buck converter sizing

[davros]

I like ProDCtoDC, but their shipping is outrageous. For 15 converters, they want $40 to ship to me in Nevada. Because they are based out of China?

All looks like the regular asian products, why use them as a vendor?

[scott4864]

Is the amperage rating of buck DC to DC converters the input amperage or the output amperage?

[ukewarrior]

Ouput
Of course, it can't produce more amps than are input. It only changes the voltage.

Is the amperage rating of buck DC to DC converters the input amperage or the output amperage?

[plasmadrive]

I like ProDCtoDC, but their shipping is outrageous. For 15 converters, they want $40 to ship to me in Nevada. Because they are based out of China?

I email them directly and ask for a quote.. shipping is cheaper that way.. but still not low cost.. adds about $1.50 to $2 per converter.. but still cheaper than Amazon.. and I wouldn't buy any of that stuff off of ebay.. been burned too many time with ebay and Chinese "stuff"...

[plasmadrive]

Ouput
Of course, it can't produce more amps than are input. It only changes the voltage.

Well.... I'm pretty sure you didn't mean it the way you wrote it. For example, if you use 24v input and 12v output, the input current will be a bit over half the output current.

[MartinMueller2003]

Is the amperage rating of buck DC to DC converters the input amperage or the output amperage?

Most are specked using output current since the output winding on the transformer and the bridge rectifier are the gating items. Specking input current means you can control the efficiency to a very tight tolerance.

[ukewarrior]

Duh on my part. Correct

Well.... I'm pretty sure you didn't mean it the way you wrote it. For example, if you use 24v input and 12v output, the input current will be a bit over half the output current.

[LightUp]

The better view is "power transfer", minus some power loss, then the voltages or currents can be "re-arranged" to match the output power.

[MartinMueller2003]

Look at it this way: For a DC-to-DC converter

Input voltage = 12v
Output voltage = 5v
Conversion efficiency = 90%

Load "Power" is 10W
Load current = load power / load voltage
That means load current = 10W / 5V = 2A

Input power = output power / conversion efficiency / 100
Input Power = 10W / .90 = 11.11W
Input current = Input power / input Voltage
Input current = 11.11W / 12V = 0.926A

Where does that extra power go? It is dissipated as heat. The loss is 1.11W

For a voltage regulator (Like you find in 12v pixels and in some devices that call themselves converters)
Input voltage = 12v
Output voltage = 5v
Load "Power" is 10W
Load current = load power / load voltage
In our example load current = 10W / 5V = 2A
Input current = output current = 2A (this is the big difference in the calculations)
Input power = Input Voltage * Input current
Input power = 12V * 2A = 24W

Where does that extra power go? It is dissipated as heat. The loss is 14W

[Utimgr]

And thus why those of us that design power supplies are always looking at increasing the efficiencies to eliminate heat and wasted power. Martins excellent explanation also shows why its better to use a switching regulator over a linear.