Common emitter amplifier with collector current source bias

Andrei Suditu

Joined Jul 27, 2016
52
Hi,
I'm trying to find out the voltage gain and the input/output impedances for such a common emitter amplifier.
My resulting expression is something like Av=hfe*RE/(hie+RE)
hie is the base emitter resistance calculated from the formulas on paper.gm is an approximation for a collector current of 1mA(static operating point).Iset current source is also 1mA.I considered that the Vout is equal to the Collector current(Ib*hfe where Ib is also known ....1uA) times the RE resistance(Ib is small so i got rid of it).
I expect Zout (output impedance ) to depend mainly on RE.Is this true?
Input one i'm pretty sure is gonna be hie + RE.
Thx.

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Jony130

Joined Feb 17, 2009
5,419
My resulting expression is something like Av=hfe*RE/(hie+RE)
From the small-signal point of view any ideal constant current source must be treated as a open circuit.
We replace the ideal constant current source with a open circuit.
I expect Zout (output impedance ) to depend mainly on RE.Is this true?
No. How it can be true?

MrAl

Joined Jun 17, 2014
9,549

Hi,
I'm trying to find out the voltage gain and the input/output impedances for such a common emitter amplifier.
My resulting expression is something like Av=hfe*RE/(hie+RE)
hie is the base emitter resistance calculated from the formulas on paper.gm is an approximation for a collector current of 1mA(static operating point).Iset current source is also 1mA.I considered that the Vout is equal to the Collector current(Ib*hfe where Ib is also known ....1uA) times the RE resistance(Ib is small so i got rid of it).
I expect Zout (output impedance ) to depend mainly on RE.Is this true?
Input one i'm pretty sure is gonna be hie + RE.
Thx.
Hello,

What you could do is write the expressions for all the nodes, like input and base voltage with added small value resistor, then you can just use the expressions to calculate anything youwant.

For example, say you get:
Vout=f1(Vin)

then since the gain is Vout/Vin with a perfectly linear circuit, the gain is:
Av=f1(Vin)/Vin

f1(Vin) a function of Vin, will be semi complicated but not too bad.

A really simple example with a voltage divider with R1 on top and R2 on bottom taking the output from the junction and exciting with a DC source Vin volts...
The output is:
Vout=Vin*R2/(R1+R2)

and since the gain is Vout/Vin we get:
Av=[Vin*R2/(R1+R2)]/Vin=R2/(R1+R2)

So the gain is R2/(R1+R2)

To get input impedance you could calculate the input current 'Iin' and then:
Rin=Vin/Iin

Another very simple example with the voltage divider above...
Iin=Vin/(R1+R2)

and since the Rin is Vin/Iin we have:
Rin=Vin/[Vin/(R1+R2)]=R1+R2

so the input impedance is pure resistance R1+R2

Last edited:

MrAl

Joined Jun 17, 2014
9,549
Hi,
I'm trying to find out the voltage gain and the input/output impedances for such a common emitter amplifier.
My resulting expression is something like Av=hfe*RE/(hie+RE)
hie is the base emitter resistance calculated from the formulas on paper.gm is an approximation for a collector current of 1mA(static operating point).Iset current source is also 1mA.I considered that the Vout is equal to the Collector current(Ib*hfe where Ib is also known ....1uA) times the RE resistance(Ib is small so i got rid of it).
I expect Zout (output impedance ) to depend mainly on RE.Is this true?
Input one i'm pretty sure is gonna be hie + RE.
Thx.
Hello again,

Hey your calculation of hie does not look right.
You have:
hie=gm/hfe

however looking at the dimensions we have:
R=(I/V)/K

where K is a dimensionless constant, so we end up with:
R=I/V

which is NOT correct because I/V is a conductance not a resistance.
We must end up with:
R=V/I

so that means that:
hie=hfe/gm

Now do the same with the dimensions:
R=K/(I/V)

which since K is a dimensionless constant:
R=1/(I/V)

so we end up with:
R=V/I

and that is correct because Ohm's Laws tells us that R=V/I.

So the relationship must be:
hie=hfe/gm

A typical value for gm at low current would be 0.5 S, and for hfe would be 50.so we would have:
hie=50/0.5=100 Ohms

With gm=0.1 S we would have:
hie=50/0.1=500 Ohms

That's at least reasonable.

Now if we did it the other way around:
hie=0.1/50

we would get:
hie=0.002 Ohms

which looks unreasonable so the equation:
hie=hfe/gm

must be the right approximation.

Note:
The value of hie for a 2N2222 transistor can vary from about 2 to 5k depending on bias.
The value of hie for a 2N3906 according to the OnSemi data sheet can range from 2k to 12k for a collector current of 1ma. It's never as low as 0.002 however.
Using a spice model for the 2N2222 the measured values of hie vary from about 12 Ohms on up into several kilohms like 10k for Ic=1ma, however very low bias current can raise it even higher.

Last edited:

Picbuster

Joined Dec 2, 2013
1,040
Hi,
I'm trying to find out the voltage gain and the input/output impedances for such a common emitter amplifier.
My resulting expression is something like Av=hfe*RE/(hie+RE)
hie is the base emitter resistance calculated from the formulas on paper.gm is an approximation for a collector current of 1mA(static operating point).Iset current source is also 1mA.I considered that the Vout is equal to the Collector current(Ib*hfe where Ib is also known ....1uA) times the RE resistance(Ib is small so i got rid of it).
I expect Zout (output impedance ) to depend mainly on RE.Is this true?
Input one i'm pretty sure is gonna be hie + RE.
Thx.
A transistor is a current amplifier
Start with drawing the currents in B C E and define the relation like IC= Ib * amplification
Do the same for all.
However; there must be a point of no conductance find that point expressed in transistors base connection.
Express the voltage over the resistors Ohm law.( just the formulae)

Find an expression for output impedance.

Show us.

Picbuster

Andrei Suditu

Joined Jul 27, 2016
52
Hi.
Sorry for the delayed response and thx for the suggestions.I applied the suggested methods and it works.Forgot about the current sources.In the end i got somethin like gm*Rload since the current source goes open as someone pointed out.
I'll go talk with my lab professor on Monday and ask him a bit more.
I'll use this as the second stage after a differential pair (with a single ended output).
Thx guys.
PS:I'll simulate my whole circuit in LTSpice after too cross check my solutions.