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I have this circuit:enter image description here and I have to find the voltage gain at the middle of the band (with all capacitors shorted). $$ G_v=\frac{Vo}{Vi} $$ I've drawn the small signal model in which: $$ Veq = \frac{Rb1//Rb2}{Rb1//Rb2 + Rs}Vin; Req = RB1//RB2//Rs $$ 1 - Is it correct?

schematic

simulate this circuit – Schematic created using CircuitLab

2 - How to solve it and find Vo/Vi? By now I have: $$ Vo = -g_{m2}V_{pi2}(RC//RL) $$ $$V_{pi2} = V_{e1} -V_{e2} $$ But I don't know how to find Ve1 and Ve2. Can you please give me and advice?

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  • \$\begingroup\$ Because of the data you were given, it seems like you are looking for a numeric answer. Are you trying to find a symbolic result? You need to know how to calculate the small signal model parameters form the large signal values. Req needs to have Rs added and Veq needs to multiplied by Vsine. As far as how to solve it, I'd go with node equations. \$\endgroup\$ – owg60 Nov 5 '16 at 11:37
  • \$\begingroup\$ Do you know the answer they are looking for because I visually calculate it as (Rc||Rload)/Re2 assuming Rs is negligible (which it is) = 6.6. \$\endgroup\$ – Andy aka Nov 5 '16 at 11:48
  • \$\begingroup\$ Since it's a double stage amplifier, can I just multiply the single stage gains? \$\endgroup\$ – FataMadrina Nov 5 '16 at 11:49
  • \$\begingroup\$ No I don't know the correct answer. \$\endgroup\$ – FataMadrina Nov 5 '16 at 11:50
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    \$\begingroup\$ Yes, your small signal topology is correct. \$\endgroup\$ – owg60 Nov 5 '16 at 15:59
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I have tried to analyze the circuit (without using the small-signal equivalent diagram). At first, it is necessary to calculate the DC quiescent currents in order to find the corresponding transconductances gm=Ic/Vt for both transistors (gm1 and gm2)

Here is the set of equations I have found:

Vcc-VB1=RB1(IB1+I1) with I1=VB1/R2

Vcc-VB1=RB1(IB1+VB1/R2).

VB1+0.56+0.71=VE2 >>> VB1=VE2-1.27volts.

VE2=Vcc-IE2*RE2

IE2=IB2*200

VB1=Vcc - IE2*RE2 - 1.27volts.

IE2=IB2*200

VB1=Vcc - IB2*200*RE2 - 1.27volts.

with IB2=I1-IE1 with I1=(Vcc-VE1)/RE1

IB2=I1-IE1= (Vcc-VE1)/RE1 - IE1

IB2=(Vcc-VE1)/RE1 - IB1*200

VE1=VB1-0.56volts

IB2=(Vcc-VB1+0.56)/RE1 - IB1*200

Result: Three equations (in bold)for three Unknowns: VB1, IB1, IB2

(I hope, I didn`t make any errors)

After Ic1 and Ic2 have been found (and the corresponding values for gm1 amd gm2) it shouldn`t be a problem to apply the well-known gain formulas.

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  • \$\begingroup\$ Ok. This steps are clear and correct. But, instead of using the gain formulas I have to take in account the load effect. In your opinion, should be correct to: -Use the small signal model of the CE (second stage) in order to find out the input resistance of the stage. -Use the input resistance just found as it was the load resistance of the CC (first stage) stage and calculate the gain of the first stage (loaded). - Calculate the gain of the Ce stage -Multiply the gains? – \$\endgroup\$ – FataMadrina Nov 5 '16 at 16:12
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    \$\begingroup\$ FataMadrina, when I say "classic gain formulas" I mean, of course, the correct formulas which take into account the loading effects. That`s quite normal. Once the DC currents (and the gm values) are known, it is rather simple to find the total gain. Remember: Base-emitter resistor rbe=beta/gm. \$\endgroup\$ – LvW Nov 5 '16 at 16:30
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Ok, let's do this in steps. Since we are looking for a numeric solution the first step is to solve for the bias condition. They gave you the VBEs so you could create a large signal model that looks like this; enter image description here This may look intmiadating but you can do it. If you know how to use the supernode methode at the Vbe it will be pretty easy for you. When you say you can't make any assumpion does that include you can't assume the first base current is 0? You have to get the collector curents so you can calculate gm, rpi, adn ro.

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