# Common emitter amp input impedance

#### nik2009

Joined Jan 17, 2010
13
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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#### t_n_k

Joined Mar 6, 2009
5,455
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.

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#### Jony130

Joined Feb 17, 2009
5,475
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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#### nik2009

Joined Jan 17, 2010
13
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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#### nik2009

Joined Jan 17, 2010
13
Thank you, Jony130.

You must remember that beta is not a constant.
The BJT current gain is very heavily collector current dependent.
I'm going to plot β with the help of ACSweep.

#### Jony130

Joined Feb 17, 2009
5,475
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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#### MrChips

Joined Oct 2, 2009
29,809
The circuit as drawn has poor temperature stability. The top end of Rb should be connected to the collector of Q1.

#### nik2009

Joined Jan 17, 2010
13
The circuit as drawn has poor temperature stability. The top end of Rb should be connected to the collector of Q1.
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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#### t_n_k

Joined Mar 6, 2009
5,455
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|.

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#### Jony130

Joined Feb 17, 2009
5,475
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?)
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