# Input and Output Impedance of the Common Emitter Amplifier

#### cikalekli

Joined Dec 10, 2020
103
hi,
You have already correctly calculated the Base current as 23uA.
With a beta of 100 , Ic = 100 * Ib.

The Is ranges from 10^-12 thru 10^-15... I would use 10^-12.

What is the problem.?
E

For self study there are number of video's and tutorials on the web
Oh okay, I haven't known that there is a range for that particularly, now it's okay. Thank you again eric.

Actually, I do look at a lot of places but I gradually confused. Because there are many different model designs and each of them has a particularly different way to find impedances. That is why I haven't seen such a place where explained it with totally. Sorry...

#### ericgibbs

Joined Jan 29, 2010
19,078
hi,
Look at this LTSpice simulation of your circuit showing the Zin with 100K and direct Base input.

The plots show the Zin impedances for the two circuits.

E

Updated:

another image.

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#### cikalekli

Joined Dec 10, 2020
103
hi,
Look at this LTSpice simulation of your circuit showing the Zin with 100K and direct Base input.

The plots show the Zin impedances for the two circuits.

E

Updated:
another image.
hi,
Look at this LTSpice simulation of your circuit showing the Zin with 100K and direct Base input.

The plots show the Zin impedances for the two circuits.

E

Updated:
another image.
Actually, that's impeccably perfect to see this circuit inside the LTspice with the variables.
That's amazingly beneficial to understand the concept, thank you again eric...

#### ericgibbs

Joined Jan 29, 2010
19,078
hi,
How do those plot simulations compare with your calculated results.?

E

BTW:
If you intend to become a electronics design engineer, LTSpice can help test your designs.
So learning LTS would be a big advantage.

#### cikalekli

Joined Dec 10, 2020
103
hi,
How do those plot simulations compare with your calculated results.?

E

BTW:
If you intend to become a electronics design engineer, LTSpice can help test your designs.
So learning LTS would be a big advantage.
Actually, I tried to follow the rule here:
https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

Maybe nowadays, I find it hard to learn sometimes as a self-learner. However, just be sure that this summertime I will spend all of my time for LTspice and Matlab to test my earlier circuits related to my lectures.

I have found unfortunately absurd values. I could not find the input and output results that you have found inside the LTspice, I will try it again to understand the hand calculation.

Again I'm sorry Eric, I bet from your view, I look so miserable but this whole year, I needed to cover up everything just on my own. Just because, my academicians just do 1-2 examples inside the presentation and finish during the online lessons. They tell us that, we need to research but I wish at least they could supply regular lecture notes as well...

#### ericgibbs

Joined Jan 29, 2010
19,078
hi cika,
I do not consider you to be miserable, I respect people who make the effort to understand their subject of interest.
You will get some people who seem to enjoy putting you down as a learner, but let that make you more determined to succeed in your studies.

My Institution Motto is: DISCE DOCE, translated means Learn to Teach.

So hopefully one day you can teach others.

E

#### cikalekli

Joined Dec 10, 2020
103
hi cika,
I do not consider you to be miserable, I respect people who make the effort to understand their subject of interest.
You will get some people who seem to enjoy putting you down as a learner, but let that make you more determined to succeed in your studies.

My Institution Motto is: DISCE DOCE, translated means Learn to Teach.

So hopefully one day you can teach others.

E
I hope as well, I loved the motto. Simple and effective.

I'm considerably honored to hear those words from you, sir, thank you so much again.

Edit: Ah I see that motto is also written in here:
It is mentioned in this latin poet:
490 of Book 2 of the "Georgics" (29 BC), by the Latin poet Virgil (70 - 19 BC).
Felix, qui potuit rerum cognoscere causas ( Fortunate, who was able to know the causes of things)
That's lovely.

Last edited:

#### neonstrobe

Joined May 15, 2009
191
Looks like Eric has got you sorted. Just some additional comments - the forward voltage of a silicon transistor or diode is "about" 0.7V (it used to be 0.6V when I first studied electronics). As you have seen it is not a fixed value but depends on the saturation current of the transistor (and temperature). That voltage is needed for any D.C. or circuit operating conditions.
When you calculate the signal gain it is an A.C. parameter that is of interest. Instead of a diode, an equivalent impedance is used, so as you have seen, the D.C. model of a diode with "0.7V" drop is replaced by the small signal Rpi. That value is also strongly dependent on the transistor and operating current.
The small signal impedance can be determined by using a small signal added to the D.C. conditions, but it has to be small. like 1mV. If you simulate that you add a low frequency sinewave (e.g. 1kHz) of 1mV (rms or peak, up to you as long as you remember when you look at the output) in series with the D.C. bias.
SPICE simulators usually calculate the operating conditions from the D.C. and then use that for the A.C. analysis. That is also how many hand calculations of AC gain are done.
The impedance of a diode or transistor emitter-base junction can be* Vt/Ie which gives a gm of Ie/Vt. The input impedance to the base is beta or hfe times higher.
So in your example, if your collector current is 2.3mA the input impedance (if you were looking into the emitter with the base grounded) is approximately 11 ohms. Multiplied by 100 gives you the rpi of 1100 ohms. That is the value you need to calculate the circuit gain. You then ignore the diode voltage drop as that does not apply to a small signal or A.C. calculation, even if you (or SPICE) uses 1V on the input to the circuit. Also in this case the D.C. bias is not relevant, so you just use 1V on the input to your 100k resistor which is in series with Rpi. Then you can either use the current flowing as the base current to calculate the collector current, or the voltage across Rpi and use gm to get the collector current, then use that and the collector load (ohm's law) to get the output voltage. The answers should be the same.
* "can be" because it depends on how ideal the diode or transistor is, and Vt depends on temperature.

#### cikalekli

Joined Dec 10, 2020
103
Looks like Eric has got you sorted. Just some additional comments - the forward voltage of a silicon transistor or diode is "about" 0.7V (it used to be 0.6V when I first studied electronics). As you have seen it is not a fixed value but depends on the saturation current of the transistor (and temperature). That voltage is needed for any D.C. or circuit operating conditions.
When you calculate the signal gain it is an A.C. parameter that is of interest. Instead of a diode, an equivalent impedance is used, so as you have seen, the D.C. model of a diode with "0.7V" drop is replaced by the small signal Rpi. That value is also strongly dependent on the transistor and operating current.
The small signal impedance can be determined by using a small signal added to the D.C. conditions, but it has to be small. like 1mV. If you simulate that you add a low frequency sinewave (e.g. 1kHz) of 1mV (rms or peak, up to you as long as you remember when you look at the output) in series with the D.C. bias.
SPICE simulators usually calculate the operating conditions from the D.C. and then use that for the A.C. analysis. That is also how many hand calculations of AC gain are done.
The impedance of a diode or transistor emitter-base junction can be* Vt/Ie which gives a gm of Ie/Vt. The input impedance to the base is beta or hfe times higher.
So in your example, if your collector current is 2.3mA the input impedance (if you were looking into the emitter with the base grounded) is approximately 11 ohms. Multiplied by 100 gives you the rpi of 1100 ohms. That is the value you need to calculate the circuit gain. You then ignore the diode voltage drop as that does not apply to a small signal or A.C. calculation, even if you (or SPICE) uses 1V on the input to the circuit. Also in this case the D.C. bias is not relevant, so you just use 1V on the input to your 100k resistor which is in series with Rpi. Then you can either use the current flowing as the base current to calculate the collector current, or the voltage across Rpi and use gm to get the collector current, then use that and the collector load (ohm's law) to get the output voltage. The answers should be the same.
* "can be" because it depends on how ideal the diode or transistor is, and Vt depends on temperature.
Thank you su much for you detailed well covered explanation.

Actually I the thing what you have written really made a great ssummary of my question.

Thank you again.