Hi for the question in the attached image below, would

Ve/Vs = [hfe*Re]/[Rb + hie + (hf'e*Re)], with hf'e being the value I found for Vc/Vs?

 

I would rather have:

\(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\)

where hfe is the value per the original problem statement.

 

Hi for the question in the attached image below, would

Ve/Vs = [hfe*Re]/[Rb + hie + (hf'e*Re)], with hf'e being the value I found for Vc/Vs?

What's the difference between hfe and hf'e? I never did see an explanation for that in the other thread.

 

t_n_k is right.

\(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\)

 

I would rather have:

\(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\)

where hfe is the value per the original problem statement.

Ok but would Vc/Vb stay as (-hfe*Rc)/{Rb+hie + (Re*hfe)} or would this also be Re*(hfe + 1)?

 

Do you know why we use (hfe+1) ??

 

Do you know why we use (hfe+1) ??

Not entirely, I think it's because it's the resistance Re + the resistance of the emitter Re affected by the hfe of the junction?

 

Not entirely, I think it's because it's the resistance Re + the resistance of the emitter Re affected by the hfe of the junction?

We use hfe+1 because the emitter current is equal to
Ie = Ib + Ic = Ib + Ib*hfe = (hfe + 1)*Ib
And this is why Re resistor is seen at the base as (Hfe +1)*Re

 

We use hfe+1 because the emitter current is equal to
Ie = Ib + Ic = Ib + Ib*hfe = (hfe + 1)*Ib
And this is why Re resistor is seen at the base as (Hfe +1)*Re

So to clarify my initial question is yes it would be (Re*hfe+1)/{Rb+hie+(Re*hfe+1)}?

 

So to clarify my initial question is yes it would be (Re*hfe+1)/{Rb+hie+(Re*hfe+1)}?

Check your parentheses

 

Check your parentheses

Sorry just spotted it (Rc*hfe)

 

But you in post 9 show equation for the voltage gain

Ve/Vb = (Re*hfe+1)/{Rb+hie+(Re*hfe+1)

The correct one is Ve/Vb = (Re*(hfe+1))/{Rb+hie+(Re*(hfe+1))

But now Vc/Vb = ?

 

Sometimes you are asking about Ve/Vs and Vc/Vs, but other times it seems to morph into Ve/Vb and Vc/Vb.

What do you understand to be meant by Ve/Vb as opposed to Ve/Vs? Wouldn't Vb mean the signal voltage at the base directly, whereas Vs is the signal at the left end of Rb? So, Ve/Vb is a different gain ratio than Ve/Vs?

Please be absolutely clear which you are asking about.

 

But you in post 9 show equation for the voltage gain



The correct one is Ve/Vb = (Re*(hfe+1))/{Rb+hie+(Re*(hfe+1))

But now Vc/Vb = ?

It's supposed to be Vc/Vs, it's just a typo.

Vc/Vs= (hfe*Rc)/{Rb + hie +(Re*1+hfe)}

Although technically wouldn't Vc/Vs = Vc/Vb since R1, and R2 are ignored?

 

t_n_k is right.

\(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\)

If there were a capacitor across Re that grounded/shorted Re, would Ve/Vs still = \(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\) ?

 

It's supposed to be Vc/Vs, it's just a typo.

Vc/Vs= (hfe*Rc)/{Rb + hie +(Re*1+hfe)}

Although technically wouldn't Vc/Vs = Vc/Vb since R1, and R2 are ignored?

OK, now every thinks looks good except one small detail.
The voltage gain Vc/Vs is negative because transistor in common emitter configuration have 180 degree phase shift.
When the base current increases the collector current also increases.

As you can see when input voltage is increasing ( base current also) the Vce voltage decrease. So we have 180 degrees out of phase (negative gain).

 

OK, now every thinks looks good except one small detail.
The voltage gain Vc/Vs is negative because transistor in common emitter configuration have 180 degree phase shift.
When the base current increases the collector current also increases.

As you can see when input voltage is increasing ( base current also) the Vce voltage decrease. So we have 180 degrees out of phase (negative gain).

Ahh ok, I see. Thanks

 

If there were a capacitor across Re that grounded/shorted Re, would Ve/Vs still = \(\text{\frac{V_e}{V_s}=\frac{R_E(1+h_{fe})}{R_B + h_{ie} + R_E(1+h_{fe})}}\) ?

No ,try to use small signal analysis and you will know.

 

No ,try to use small signal analysis and you will know.

Would there be no gain since Re is grounded meaning Ve/Vs would = Vs?

 

Yes, how can we talk about the voltage gain in circuit in which we short our output terminal to ground.