How can I find the input impedance of this CB stage?

[giorgi3092]

I am given this circuit:

I draw small signal model like this:

Then, I am not sure how to write the KCL at the emitter.
A good samaritan online told me that the correct KCL would be:

v_pi/r_pi + gm*v_pi + i_x = 0

which yields R_in = v_x/i_x = r_e = V_T/I_C*alpha. which seems to be correct because I simulated in LTspice with typical values and calculations and simulation results match. Both gave 25 Ohm for I_c = 1mA, B=100 transistor.

What I cannot understand at all is why we leave out the current that flows thru r_o??
My naive KCL would be:
v_pi/r_pi + gm*v_pi - gm*v_pi + i_x = 0 so, I would also write the current that flows out of the node thru r_o, simplifying to v_pi/r_pi + i_x = 0, which gives R_in = v_x/i_x = r_pi which is totally WRONG!

Can anyone explain the weeds of this circuit? and why we leave r_o out of the KCL equation?

Thank you

 

[danadakk]

Typically the current thru r0 is quite small compared to gm*Vr, so ignored.

https://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/cbamp.pdf

Regards, Dana.

 

[giorgi3092]

Typically the current thru r0 is quite small compared to gm*Vr, so ignored.

https://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/cbamp.pdf

Regards, Dana.

But the current thru r_o is the same current that is gm*v_pi, so why don't we omit gm*v_pi as well?

 

[danadakk]

But the current thru r_o is the same current that is gm*v_pi, so why don't we omit gm*v_pi as well?

Not true typically. Collector current typically in ma area, R0 in common base around 100k.

http://www.ece.sunysb.edu/~oe/Leon/ESE372/Lecture16.pdf

Regards, Dana.

 

[The Electrician]

I'm sure you can see that if the emitter is grounded and you measure Rin at the base, it will be r_pi. But since Vout (the collector) is floating, the resistance between base and emitter is the same no matter which is grounded. Add some load, such a resistor Rc from the collector to ground and the input resistance will be different.

--- Updated Aug 1, 2021 ---

You say:
"v_pi/r_pi + gm*v_pi - gm*v_pi + i_x = 0 so, I would also write the current that flows out of the node thru r_o, simplifying to v_pi/r_pi + i_x = 0, which gives R_in = v_x/i_x = r_pi which is totally WRONG!"

You correctly have "v_pi/r_pi + i_x = 0" and you have the correct Rin (it's rpi)at the emitter.

 

[giorgi3092]

I'm sure you can see that if the emitter is grounded and you measure Rin at the base, it will be r_pi. But since Vout (the collector) is floating, the resistance between base and emitter is the same no matter which is grounded. Add some load, such a resistor Rc from the collector to ground and the input resistance will be different.

--- Updated Aug 1, 2021 ---

You say:
"v_pi/r_pi + gm*v_pi - gm*v_pi + i_x = 0 so, I would also write the current that flows out of the node thru r_o, simplifying to v_pi/r_pi + i_x = 0, which gives R_in = v_x/i_x = r_pi which is totally WRONG!"

You correctly have "v_pi/r_pi + i_x = 0" and you have the correct Rin (it's rpi)at the emitter.

I believe it can't be r_pi because I simulated the circuit in LTSpice and got about R_in=25 Ohms for typical circuit values (Ic=1ma, Vbe=0.8V, Beta=100).

But, computing r_pi, we get \[ r_{\pi} = \frac{100}{\frac{1mA}{26mV}}=2.6k\Omega \]

And this value of r_pi does not correspond to the simulation result.

sim:

 

[giorgi3092]

Not true typically. Collector current typically in ma area, R0 in common base around 100k.

http://www.ece.sunysb.edu/~oe/Leon/ESE372/Lecture16.pdf

Regards, Dana.

Wouldn't that be true only if the Ideal current source were not replaced by the open circuit? But, since the ideal current source is left open, that signal current has to flow thru that 100k resistor too...

 

[The Electrician]

Check the DC conditions of your simulated transistor. The Vb source voltage of .8 volts seems a little high. What is the current out of the Vb supply into the base of the transistor? What is the DC voltage at the emitter? You may have saturated your transistor.

 

[giorgi3092]

Check the DC conditions of your simulated transistor. The Vb source voltage of .8 volts seems a little high. What is the current out of the Vb supply into the base of the transistor? What is the DC voltage at the emitter? You may have saturated your transistor.

DC Report:
Semiconductor Device Operating Points:
--- Bipolar Transistors ---
Name: q1
Model: mynpn_1
Ib: 1.59e-03
Ic: 1.00e-03
Vbe: 8.00e-01
Vbc: 7.87e-01
Vce: 1.26e-02
BetaDC: 6.29e-01
Gm: 9.52e-02
Rpi: 3.60e+02
Rx: 1.00e+01
Ro: 5.90e+00
Cbe: 1.47e-10
Cbc: 1.76e-08
Cjs: 0.00e+00
BetaAC: 3.43e+01
Cbx: 0.00e+00
Ft: 8.52e+05


\[ V_{bc}=787mV \], so yes, you are right, the transistor \[Q_1\] is in deep saturation. For better visualization, I placed \[.op\] data on the schematic:

...
(fiddling with the ckt)
...
Yes, solution reached!
I placed a 10uA current source at the base to find out what voltage has to be established at the base for the transistor to operate in FA region. I chose 10uA because 1mA/beta = 10uA. It came out at 732.9mV and the collector voltage was at a higher potential, so all good for the bias.

Then I replaced the current source with 732.9mV voltage source and here is the result:

Yes, \[R_{in}=r_{\pi}\] which I hand calculated to be around \[2.6k\Omega\] as well. So, my naive solution was correct!

Bias is cool too:
Semiconductor Device Operating Points:
--- Bipolar Transistors ---
Name: q1
Model: mynpn_1
Ib: 1.00e-05
Ic: 1.00e-03
Vbe: 7.33e-01
Vbc: -1.76e-01
Vce: 9.09e-01
BetaDC: 1.00e+02
Gm: 3.85e-02
Rpi: 2.59e+03
Rx: 1.00e+01
Ro: 5.02e+04
Cbe: 5.67e-11
Cbc: 7.46e-12
Cjs: 0.00e+00
BetaAC: 9.96e+01
Cbx: 0.00e+00
Ft: 9.55e+07

Thank you!

 

[The Electrician]

So next you need the input resistance at the collector, and the voltage gain from emitter to collector. Have you got any result?

 

[giorgi3092]

So next you need the input resistance at the collector, and the voltage gain from emitter to collector. Have you got any result?

Yes,

The input resistance at the collector, i.e. the output resistance
\[R_{out}=r_o\]

And, I got the gain

\[A_v=1+g_m \cdot r_0\]

The work:

 

[The Electrician]

It's common when calculating the output impedance of a two-port (which is what you have) to assume the input is shorted to ground because typically the input will be driven by a voltage source. But that's not always the case. Your problem did not specify the state of the opposite port when calculating the impedance at a particular port. You chose to short the input when calculating Rout. Can you calculate Rout when the input is not shorted to ground. Also, for additional exercise , can you calculate Rin when there is a load resistance RL on the output?

Also, calculate the current gain with the output shorted. Then let gm -> infinity; this result will surprise you.

 

[FvM]

I have difficulties to understand the purpose of the problem specification in post #1 with infinite load impedance. It makes the input resistance mostly dependant on second order transistor parameters. The typical CB amplifier with reasonable finite gain has an input resistance of 1/gm.

 

[giorgi3092]

It's common when calculating the output impedance of a two-port (which is what you have) to assume the input is shorted to ground because typically the input will be driven by a voltage source. But that's not always the case. Your problem did not specify the state of the opposite port when calculating the impedance at a particular port. You chose to short the input when calculating Rout. Can you calculate Rout when the input is not shorted to ground. Also, for additional exercise , can you calculate Rin when there is a load resistance RL on the output?

Also, calculate the current gain with the output shorted. Then let gm -> infinity; this result will surprise you.

Textbook assumes that, unless otherwise stated, we are working with voltage sources, that's why I short the input in this example for Rout calculation.

Rout when the input is not shorted to ground.
Okay, I am going to assume that some type of HUGE resistance (basically open) is going to appear at the input, when I am calculating \[R_{out}\]. Let's say it's a current source, therefore it will appear as an open to very small perturbations of my test voltage \[v_x\]. Solution goes as:
\[R_{out}=\frac{v_x}{i_x}=r_o+r_{\pi}+g_m \cdot r_o \cdot r_{\pi}\]

Also, for additional exercise , can you calculate Rin when there is a load resistance RL on the output?
Solution for Rout and Rin:
(Did not bother to simplify Rin)

Current gain:

I read about this in the textbook:

 

[The Electrician]

You are a network analysis ACE. You might be interested in how to use linear algebra to quickly find everything you could want to know about your circuit. This circuit can be treated as a two-port: https://en.wikipedia.org/wiki/Two-port_network

Here is how it's done

 

[giorgi3092]

You are a network analysis ACE. You might be interested in how to use linear algebra to quickly find everything you could want to know about your circuit. This circuit can be treated as a two-port: https://en.wikipedia.org/wiki/Two-port_network

Here is how it's done
View attachment 171268

Damn, that looks like a very standardized way to solve this type of problems which I'd like to master. In my circuit theory class, we did not cover 2-port theory and I never bothered to do it myself. Time to do it. What software is that? My guess is Jupyter notebook (?)

I will go through this later and reply to this thread if something interesting comes up

 

[The Electrician]

Damn, that looks like a very standardized way to solve this type of problems which I'd like to master. In my circuit theory class, we did not cover 2-port theory and I never bothered to do it myself. Time to do it. What software is that? My guess is Jupyter notebook (?)

I will go through this later and reply to this thread if something interesting comes up

It's Mathematica, but any of the modern "mathematical assistants" such as Matlab, Mathcad, etc., can do the same.

Have a look here for how this method can greatly simplify analysis: https://forum.allaboutcircuits.com/threads/two-stage-bjt-amplifier-with-feedback.26710/