How Do You Handle the Negative Gain When Calculating Resistance?

[STEMucator]

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.

 

[milesyoung]

This is for a common-source amplifier with source resistance $R_S$? If so, then you're right about $A_v$.

 

[rude man]

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.

 

[milesyoung]

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.

 

[rude man]

I just guessed it was FET-based from the subscripts, i.e. D, S, L for drain, source, load.

Sure.