DC Report: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.
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?Typically the current thru r0 is quite small compared to gm*Vr, so ignored.
Regards, Dana.
Not true typically. Collector current typically in ma area, R0 in common base around 100k.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?
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).I'm sure you can see that if the emitter is grounded and you measure Rin at the base, it will be rpi. 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.
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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.
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...Not true typically. Collector current typically in ma area, R0 in common base around 100k.
Regards, Dana.
DC Report: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.
Yes,So next you need the input resistance at the collector, and the voltage gain from emitter to collector. Have you got any result?
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.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.
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 (?)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
It's Mathematica, but any of the modern "mathematical assistants" such as Matlab, Mathcad, etc., can do the same.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
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