Op-Amp circuit to supply constant power

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Discussion Overview

The discussion revolves around designing an operational amplifier (op-amp) circuit that can supply a constant power output to two different resistive loads. Participants explore various circuit configurations and calculations related to achieving the desired power levels for the specified resistances.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that if the power needs to be the same for two specific resistance values, a simple solution could involve adding a specific resistor in series to achieve the desired power levels.
  • Another participant proposes calculating the necessary voltage across each load to achieve 10 mW, leading to a system of equations involving the output voltage and resistance.
  • Some participants express uncertainty about how a single circuit could adjust to provide 10 mW to one load while changing the resistor value, given the different voltage requirements for the loads.
  • A participant mentions achieving a specific gain and resistance value to ensure the correct voltage across the 100-ohm resistor, which would yield the desired power output.
  • There are multiple references to circuit simulations and calculations, with some participants sharing links to their circuit designs, although access issues are noted.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the best approach to design a single circuit for the problem. There are competing views on how to achieve the desired power output, and some uncertainty remains regarding the feasibility of the proposed solutions.

Contextual Notes

Some calculations and assumptions regarding voltage and resistance values are presented, but there are unresolved mathematical steps and dependencies on specific circuit configurations that may affect the outcomes.

JeeebeZ
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Homework Statement



see image
Untitled.png


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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