Constant Power to a Varying Load

[fobos3]

Is there a way to deliver constant power to a varying load. Any ideas will be appreciated.

 

[russ_watters]

Could you elaborate please - the question doesn't seem to make much sense.

 

[fobos3]

Could you elaborate please - the question doesn't seem to make much sense.

Say you have a resistor with resistance R. What I want is to deliver say 1W to the resistor no matter it's resistance.

 

[skeptic2]

Monitor the voltage across the resistor and the current through it. In a microcontroller multiply the voltage and current and compare to desired power setting. Adjust the voltage until calculated power matches the desired power.

 

[fobos3]

I've attached my idea.

U2 is a differential amplifier with gain 1. U1 is there to make sure the voltage at + and - is the same.

 

[rbj]

I've attached my idea.

U2 is a differential amplifier with gain 1. U1 is there to make sure the voltage at + and - is the same.

the idea is good. i think R1 should be lower resistance and U2 gain be larger.

what do you plan to use for your analog multiplyer? you might want to look into something from Analog Devices, say the AD633 or AD734 or something like that. you can get some of them pretty cheap ($8) at Digi-Key.

you also want to have some pot in there to set and control the power.

 

[fobos3]

the idea is good. i think R1 should be lower resistance and U2 gain be larger.

what do you plan to use for your analog multiplyer? you might want to look into something from Analog Devices, say the AD633 or AD734 or something like that. you can get some of them pretty cheap ($8) at Digi-Key.

you also want to have some pot in there to set and control the power.

I was thinking with two log amplifiers then a summing amplifier and exponential amplifier.

$$e^{\ln a+\ln b}=ab$$

With something like LM324(quad) which costs 10p(£0.10). I'll post a circuit tomorrow because I'm too tired now.

P.S.
I just found that it costs $1.69 on Radio Shack. You've got to love the uni labs.

 

[rbj]

I was thinking with two log amplifiers then a summing amplifier and exponential amplifier.

$$e^{\ln a+\ln b}=ab$$

With something like LM324(quad) which costs 10p(£0.10).

you still need a logarithmic element (actually 3 of them). run-of-the-mill diodes don't follow the curve good enough. what do you plan to use for that?

 

[fobos3]

you still need a logarithmic element (actually 3 of them). run-of-the-mill diodes don't follow the curve good enough. what do you plan to use for that?

Transistors. Here's my final schematics. I haven't calculated the values of the resistors so all are 10k.

Edit
Still need to figure out the gain for each op amp taking into account the offset voltage. In particular the log amplifiers.

 

[rbj]

Transistors.

rots 'o ruk finding transistors true enough to the Ebers-Moll equation to be accurate loggers. this is how Analog Devices makes their money, by somehow designing accurate transistors for their logging functions.

Here's my final schematics. I haven't calculated the values of the resistors so all are 10k.

Edit
Still need to figure out the gain for each op amp taking into account the offset voltage. In particular the log amplifiers.

also, make sure you're getting your polarities right. it looks like you are (with the inverting summing amp), but I'm a little concerned about the output of the exp amp.