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Constant Power Load

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akahrim

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Hello All,

I am currently working on throwing together a constant power load to test batteries. I can not simulate the circuit because I can not get a spice model of the AD835 multiplier. I was wondering if you all could provide feedback on if this circuit since I can not simulate it. Any feedback and suggestions would be greatly appreciated. The circuit must draw a constant 2.25W from the Battery Under Test. The battery's voltage will range from 1.8V down to 1.3V. I have attached a drawing of the circuit. Any feedback would be greatly appreciated.

Amir
 

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I did not look at the logic of how you wired the AD835. I assume the output is related to power.

I think your circuit will oscillate. The problem is the error amplifier it is too fast. Look at error amplifiers in power supplies.
I would put a capacitor from (-) to output of the error amplifier. Add a resistor from (-) to the AD835. The RC will cause the amplifier to be slow.
 
Hello Ron,

Thank you for your feedback. The output of the AD835 (W) is related to power. It is equal to (Y1-Y2)*(X1-X2)/U + Z. Where U is around 1V-1.05V. I will take a look into the error amplifiers in power supplies. Thank you for your feedback.

Amir
 
I can not simulate it.
Just use a behavioral voltage source with V(out) = V(battery) x I(sense) x Scale-factor.
 
The AD835 is a VERY FAST multiplier...you don't need 250 Mhz for a battery load. Use something slower AND cheaper, like the AD633.

Additionally, your input values will be very low voltage, and thus the device's input offset error will be a significant percentage error.
Use the offset nulling circuits described in the data sheet.
 
More on why the op-amp needs to be made into an integrator.

As shown in the schematic in post one;
>the output of the amplifier will be full off if the power is above .25
>he output of the amplifier will be full on if the power is below .25
There is no point where the output of the amplifier is half way on.
So the MOSFET is either full on or full off. (It oscillates)

If the amplifier is an integrator;
>If the power is above .25 the gate voltage on the MOSFET will be slowly reduced.
>If the power is below .25 the gate voltage on the MOSFET will be slowly increased.
So the MOSFET will balance at a point where the power is 0.25
 
There is a serious error in the circuit. The nmosfet source should be directly connected to ground. Load should be in the drain and the positive of the battery. Otherwise your circuit will not operate correctly.
 
There is a serious error in the circuit.
No, there isn't. The FET is acting as an active load (current source). It simulates fine according to LTspice, although parasitics in the circuit might cause oscillation (see ronsimpson's post above).
 
Yes. If the opamp is a rail-to-rail output type then its 5V ouput exceeds the gate threshold voltage of the vast majority of N-MOSFETs.
 
Hello All,

I am currently working on throwing together a constant power load to test batteries. I can not simulate the circuit because I can not get a spice model of the AD835 multiplier. I was wondering if you all could provide feedback on if this circuit since I can not simulate it. Any feedback and suggestions would be greatly appreciated. The circuit must draw a constant 2.25W from the Battery Under Test. The battery's voltage will range from 1.8V down to 1.3V. I have attached a drawing of the circuit. Any feedback would be greatly appreciated.

Amir
I do not want to distract you, but if you like I will be happy to post a circuit of on I built several years ago that uses a transistor, op amp, fixed resistor and a voltage reference and a DVM to do exactly as you want.
 
k7elp60:

Please do post it. I believe that not only the original poster but everyone else following this thread would appreciate it.
 
schmitt trigger here it is. I have been using this for more years than I can remember. I have used it for testing batteries, power supplies etc. The DVM is adj to show current with the pot. I realize that the true current does not include the base current of the darlington transistor that is measured with the DVM connected to the emitter resistors, but when you are talking about amps the base current is not important.
 

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schmitt trigger here it is. I have been using this for more years than I can remember. I have used it for testing batteries, power supplies etc. The DVM is adj to show current with the pot. I realize that the true current does not include the base current of the darlington transistor that is measured with the DVM connected to the emitter resistors, but when you are talking about amps the base current is not important.
There is one resistor I failed to put in the schematic that is a 1k from the output of the op-amp to ground. The circuit requires a darlington transistor as the variable resistor to work properly.
 
a circuit ......to do exactly as you want.
:confused: The circuit in post #13 is undoubtedly useful for testing batteries, but it provides a constant current load, not a constant power load as the OP specified.
 
Have you thought about using constant current instead of constant power? Most batteries are spec'd in amp hours anyway.

Hmm. missed 2 posts.
 
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