Hello,

I'm thinking of using a USB-C trigger board as a means of powering up a power amplifier and a signal generator. Both devices need an input voltage of 12V, but their current consumption is 0.05A for the signal generator and 1.3A for the power amplifier.

From what I can see, on every single merchant's details on the trigger board, the current output is 5A at most and the actual current is dependent on the load itself. Does this mean that when I connect the trigger board to my signal generator for example, the output current will be 0.05A? Or is the trigger board's output current fixed at 5A?

Here's the trigger board I'm looking at:
https://www.aliexpress.com/item/100...f1277204c01a9940f24209e1e30&afSmartRedirect=y

 

Hello,
Does this mean that when I connect the trigger board to my signal generator for example, the output current will be 0.05A? Or is the trigger board's output current fixed at 5A?

Welcome to AAC.

Yes, the output current will be ≤5A (if the board can really supply it), but it will be no more than the device draws. It’s worth understanding a little about why. I promise this is dead simple, even if there is math in the explanation. Please read the following and ask any questions. You will be able to figure out a lot of things if you spend just a little time working this out in your mind…

While voltage is very roughly analogous to pressure, current is more like volume. Ohm’s Law is a very simple formula that describes the relationship among voltage (V or E), current (I), and resistance (R):

\[ \mathsf{V=I\times R} \space \space or \space\space \mathsf{I = {V\over{R}}} \space \space or \space\space \mathsf{R = {V\over{I}}} \]
​

Since we are interested in current we want the second form: \( \mathsf{I = {V \over{R}}} \). From this you can see that current (in amps or A) is directly proportional to voltage (in volts or V) and inversely proportional to resistance (in ohms or Ω).

So while the voltage of a power source ”pushes”, current is a result of that pushing, and resistance is how hard the circuit resists the pushing. A power supply can underrated for current (e.g.: you need 2A and it can provide only 1A) but it can’t be overrated*. The load will only use as much current as the device needs.

As long as the voltage provided by the supply is correct, the current will be too. This is why things burn up if you use a higher voltage than the device can handle. The extra current has to be dissipated by the device which happens as heat. This literally burns up the device. Sometimes the only reason a device can’t use a higher voltage is the inability to deal with this extra heat.

One more thing: a nice trick for using Ohm‘s Law is to organize it in a triangle:

​

To work out which form of the law you need, cover the unknown and read the result. For example, to find voltage when we know current and resistance, cover the V and read “IR” or I times R. For current, cover the I and read “V/R” or V over R, etc.

*The exceptions to this are devices that require a constant current supply. A very common case of this is powering LEDs. When dealing with this there must be a way to limit current. This is a different topic but keep it in mind for the future.