# Common emitter amp input impedance

Discussion in 'Homework Help' started by nik2009, Oct 4, 2011.

1. ### nik2009 Thread Starter New Member

Jan 17, 2010
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0
Hello guys,

I'm learning how this simple circuit works just for a hobby.
The circuit is a common emitter amp with fixed bias and grounded emitter.

I'm trying to calculate its input impedance. As I understand the input signal current flows through Q1 base-emitter (re) to the ground and through R2 to the ground.

So:
Zin = R2 || βre

β = 100
re = 25/IE Ohm ≈ 25mV/50mA = 0.5 Ohm

Then:
Zin = 50KOhm || 50Ohm = 49.95Ohm

But when I run a simulation the current through the input voltage source V1 is
56μA PP.

As V1 voltage is 5mV this gives
Zin = 5mV / 56μA ≈ 89.29Ohm

which is far away from 49.95Ohm I get from my calculations.

Could anyone help to figure out what am I doing wrong?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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That's a hefty emitter current. Reduce the current by a factor of 10 and review the results. Set Rb=500K and Rc=2.5k. And put some resistance (say 1k) in series with the source to improve the linearity.

Last edited: Oct 4, 2011
3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Simply the current gain "beta" of your BJT is not equal 100.
You must remember that beta is not a constant.
The BJT current gain is very heavily collector current dependent.

So the Zin = Xc + RB||(β+1)*re

In you case we can plot the input impedance with the help of a ACsweep analyze

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4. ### nik2009 Thread Starter New Member

Jan 17, 2010
13
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Hello t_n_k,

I changed the circuit in the way you suggested.

For this circuit I calculate the input impedance like this
Zin = Rs + (Rb || βre)

β = 100
re = 25mV/IE ≈ 25mV / 5mA = 5 Ohm

Zin = 1K + (500K + 100 * 5) = 1499.5 Ohm

But when I run a simulation it show 6.4μA flowing through the input source V1. The V1 voltage amplitude is still 5mV so this gives

Zin = 5mV / 6.4μA = 781.25 Ohm.

Almost two times less than the result of my calculations. I think there's a mistake in my calculations by I can't see where exactly.

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5. ### nik2009 Thread Starter New Member

Jan 17, 2010
13
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Thank you, Jony130.

I'm going to plot β with the help of ACSweep.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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For Vin = 5mV and F = 100Hz the capacitor reactance is equal to
Xc = 1/( 2 * pi * F * C1) = 159.154943Ω ≈ 160Ω
And the collector current is equal Ic = 4.837mA and Ib = 48.37μA
For Rb = 500K; Rc= 2.5K and Vcc = 25V

β = 4.837mA/48.37μA = 100

Zin = Rs + Xc + (Rb|| (β+1)*re = Xc + 1K + 500k||101*5.16Ω =
= Xc + 1KΩ + 500KΩ||522Ω = Xc + 1.521KΩ

so the final

Zin = √(1.52K^2 + 160^2) = 1.52KΩ

The LTspice give me 1.54KΩ
And form the transient analyze
The input current is equal to 3.2μA
Zin = 5mV/3.2μA = 1.56KΩ

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7. ### MrChips Moderator

Oct 2, 2009
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The circuit as drawn has poor temperature stability. The top end of Rb should be connected to the collector of Q1.

8. ### nik2009 Thread Starter New Member

Jan 17, 2010
13
0
Done it.

For this circuit I estimated
Zin = √(Xc^2 + (Rs + (Rb || (β + 1)re))^2)
β = 100, Zin = 1527

But the results from a simulation

give 5mv / 3.9μA = 1282 Ohm

I've read that negative feedback decreases the input impedance of a circuit. And there's a negative feedback in this circuit from the collector to the base through Rb.

So I guess the actual Zin is less than estimated because of this feedback. Is it correct?
(If it's correct that Zin is decreased because of the feedback, where can I find an explanation of why this happens? I know chapter 4 of the Art of electronics talks about feedback, should I search there?)

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9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Ignoring the capacitive reactance and the source resistance for the moment, a reasonable estimate for the resistance "looking into" the base would be

R_base = (1+β)re||Rb/(|Av|)

Where |Av| is the small signal voltage gain magnitude going from from base to collector.

In this circuit configuration |Av|≈Rc/re if Rc<<Rb {More accurately |Av|=(Rc[1/(1+Rc/Rb)])/re if β>>1}

So yes, the input resistance looking into the base decreases due to the negative feedback provided by Rb, with the effective AC equivalent resistance of Rb being reduced by a factor of 1/|Av|.

Last edited: Oct 5, 2011
10. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Yes , you right.
Zin is smaller because you use Rb as a feedback resistor.
Output voltage sampling - shunt (parallel) connections to the input--> shunt-shunt topology.

And this imagine will help you understand why this type of feedback reduce the input impedance

I1 = (1V - (-10V)/10Ω = 1.1A
So the Zin = 1V/1.1A = 0.9Ω

Or if we use |Ao|

Zin = Rb/(Ao+1)

Where Ao is the voltage gain.

Your BJT amp has the voltage gain equal to:

Ao ≈ Rc/re = 500V/V

Zin ≈ Rs + Rb/(Ao+1)||(β+1)*re = 1KΩ + 500Ω||500Ω ≈ 1.25KΩ

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11. ### nik2009 Thread Starter New Member

Jan 17, 2010
13
0
Thanks, t_n_k and Jony130!