Op-Amp circuit to supply constant power

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SUMMARY

The discussion focuses on designing an op-amp circuit to supply constant power to two loads with specific resistance values of 100 ohms and 100K ohms, each requiring 10 mW. The solution involves using a series resistor to achieve the necessary voltage across each load, specifically 31.6V for the 100K ohm resistor and 1V for the 100 ohm resistor. The circuit configuration requires an op-amp capable of handling approximately 40V between its power supply pins, with a gain of 32.62 and a resistor value of 3162 ohms to maintain the desired power output.

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JeeebeZ
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Homework Statement



see image
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Homework Equations





The Attempt at a Solution


I can make 2 different circuits to solve this. However, Using just op amps and resistors I am not sure how to make a single circuit that could do this
 
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I am not sure if that is intended, but if power just has to be the same for those two specific resistance values, there is an easy solution: Add a specific resistor in series, and find a pair (voltage, resistance) where both resistors get the same power.
 
I would consider an op amp circuit set up so that the op amp output voltage is Vo. Then I would run a resistor R from Vo to each of your loads, one at a time of course.

Given the 10 mW for each load, compute the voltage needed across each load.

You now have 2 equations and 2 unknowns: R and Vo. Solve for both.

(You will obviously need about 31.6V across the 100K resistor so figure on an op amp that can handle about 40V between its power supply pins.)
 
It is really easy to do with 2 difference circuits. But using one circuit I don't see how changing one resistor could allow it to get 10mW just on the one resistor that has been changed.

In order for the 100k ohm, it needs ~31V, while the 100 ohm only needs 1V.
 
JeeebeZ said:
It is really easy to do with 2 difference circuits. But using one circuit I don't see how changing one resistor could allow it to get 10mW just on the one resistor that has been changed.

In order for the 100k ohm, it needs ~31V, while the 100 ohm only needs 1V.

That's what the series resistor is there for! It puts ~ 31V across the 100K but only 1V across the 100 ohm.

Have you considered doing what I suggested?
 
h... and I got it to work.

https://www.circuitlab.com/circuit/283w8y/screenshot/1024x768/
 
JeeebeZ said:
h... and I got it to work.

https://www.circuitlab.com/circuit/283w8y/screenshot/1024x768/

Can't access the link. What did you come up with?
 
32.62 gain, 3162 R
That way you have 1V over the 100ohm = 10mW
and 32.62(100000/(3162+100000)=31.62V -> 31.62^2 / 100000 = 9.998mW
 
JeeebeZ said:
32.62 gain, 3162 R
That way you have 1V over the 100ohm = 10mW
and 32.62(100000/(3162+100000)=31.62V -> 31.62^2 / 100000 = 9.998mW

Superb! Congrats!
 
  • #10
JeeebeZ said:
h... and I got it to work.

https://www.circuitlab.com/circuit/283w8y/screenshot/1024x768/
That link is dead. Can you attach your sketch to a post here?
 
  • #11
the link is linked to my account, its forbidden i guess\
1024x768.png
 

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