power rating of that resistor, use the maximum input voltage. The power rating should be (48V-5.1V)*((48V-5.1V)/3780Ω) = .48W, which you should probably increase to either .75W or 1W for adequate margin.

Incidently, if I were you I would use a 9.1V zener instead of 5.1V, because the MOSFETs perform better with a higher gate voltage.

Edit: You should make the decision whether or not to make it run at a lower voltage as well as 48V. This is all just an example.

[christmas-light]

Okay I understand your point, but I can´t see why, because:

(I will change the Zener diode to 1N5999)

If eg we say that the power supply is 24 VDC

Then the resistor value should be 2980 Ohm (based on the 1N5999)

And the resistor watt value should be 0,075 Watt but then we add 50% extra watt of that value then it will be 0,112 Watt, just to be sure that it can handle enough watt

The thing I am trying to say is that I can´t see why I shall calculate the watt at eg 48 V when the power supply only can give 24 VDC ?

[P. Short]

Because it is a design that should work at both 24V and 48V, so that you don't need to change the resistor when you change the voltage.

[christmas-light]

AAAA Now I see that is smart THANKS

You and ErnieHorning great to this ;)

[christmas-light]

okay so fare so good.

I am done with the designing of the PCB

Pls comment it if there is some thing wrong og ask if you have any question ? ;)

Best regards

[ErnieHorning]

The first thing that I see is the huge trace on the output. Remember, what comes out of the FET's, must also come in. Also, you get almost twice the trace width by putting it on both sides.

If you use the Q1 circuit as an example, swapping R12 & R20 locations you can move the 10K closer to Q1 so that it top pin is centered on the 470Ω next to it, you can run a slightly wider trace under the 470Ω.

I know some people like to have both sides of a channels output right next to each other but beign close to circuit board manufacturing, I find that two pin connectors are weak connections. The connector will easily twist when turning a screw and rip the pins right out the holes. I would recomend consolidating some or all of the connection on fewer connectors.

I normally leave the common connection off of the board. This makes the connector cheaper, the board is smaller and the board heat rise is less.

If you take another look at the layout, you can get rid of the via's going to the the two 470Ω (R18 & R21) resistors.

I would recommend that you try to make the ground trace as large as possible to lessen power noise. A ground flood is better though.

I'm assuming that you intend to clean up the routing on RJ45-1 & RJ45-2.

[christmas-light]

The first thing that I see is the huge trace on the output. Remember, what comes out of the FET's, must also come in.



If you use the Q1 circuit as an example, swapping R12 & R20 locations you can move the 10K closer to Q1 so that it top pin is centered on the 470Ω next to it, you can run a slightly wider trace under the 470Ω.



If you take another look at the layout, you can get rid of the via's going to the the two 470Ω (R18 & R21) resistors.

Ok have updated that, and actually I didn´t see that option with R18 and R21 before you said it :D



Also, you get almost twice the trace width by putting it on both sides.



I know some people like to have both sides of a channels output right next to each other but beign close to circuit board manufacturing, I find that two pin connectors are weak connections. The connector will easily twist when turning a screw and rip the pins right out the holes. I would recomend consolidating some or all of the connection on fewer connectors.

Well that can be difficult because I am going to make it all most one site pcb, it is also only jumbers I have added at the top site of the pcb ;)



I would recommend that you try to make the ground trace as large as possible to lessen power noise. A ground flood is better though.

I'm assuming that you intend to clean up the routing on RJ45-1 & RJ45-2.

Can you explain that a little more because I don´t know what it is you are saying.

I don´t know how much I can clean up the routing on the RJ45-1 and RJ45-2 ?

I have uploadet a new .pdf of the pcb

And I am MUCH thanks full that you still want to help me ;) Thanks

Best regards

[n1ist]

I just found this project, and have a few comments:
Schematic:
- C3 should connect after the D1; the way it is now, if the polarity on the RJ45's power is backwards, C3 will blow
- The 7805 needs a 330nF cap on the input and a 100nF cap on the output for stability (but I have used 100nF for both with no problems)
- Add a 100nF bypass cap for the oscillator
- Put the status LEDs in series with the opto LEDs; that will eliminate the resistor networks and halve the amount of current the PIC needs to supply. The 680R values will have to be changed to account for the higher drop across the two diodes
- Something doesn't look right in with the output stage. Right now, you take the supply rail, subtract 5.1v with the zener in series, then divide it and feed 0.8 * that voltage to the gate of the FET. If the output stage supply is more than 30V, you will exceed the VGS breakdown of the FET. Wouldn't it be better to create a single 5.1V zener-based regulated voltage (series resistor, zener to ground) and feed that to all of the optos?

PCB:
- Try to avoid 90 degree bends in traces; two 45-degree bends are better
- Double-check the footprint of the oscillator, most don't have the two mounting pins, and do have a pin at the pin 1 location
- The traces feeding the optos from the output supply, and to the gates of the FETs don't carry much current, so they can be quite a bit thinner
- The traces around the 7805 should be wider to reduce their inductance
- There are a number of places where you have two jumpers in series; I would use a single jumper and eliminate some solder connections. Jumpers don't have to be straight. Of course, if this will truly be a double-sided board, then it's no problem at this speed.
- Traces should enter pads straight on. The trace connecting to pins 7 and 8 on the RJ45s, for example, should go to the middle of pin 8, then diagonally to pin 7, then on to the '485 chip.

/mike

[P. Short]

In terms of combining the eight two-pin connectors into one 16-pin connectors (or two 8-pin connectors), I agree with Ernie. This is especially important for a home-etch PCB. When you use two-pin connectors, there isn't enough mechanical strength, and it will be easy to accidently twist or lift the connectors and thus damage either the solder joints or to lift and break the PCB traces.

[ErnieHorning]

Can you explain that a little more because I don´t know what it is you are saying.
Ground is everything that is connected to pin 2 of IC1. A good ground is very important for a low noise circuit. A poor ground is like throwing a rock in a swimming pool. It will create waves and everything in the pool will jump around violently. Through the same rock into a lake and you'll have little or no effect.

Believe it or not, a large copper area acts like a capacitor to noise. This is the reference for all or the capacitors that are connected to the power supply. If ground and +5V have the same size trace, noise will cause ground and +5V to bounce around with the noise. This may work for a single IC but when it's connected to other circuits which will also be bouncing around at different times and rates, it will act as though the noise is worse.

As part of my day job, I generate several thousand volts of noise around circuit boards and if properly designed they will work flawlessly.


- Try to avoid 90 degree bends in traces; two 45-degree bends are better
The reason for this is that the etching acid will tend to remove more of the inside corner of a 90° angle then it will of a 45° angle. A rounded corner will work even better but a lot of CAD systems don't support it. This is only a problem if you're etching your own circuit boards. This usually isn't an issue with commercially made boards.

I also agree with everything else that Mike mentioned.

[christmas-light]

I just found this project, and have a few comments:
Schematic:
- C3 should connect after the D1; the way it is now, if the polarity on the RJ45's power is backwards, C3 will blow
- The 7805 needs a 330nF cap on the input and a 100nF cap on the output for stability (but I have used 100nF for both with no problems)
- Add a 100nF bypass cap for the oscillator



I have just udated it ;)




- Put the status LEDs in series with the opto LEDs; that will eliminate the resistor networks and halve the amount of current the PIC needs to supply. The 680R values will have to be changed to account for the higher drop across the two diodes



Well I don´t think that I will put the LEDs in serie with the opto LED because If the LED burn out then I have one channel there don´t work. But I was thinking, can I not just connect the + site of the LEDs to the resistor to the opto LED resistor, they is sharing the same resistor ?

Then I dont have to have the resistor network to the leds ?




- Something doesn't look right in with the output stage. Right now, you take the supply rail, subtract 5.1v with the zener in series, then divide it and feed 0.8 * that voltage to the gate of the FET. If the output stage supply is more than 30V, you will exceed the VGS breakdown of the FET. Wouldn't it be better to create a single 5.1V zener-based regulated voltage (series resistor, zener to ground) and feed that to all of the optos?



I can´t tell you, I am bot so good at this with the Zener diode ? ;)

But to me it sounds as a good idea




PCB:
- Double-check the footprint of the oscillator, most don't have the two mounting pins, and do have a pin at the pin 1 location
- The traces around the 7805 should be wider to reduce their inductance
- Try to avoid 90 degree bends in traces; two 45-degree bends are better
- The traces feeding the optos from the output supply, and to the gates of the FETs don't carry much current, so they can be quite a bit thinner



Okay I have also updated that, THANKS

I have also uploadet the new secmatick and pcb layout

Best regards

[christmas-light]

Ground is everything that is connected to pin 2 of IC1. A good ground is very important for a low noise circuit. A poor ground is like throwing a rock in a swimming pool. It will create waves and everything in the pool will jump around violently. Through the same rock into a lake and you'll have little or no effect.

Believe it or not, a large copper area acts like a capacitor to noise. This is the reference for all or the capacitors that are connected to the power supply. If ground and +5V have the same size trace, noise will cause ground and +5V to bounce around with the noise. This may work for a single IC but when it's connected to other circuits which will also be bouncing around at different times and rates, it will act as though the noise is worse.

As part of my day job, I generate several thousand volts of noise around circuit boards and if properly designed they will work flawlessly.



Okay I see :)

I have just added it, (had to finde out how to ;) ) But here it is:

If there is some thing there need to be changed then please tell :D

I have also made a boom for the controller, that have I also uploaded

There is just one "little" problem, I can´t find the Zener dioder on mouser ? and I still not sure witch resistor I need to them ? :oops:

Best regards

[n1ist]

The pads that connect to the ground pour should be thermals (ie, the pad connects to the pour with four "spokes". Otherwise, it will be hard to solder to it. Also, the 330n and 100n caps by the regulators are ceramics, not electrolytics.
/mike

[ErnieHorning]

Two things I see...

First, see if there is away in your drawing program to turn off islands or orphans. Leaving floating copper isn't good either. Then see if you can move some traces around to get the ground plane to fill-in some of those open areas. You don't need to get too carried away though.

Run the ground plane all of the way over to the reculator and low voltage supply capacitor.

Though its not as bad with a single sided board, as Mike pointed out, you should turn on thermal isolators. This makes it easier to solder the pads that are connected to the ground plane copper. With bare copper, it also helps to keep the solder from spreading out when you heat a pad on the ground plane. If you ordered a double sided board without thermals, it would be almost imposible to solder these pads with low cost soldering iron that a lot of people here use.

[christmas-light]

Also, the 330n and 100n caps by the regulators are ceramics, not electrolytics.
/mike

Sorry didn´t know , I have updated that :D

Now I just need to finde the 330 nF part on mouser :)


Two things I see...

First, see if there is away in your drawing program to turn off islands or orphans. Leaving floating copper isn't good either. Then see if you can move some traces around to get the ground plane to fill-in some of those open areas. You don't need to get too carried away though.

Run the ground plane all of the way over to the reculator and low voltage supply capacitor.

Though its not as bad with a single sided board, as Mike pointed out, you should turn on thermal isolators. This makes it easier to solder the pads that are connected to the ground plane copper. With bare copper, it also helps to keep the solder from spreading out when you heat a pad on the ground plane. If you ordered a double sided board without thermals, it would be almost imposible to solder these pads with low cost soldering iron that a lot of people here use.

I have also updated that, and now it´s begining to look like a real PCB

Thanks

Here is the updated pcb and secmatick (In the secmatick the only thing there is updatede is the 100 nF and 330 nF ceramics) pls tell me if there is other debugs ;)
9625

9626

[ErnieHorning]

The only thing I see is the mounting hole on RJ45-1 thats cutting into a trace.

[christmas-light]

Oh didn´t see that :oops:

I have edited that :)

Best regards

[n1ist]

What about 810-FK24X7R1H334K for the 330n cap?
/mike

[christmas-light]

What about 810-FK24X7R1H334K for the 330n cap?
/mike

THANKS ! :D

Now there is only missing two thing the zener diode and the resistor to the zener diode, but I can´t find them on mouser :(

[n1ist]

You don't want to order 30000 of the zeners? Would 771-NZX5V1D,133 or 78-1N5231B work? I haven't checked all of the specs but at quick glance, I think it will be OK.
/mike

[christmas-light]

You don't want to order 30000 of the zeners?

Why not, it always good to have some extra :lol:



Would 771-NZX5V1D,133 or 78-1N5231B work? I haven't checked all of the specs but at quick glance, I think it will be OK.
/mike

I am not sure, which one, but I think that the 78-1N5231B will work best, but I am going to order both when I am ready to test the controller.

There is only one thing that we / I need, and it is the resistor to the zener diodes and I am not sure what the value shall be :oops:

Best regards

[griffixdc]

You don't want to order 30000 of the zeners? Would 771-NZX5V1D,133 or 78-1N5231B work? I haven't checked all of the specs but at quick glance, I think it will be OK.
/mike

if you have the money then why not..they can be used for all kinds of things heck at the amount you can afford to waste a few and still be saving money per diode cost

[nwcworld]

ok with the drawing in the PDF file do you have a set that separates the top and bottom views and the skin? Or by chance any one have any of the DC renard boards already made that i can purchase? i can get the parts and solder i just need the boards or the files to send out and have them made. :( i'm not good when it comes to CAD and the program i have doesn't like to play ball with me. Let me know please and i hope every one is having a good day.:cool:

[christmas-light]

At this point I am the only one there have the pcb files, top and bottom view.

I am still waiting on the part to test the controller, and I will not release the pcb files before the controller is tested.

[christmas-light]

UPDATE ! UPDATE ! UPDATE ! :D

I have got all the parts, the PCB and I have VACATION :D

Now I am going to solder :)

[dirknerkle]

UPDATE ! UPDATE ! UPDATE ! :D

I have got all the parts, the PCB and I have VACATION :D

Now I am going to solder :)

Breathe deeply... in... out... in... out... let the flux fumes fill your lungs with blinky-flashy goodness... in.... out... :lol:

[christmas-light]

Breathe deeply... in... out... in... out... let the flux fumes fill your lungs with blinky-flashy goodness... in.... out... :lol:

ROLF :lol:

[Mactayl]

Want to see pictures when it's done..:D

[christmas-light]

okay now it is soldered :)

There was only a few parts there didn´t fit into the pcb, eg the fuseholders

I will test it tomorrow

Here is some pictures, sorry for the bad quality but I had to take the pictures with my Nokia N95 :(

[christmas-light]

one more pricture

[christmas-light]

HELP !

It don´t work correct :(

when I power it up with the external power supply for the board (12v) it power up as it should.

when I add 24 VDC for the lights then all the lights is on ?

I have tryed to remove the opto and then the lights is off

I have also tryed to use my volt meter to see if there is any output from the pic16f688 and ofcourse I have told vixen to set the channels to "on", but there isn't any output :(

any idea ?

christmas-light

[christmas-light]

hmm okay I have found out that it is not the pic16F688

any idea what it can be ? :)

[dirknerkle]

The very first picture (15042011053.jpg‎) shows the bottom side of the board. Since a solder mask is not there to offer some protection, the first thing I'd do is take a strong magnifying glass and examine everything on the bottom -- even the traces where there are not solder points -- to make sure nothing is bridged. It looks like it would be very easy to create a bridge on that prototype board.

[christmas-light]

okay, that is true, I have just checked it and there is nothing there is bridged :/

The strange part is that it is every second there is on and every second there is off

[ErnieHorning]

... I have just checked it and there is nothing there is bridged
I'd recommend using an ohmmeter. It's amazing how small a metal sliver can be to still cause a problem.

[budude]

okay, that is true, I have just checked it and there is nothing there is bridged :/

The strange part is that it is every second there is on and every second there is off

Can you clarify what you mean by this? What is on and off? I'm a bit confused because it sounds like the outputs are solid ON?

Assuming the PIC is programmed properly you'll want to check the oscillator output and ensure it's not tied to ground or 5v - also check the orientation of it.

[dirknerkle]

It doesn't look like you have an H11AA1 on the board so you must be using PWM and a DC build. Do you have the firmware set right? There's also the OUTPUT_NEGATIVE TRUE setting which can be remmed-out to make the PIC's outputs positive instead of negative polarity.

[christmas-light]

Here is what I found out

ch 1 is constantly off
ch 2 is constantly on
ch 3 is constantly off
ch 4 is constantly on
ch 5 is constantly off
ch 6 is constantly on
ch 7 is constantly off
ch 8 is constantly on

Yes there is no H11AA1 on the board, I have added a 10k ohm resistor to the pin on the pic16f688 to + on the board.

Here is what I have set the firemware to:



;; PWM_build can be either 0 or 1
#define PWM_build 1
;; DC_build can be either 0 or 1 (it should be 0 if PWM_build is 0)
#define DC_build 1

#define CLOCKRATE 18432000
#define BAUDRATE 57600

;; CTR_LOCKOUT should be 1 for the DC build, can be a somewhat larger number
;; for non-PWM or non-DC builds.

#define CTR_LOCKOUT 15

;; For positive_true outputs, remove the following define (normally, this
;; should be left in place).
#define OUTPUT_NEGATIVE_TRUE


The pic16f688 have to have a negative output because I am adding a positiv (+5) through a resistor to the opto, so negative form the pic and posetive from the volt reg to the opto.

I have also found out that there is a output from the pic16f688 because when I take my voltmeter and add the + pin on the multimeter to the board after the volt reg and then I take the negative pin and add to the output pin on the pic16f688, and then start a sequence in vixen, the multimetere goes form 0 - 5 volt (when vixen tell the pic to get that pin on) on each output on the pic, so it is working

The poblem is ether in the firmeware or at the opto, I not sure

[Mactayl]

CL, what voltage do measure on the collector of each opto in reference to the ground point on X9-1 and is that your latest scuzzy (schematic) in post #71 ?

[budude]

The poblem is ether in the firmeware or at the opto, I not sure

It doesn't sound like the firmware since the PIC outputs seem to be working OK. I think Mac has you on the right track - it sounds like the optos. I would double-check the datasheet to ensure you have the correct inputs/outputs for your PCB. Your FET layout is identical so it can't be that so that only leaves the optos... Exactly which opto did you use (and did you actually get the correct one with your order)?

[christmas-light]

I think I have found the problem..

When I take my multimeter and connect the ground to x9-1 on and take the + to my multimeter and put it before the resistors to the optos (R32 - 39) here I have these resistors


NO NO NO NO NO NO NO NO NO NO !!!!!!!!

I have found the problem !!!! :(

The resistors I should have ordred for the optos (R32 - 39) should have been 3 Kohm 1 watt resistors, I ordred these:

http://dk.mouser.com/ProductDetail/Vishay/CPF13R0100FKB14/?qs=au%252bTo7XMJ2I8o%2fCGq551Lg%3d%3d

this is 3 ohm resistors :evil:

[Mactayl]

I think I have found the problem..

When I take my multimeter and connect the ground to x9-1 on and take the + to my multimeter and put it before the resistors to the optos (R32 - 39) here I have these resistors


NO NO NO NO NO NO NO NO NO NO !!!!!!!!

I have found the problem !!!! :(

The resistors I should have ordred for the optos (R32 - 39) should have been 3 Kohm 1 watt resistors, I ordred these:

http://dk.mouser.com/ProductDetail/Vishay/CPF13R0100FKB14/?qs=au%252bTo7XMJ2I8o%2fCGq551Lg%3d%3d

this is 3 ohm resistors :evil:


Look at this way, you were only 2,997 ohms away:shock:
kinda thought something might not be correct in that area

[christmas-light]

I have just found out that it is not just the resistors there is wrong, there is also a BIG bug around the opto design, if you take a look at lab rats dc ssr (the pcb):
http://doityourselfchristmas.com/forums/dynamics/showentry.php?e=36&catid=8

and my pcb here (uploaded file)

then you can see that all my wires is wrong connected :S

[christmas-light]

And we got light :D

I have fixed the last 4 ch, you can see in the picture how I fixed it, now I am going to edit the pcb design and the bom ;)

I made a 3 kohm resistor out of 2 resistors

[budude]

Well - she ain't pretty but at least you got it working! Congrats...

[christmas-light]

yes that is true, I am also glad that it is only a test version :)

[Mactayl]

Got it working,,, great.. now your next PCB should have all your corrections on it and you'll be good to go...:)