How Do You Handle the Negative Gain When Calculating Resistance?

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Discussion Overview

The discussion revolves around the calculation of source resistance ##R_S## in the context of a common-source amplifier, particularly focusing on the implications of a negative gain in the expression for voltage gain ##A_v##. Participants explore how to handle the negative sign in the formula and its impact on the physical interpretation of resistance values.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions whether to disregard the negative sign in the gain expression when calculating ##R_S##, noting that leaving it in results in a negative resistance value, which is non-physical.
  • Another participant confirms that the circuit is a common-source amplifier and agrees with the interpretation of the gain as negative.
  • There is a suggestion that the circuit could be an inverting stage, with speculation about the type of transistor used based on the notation in the parameters.
  • Some participants express uncertainty about the correct approach to calculating ##R_S##, with one suggesting that the gain should be considered as ##A_v = -5 V/V## instead of 5.

Areas of Agreement / Disagreement

Participants generally agree that the gain is negative and that this affects the calculation of ##R_S##. However, there is no consensus on the proper method to handle the negative sign in the calculations, and multiple viewpoints on the interpretation of the circuit remain unresolved.

Contextual Notes

Participants have not reached a definitive conclusion regarding the treatment of the negative sign in the gain expression, and assumptions about the circuit type and parameters are based on interpretation rather than established definitions.

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Homework Statement



Calculate ##R_S## given the expression:

$$A_v = - \frac{g_m (R_L || R_D)}{1 + g_m R_S}$$

The parameters ##g_m = 1.292 \frac{mA}{V}##, ##R_L = 180 k \Omega##, ##R_D = 18 k \Omega##, ##A_v = 5 \frac{V}{V}##.

Homework Equations

The Attempt at a Solution



My question is about the negative sign. When I go to compute ##R_S##, do I pretend the negative sign isn't there or something? If I leave the negative sign in the calculation I get:

##R_S = - 4.05 k \Omega##

Resistances must be positive, so the above does not make physical sense.

If I pretend the negative sign isn't there then I get:

##R_S = 2.5 k \Omega##

What is the proper way to compute the resistance?

EDIT: I believe the gain should be ##A_v = -5 V/V## even though it was given as 5, so the resistance ##R_S = 2.5 k \Omega## should be correct.
 
Last edited:
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This is for a common-source amplifier with source resistance ##R_S##? If so, then you're right about ##A_v##.
 
Zondrina said:

Homework Statement



Calculate ##R_S## given the expression:

$$A_v = - \frac{g_m (R_L || R_D)}{1 + g_m R_S}$$

The parameters ##g_m = 1.292 \frac{mA}{V}##, ##R_L = 180 k \Omega##, ##R_D = 18 k \Omega##, ##A_v = 5 \frac{V}{V}##.

Homework Equations

The Attempt at a Solution



My question is about the negative sign. When I go to compute ##R_S##, do I pretend the negative sign isn't there or something? If I leave the negative sign in the calculation I get:

##R_S = - 4.05 k \Omega##

Resistances must be positive, so the above does not make physical sense.

If I pretend the negative sign isn't there then I get:

##R_S = 2.5 k \Omega##

What is the proper way to compute the resistance?

EDIT: I believe the gain should be ##A_v = -5 V/V## even though it was given as 5, so the resistance ##R_S = 2.5 k \Omega## should be correct.
Generalizing post 2 a bit, I'd venture that the circuit is an inverting stage of some kind, be it MOSFET, JFET, BJT, vacuum tube, etc. - based.
 
rude man said:
Generalizing post 2 a bit, I'd venture that the circuit is an inverting stage of some kind, be it MOSFET, JFET, BJT, vacuum tube, etc. - based.
I just guessed it was FET-based from the subscripts, i.e. D, S, L for drain, source, load.
 
milesyoung said:
I just guessed it was FET-based from the subscripts, i.e. D, S, L for drain, source, load.
Sure.
 

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