# small signal circuit

Joined Feb 3, 2016
42
Hi
I am trying to solve this small signal problem

Assume
VC = VCC / 2
β= 80
and VT = 26 mV
Find the small-signal voltage gain Av = Vo / VS

I am not sure how to find the voltage Vb,Vs and draw the small signal model

For the DC part
Vcc=Ic*Rc+VC+IERE
8=β*Ib*2.5k+4+(1+β)*Ib*600
Ib=0.01609mA

For the AC part
Rπ=VT/ib
=1540ohm

I am not sure how to find the voltage Vb,Vs

Vo=-βib*Rc
Vb=ib*Rπ + ib*(1+β)*Re
Vb=Vs ?

Av=Vo/Vs
Av=Vo/Vb*Vb/Vs
Av=[-βib*Rc]/[ib*Rπ+(1+β)*ib*Re]*1
Av=-4

#### shteii01

Joined Feb 19, 2010
4,644
Vb=ib*Rπ + ib*(1+β)*Re

That does not look right.
The current entering Re from the right is the $$\beta\times{I_b}$$. Where is 1 in $$1+\beta$$ coming from?

From my textbook:
vs=ib*Rb+vbe

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Joined Feb 3, 2016
42
There is another current entering Rπ
So I guess the current entering Re which is ib+β*ib

#### MrAl

Joined Jun 17, 2014
10,087
Hello there,

Let's look at this part of your result:
For the DC part
Vcc=Ic*Rc+VC+IERE
8=β*Ib*2.5k+4+(1+β)*Ib*600
Ib=0.01609mA
In line 2 you are summing the collector resistor voltage drop, the collector to ground voltage, AND the emitter to ground voltage. But that is not a valid summation because if you already know the collector voltage then you dont have to add the emitter voltage. Doing that leads to a voltage that is too high from Vcc to ground. If you did know the collector to emitter voltage then that would be a valid calculation (using Vce instead of Vc), but you dont know that right away and that's too much work anyway.
To look at this another way, check your result by using the current you calculated to compute the voltage drop across the collector resistor. If Vcc=8 and Vc=4, the voltage drop MUST come out to 4 volts.

Knowing the DC collector voltage and Vcc and Rc you can find the collector current very easily, and looking at your second diagram you can see that the only source of current which we call iC=ib*B is the DC collector voltage and the collector resistor.

With this in mind, try again and see what you get for ib, then check your result knowing that iC=ib*B and what the drop across Rc should be.

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Joined Feb 3, 2016
42
Thank you
It means that I can use the voltage drop between Vcc and Vce find Ic?

Ic =Vcc-Vce/Rc
Ic=8-4/2.5k
Ic =1.6mA

Then substitute it to the ac part?

#### MrAl

Joined Jun 17, 2014
10,087
Thank you
It means that I can use the voltage drop between Vcc and Vce find Ic?

Ic =Vcc-Vce/Rc
Ic=8-4/2.5k
Ic =1.6mA

Then substitute it to the ac part?
Hello again,

That's much better. I was assuming that we were doing the DC part first though, so you could find the DC base current quite easily now, as you know is ib=Ic/B. See if you can take it farther now.

#### shteii01

Joined Feb 19, 2010
4,644
Hello again,

That's much better. I was assuming that we were doing the DC part first though, so you could find the DC base current quite easily now, as you know is ib=Ic/B. See if you can take it farther now.
Thank you
It means that I can use the voltage drop between Vcc and Vce find Ic?

Ic =Vcc-Vce/Rc
Ic=8-4/2.5k
Ic =1.6mA

Then substitute it to the ac part?
Is $$Vc=V_{CE}+V_{R_E}$$ ?

Joined Feb 3, 2016
42
Hi
Ic =1.6mA
Ib=1.6m/80
Ib=0.02mA
Then
I assume that my ac part is correct
Rπ=Vt/Ib
=26m/0.02m
=1300
Vb=ib*Rπ+(1+beta)ib*Re

Vo=-(beta)*ib*Rc
Since Vb=Vs
Vo/vb*Vb/vs
=-4

But I still get a result close to-4

Joined Feb 3, 2016
42

#### shteii01

Joined Feb 19, 2010
4,644
I think Vc Vce+Vre
You think?
Why do you think that?

At this point I assume that Vc is from Collector to Ground, which means that Vc=Vce+Vre.

Joined Feb 3, 2016
42
You think?
Why do you think that?

At this point I assume that Vc is from Collector to Ground, which means that Vc=Vce+Vre.
the question dont provide the value of Vbb
So if Vc=Vce+Vre
then Vc=Vce+Re(1+β)*Ib
There will be two unknown value (Ib and Vce)
But there is only one equation
Vcc-0=Ic*Rc+Vce+Re(1+β)*Ib

Joined Feb 3, 2016
42
So I guess the blue circle is Vc

#### Jony130

Joined Feb 17, 2009
5,445
But I still get a result close to-4
So the -4 V/V is wrong answer ?

#### shteii01

Joined Feb 19, 2010
4,644
The formula in my textbook:
$$A_v=\frac{V_O}{V_S}=-(g_m\times{R_C})\times{(\frac{r_\pi}{r_\pi+R_B})} R_B=0 R_C=2.5k\Omega r_\pi=\frac{\beta\times{V_T}}{I_{CQ}} I_{CQ}=1.6mA g_m=\frac{I_{CQ}}{V_T}$$

There you have it. You know all the pieces now, plug them into formula and solve.

Joined Feb 3, 2016
42
So the -4 V/V is wrong answer ?
I am not sure
But I get the same ans using different method

#### shteii01

Joined Feb 19, 2010
4,644
I am not sure
But I get the same ans using different method
I get Av=-153.8 or just round it up to Av=-154.

Joined Feb 3, 2016
42
I get Av=-153.8 or just round it up to Av=-154.
It may be too small
I see some examples on my book
Their voltage gain are larger than -100

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#### shteii01

Joined Feb 19, 2010
4,644
It may be too small
I see some examples on my book
Their voltage gain are larger than -100
The minus sign indicates that the output is inverted. So the actual gain is 154, it just the polarity is different. If you had a sine wave that oscillates around zero as an input, say from -1 to +1 mV, then output would have been another sine wave, but it would be from -154 to +154 mV. In this case the minus sign would work this way, when you feed -1 mV, output is +154 mV, when you feed in +1 mV, the output is -154 mV. See? The output is inverted and amplified input.

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#### Jony130

Joined Feb 17, 2009
5,445
I am not sure
But I get the same ans using different method
For this CE amplifier with emitter degeneration resistor (Re is not bypassed by a capacitor) the voltage gain is equal to :
Av =- Rc/(Re + re) * hfe/(hfe + 1) ≈ - Rc/(Re + re) and if Re>>re we have Av = - Rc/Re
Where little re is equal to Vt/Ie ≈ 26mV/Ic. And this is why your amplifier gain is around -4V/V because Rc/Re = 2.5kΩ/0.6kΩ = 4.16.

And if you add CE capacitor across RE resistor the voltage gain value increases to the new value:
Av = - Rc/re * hfe/(hfe + 1) ≈ 2.5kΩ/16Ω ≈ -156 V/V.

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#### MrAl

Joined Jun 17, 2014
10,087
Hello again,

Again we are faced with what is allowed to be assumed and what is not allowed.
In this case, we might ask if we can keep rpi constant which would be close if we assumed very small AC changes at the input. If we can not keep rpi constant, then we must include the way it changes as the input AC current changes.

Another simplification we might notice is that because rpi is defined as:
rpi=VT/ib

then that means that according to the original equations the voltage drop between base and emitter is only 0.026v and that may or may not be allowed. Probably a better assumption is 0.026+0.7 or something like that, where rpi is the dynamic resistance of the base emitter diode and the static resistance is 0.7/ib. The combined resistance would then be used to solve for the operating point.

When i use the 0.026v alone i get a gain of -4.115 and that is by considering that the input voltage is some DC voltage plus some AC voltage, and using the peak AC voltage to calculate the voltage gain, not just the DC voltage. The AC gain is different because rpi changes with base current.
I have not done it with the extra 0.7v drop, but since that is in series with the base and it is constant, then it should be able to be added to the DC input bias voltage (i assume is Vbb).

Some questions that have to be answered in order to get the right results, and some of these can only come from knowing how the course work progressed in the past or what the instructor is looking for.
This happens a lot when questions that are not simple come up because we dont know the list of assumptions and presumptions that we would have known had we been sitting in the class since day 1 ourselves.
Of course if the OP can provide a little more detail that would help. The best way is probably to show a detailed example from the past that has already been solved. The assumptions then become more transparent to any readers of the first post.

Assuming we can either use just 0.026 or just later add 0.7 to the input bias DC voltage, a way to solve this is to come up with the solution for Vc after considering all variables, then use:
Av=(Vc1-Vc0)/(Vin1-Vin0)

which is simply the difference in the collector voltages with the starting input DC bias voltage calculated with Vc=4v, and the collector voltage with an increment in DC bias of say 0.1v, and divide that by the increment in DC bias voltage.

Since calculating Vc with all variables considered is key to the solution, you can use this formula to check your results, but dont rely only on this formula rather work it out for yourself and then compare:
Vc=(B*R3*VT)/(B*R2+R2)-(B*Ein*R3)/(B*R2+R2)+(B*E2*R2)/(B*R2+R2)+(E2*R2)/(B*R2+R2)

Factored a little:
Vc=E2-B/(B+1)*(Ein-VT)*R3/R2

Note in the above that the only variables are Vc and Ein, where Ein is the bias voltage plus the AC peak voltage: Ein=Edc+Vpk.
To calculate the initial input DC bias alone (Edc), first set Vc=4 and Vpk=0 (known values of those two) then solve for Edc. Once you have Edc then you can either calculate the first derivative or just use a small increment such as 0.1v as Vpk to calculate the gain, which will then be (Vc1-Vc0)/Vpk.
Sometimes it is good to go both ways, calculate the gains with a positive excursion and then with a negative excursion, then calculate the average gain.

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