Differential Amplifier BJT

Dear sirs, i have a new question, this time regarding differential amplifier. If you look to attached image you will se the differential amplifier.

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The exercise asks to calculate the differential gain Ad and the common mode gain Ac at small signal that will be used to obtain the output Vo1 with these input V1 = 5*sin(t) and V2=4*sin(t).
In the solution of the exercise they say that


A_d = - g_m * R_c
A_c ≈ - (g_m * R_c) / ( 1+2*g_m*R_e )

How did they manage to calculate the differential gain Ad and the common mode gain Ac?

Image attached
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LINK

Draw the small-signal equivalent circuit for each case (differential,common),write down the equations and you have the answer.

Hi faustoleali,

two comments may help for calculation.

1.) To calculate diff. gain you can treat the diff. pair as a series combination of two basic stages: Common collector and common base: Because the output of the left transistor as well as the input for the right transistor is at the common emitter point. Thus, take both gain values (watch the loading of the common collector stage) and take the product of both. And don't be surprised about the "bad" gain value for the 1st stage.

2.) Speaking of gain one should first define which gain is desired (because we have 2 inputs and 2 outputs). That means, the task is not clear in formulation. However, the given expression for Ad indicates that the symmetrical gain value is meant.
(one input: Vin1 with Vin2 grounded ; and output: Vout1-Vout2).

Differential case,
as the input is differential, voltage at emitters is zero (so no AC current through the emitter resistor), Ic1 = -Ic2 (assuming hfe>>1)

Seen from the inputs, you have two transistors in series, so the overall Gm is half the value of each transistor. You can also say that each transistor experiences half the input voltage.

So Ic1 = 0.5*gm*(v1-v2), vo1 = -RC*0.5*gm*vdiff vdiff = v1-v2. Assuming hfe>>>1, ic2 = -ic1 (as there is only DC through Re),

So v02 = Rc*0.5*gm*vdiff, v01-v02 = vout = -Rc*gm*vdiff, Ad = -Rc*Gm.

the common mode case, here both inputs are connected together and you can see this as a single-transistor case with an emitter resistor and a collector resistor of 0.5*Rc.

Gm of the equivalent transistor is now twice as high as they are in parallel. The input voltage divides across both Re and 1/(2*gm).

Ic1+Ic2 = Ic = vin/( 1/(2*gm) + Re ) So vout = -0.5*Rc*vin/( 1/(2*gm) + Re ), hence the gain is Acm = -0.5*Rc/( 1/(2*gm) + Re )

Multiply with gm/gm (that is 1) gives:Acm = -0.5*Rc*gm/(0.5 + gm*Re) = -Rc*gm/(1+2*gm*Re).

When 2*gm*Re >>1, then Acm = -Rc/(2*Re) = -0.5*Rc/Re.

As you can see, this looks like the equation for a single BJT stage with emitter degeneration.

I hope if there is a typo, somebody will respond.

Thank you everybody, it's all fairly clear to me now, except the part concerning Ac, which, if i try to compute it, results equal to zero. Isn't it right to be zero since it is a differential amplifier?

For computing Ac I'm applying a constant voltage Vc on both inputs and i compute Vo = Vo1-Vo2...

Why isn't Ac equal to zero if the amplifier is ideal (both BJT with the same charateristics)?

I have a storm in my mind...

Computing, do you mean simulate or calculate based on the small signal equivalent circuit?

I mean calculating with small signal circuit

If V1 = Vc and V2 = Vc

ib1 = ib2 -> Vo1 = Vo2 -> Vo = 0 -> Ac = 0

Where am i mistaken?

Hello,

V1 = V2 = Vc (input) is the correct common mode definition.

Ib1 = Ib2 (assuming same transistors and same emitter bias current) is also correct. However Ib isn't zero, there is a path via Re. Ve follows virtually Vc (like in an emitter follower circuit), so you get an AC component through Re because of Vc. When hfe>>1, Ie (due to Vc) splits equally across the two collector resistances.

So when Re = 1 kOhm, gm = 0.1, vc = 10mVp you will get 10mV/(5+1k) = 9.95uAp AC current through Re. "5" comes from 1/(2*gm), as both gm are in parallel.

This current causes a voltage across the collector resistances (both are in phase, there is no differential output voltage).

ib1 = ib2 -> Vo1 = Vo2 -> Vo = 0 -> Ac = 0

Where am i mistaken?

Click to expand...

For common mode gain, you should calculate common mode output voltage Vo = Vo1 = Vo2, respectively Vo = (Vo1+Vo2)/2, not differential output voltage.

FvM said:

For common mode gain, you should calculate common mode output voltage Vo = Vo1 = Vo2, respectively Vo = (Vo1+Vo2)/2, not differential output voltage.

Click to expand...

Thank you FvM i think you find the problem and maybe now i have understood ..

Calculating Vo = (Vo1+Vo2)/2, in the small signal model of the circuit, from calculations i get:

ib1 = ib2 = Vc / (r_Π + 2*(1+B)*Re) _{(B=hfe= current gain)}

ic1 = ic2 = B * ib1 = Vc * B / (r_Π + 2*(1+B)*Re)

Vo1/Vc = Vo2/Vc = -Rc * B / (r_Π + 2*(1+B)*Re) ≈ -Rc * B / (r_Π + 2*B*Re)

using B = gm * r_Π

i finally get:

Vo1/Vc ≈ -Rc * B / (r_Π + 2*B*Re) = -2*Rc gm / (1+2*gm*Re)

•(Vo1+Vo2)/(2*Vc) = Ac = common mode gain = - Rc gm / (1+2*gm*Re)

Hi faustoleali, perhaps the following approach can improve your understanding.
There is no need to calculate - some thinking is sufficient.
In common mode operation both transistors do absolutely the same and, thus, both outputs are identical.
Therefore, it is sufficient to consider only one part of the circuit that operates as a classical common emitter stage with emitter degeneration (Remember: Re causes negative feedback because both currents have the same direction).

Result (derived from the formula collection for common emitter): Acommon=-gRc/(1+g*Rf).

What is Rf? Rf is the effective feedback resistor Rf=2*Re because a feedback voltage is developped across Re that is twice as much as in case of a "normal" common emitter stage (because the current is doubled).

LvW