How to design what amounts to a variable load resistor

Thread Starter

CaryInVancouver

Joined Jan 25, 2011
9
I am interested in using a power supply that provides 12A at 5V. It requires a minimum 3A load, however. I understand that could be achieved with a (5V/3A=) 1.7 Ohm load resistor placed between +5V and ground (possibly achieved by using larger-Ohm resistors wired in parallel). My question is whether there is a more efficient way of maintaining a load of at least 3A, so that the max load is not effectively reduced to 9A.

Let X be the load in amps. I would like to add some parts to the circuit that would create an additional load of 3-X amps (or 3A) when X < 3A, but add no load when X >= 3A. Any ideas?

If it's not obvious, I'm pretty new to this hobby.

Cary
 

wayneh

Joined Sep 9, 2010
17,496
There are definitely ways to accomplish this. The question is which is easiest.

I'd consider a constant current supply, where the current level is controlled by a measurement of the total load. So if total load is greater than 3A, the circuit supplies zero to its load. At any lower total load, the circuit adds current to make up the difference.

If that all makes sense, look at the constant current supply here for the general idea. It would need to be modified to vary the current depending on the total being driven by the power supply. I think you'd need a second op-amp on the LM358 to accomplish that - to convert the total current into a reference voltage to apply to the pin of the op-amp in the circuit shown. This would replace the reference voltage set with the potentiometer in the diagram.
Picture 1.png
 
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retched

Joined Dec 5, 2009
5,207
Computer PSUs need some loading in order to be drawn into regulation. Not sure that's relevant here.
It may be the exact reason.

If metered with no load, the power supply may appear to be 7v or higher.

The manufacturer may only guarantee 5v @ 3A.
 

k7elp60

Joined Nov 4, 2008
562
There are definitely ways to accomplish this. The question is which is easiest.

I'd consider a constant current supply, where the current level is controlled by a measurement of the total load. So if total load is greater than 3A, the circuit supplies zero to its load. At any lower total load, the circuit adds current to make up the difference.

If that all makes sense, look at the constant current supply here for the general idea. It would need to be modified to vary the current depending on the total being driven by the power supply. I think you'd need a second op-amp on the LM358 to accomplish that - to convert the total current into a reference voltage to apply to the pin of the op-amp in the circuit shown. This would replace the reference voltage set with the potentiometer in the diagram.
View attachment 29617
I have a load bank which essentially the same as the posted schematic, except I am using a 20A NPN darlington transistor mounted on a large heatsink. It is real handy for checking current limit of power supplies, discharging batteries for load tests etc. The beauty of the circuit the load current remains constant when discharging batteries. Using resistors the load current decreases as the battery discharges.
 
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Thread Starter

CaryInVancouver

Joined Jan 25, 2011
9
...I think you'd need a second op-amp on the LM358 to accomplish that - to convert the total current into a reference voltage to apply to the pin of the op-amp in the circuit shown. This would replace the reference voltage set with the potentiometer in the diagram.
View attachment 29617
I'm just reading about op-amps now in my electronics self-study regime. This should be a good exercise to improve my understanding of those devices.

Thanks again, Wayne.

Cary
 

wayneh

Joined Sep 9, 2010
17,496
I have simulated something similar to what wayneh proposed.
Nice. I was having trouble with the logic and thought a 2nd op-amp might be needed, but your solution makes perfect sense - a current "mixer" if you will.

Are those "10m" resistors 10mΩ resistors? I'm not familiar. Ten is too much.
 

t_n_k

Joined Mar 6, 2009
5,455
Nice. I was having trouble with the logic and thought a 2nd op-amp might be needed, but your solution makes perfect sense - a current "mixer" if you will.

Are those "10m" resistors 10mΩ resistors? I'm not familiar. Ten is too much.

Sorry Wayneh - I thought the thread had sunk to the bottom. CDRIVE pulled it up correctly - 10milli-ohms was what I intended.
 
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