You forgot that negative feedback actively reduces the output impedance, in addition to the resistors being paralleled.
Impedance for a BJT stage, with collector-base feedback resistor
- 6th February 2012 07:28 PM #1patroclus
Impedance for a BJT stage, with collector-base feedback resistor
BJT collector-base feedback impedance
Hello.
I'm getting trouble on getting (using PSPICE-like software) the input and output impedance of a common-emitter BJT amplifier with collector-base resistor, to try to see how it compares with a more common emitter degeneration feedback (or both). I attached schematic pictures of my set-up. I hope someone can help me see where am I missing something.
As far as I understand, this is a paralell-parallel negative feedback (voltage sampling, current mixing), so, in theory, both impedances should decrease in a factor of (1+AB), being A: amplifier gain, B: feedback circuit gain. Thus, this amplifier should have lower input and output impedance than and emitter-degeneration one (generally speaking, of course).
The first picture shows the circuit and simulation for input impedance, by doing an AC sweep and plotting:
Zi = (voltage at Vi) / (current through R4).
I get 186 ohms. This example is taken out of a book, where the author works out a very close 165 ohms Zi, so I think it's OK...
But, the second picture shows the simulation for output impedance, by placing a 1A test signal at output and shorting amplifier input to ground.
The book states that Zo = 495 ohm.
I get Zo = 4,27 kohms (which is just the parallel of collector and feedback resistor, which I find to be something logic. I don't know how to get the close-to 500 ohm result...
The values are significant. In fact, (1+AB) = 8,63 for this circuit.
and the value 495 ohms is got from:
Zo = 4,27 kohms / 8,63 = 495 ohms.
4,27kohms is what I find.
Any light?
Thank you so much.
- 6th February 2012 08:59 PM #2audioguru 0
Last edited by audioguru; 7th February 2012 at 12:14 PM.
Uncle $crooge
- 7th February 2012 05:21 AM #3gilmanli
You have first check the DC biase at collector of transistor,it should be 1/2 of V2 then feed a signal ( 1kHz ) to check whether the output is disortion or not.If it is serious disortion at output,this is no meaning you measure those impedance.
0
- 7th February 2012 05:45 AM #4crutschow
To measure the output impedance you need to ground the input at V1 (or just leave V1 connected since it has a zero AC impedance to ground), not ground the base (which effectively turns the transistor off).
0Zapper
Curmudgeon Elektroniker
--Inside every little problem is a big problem trying to get out.--
- 7th February 2012 10:52 AM #5patroclus
Thanks for all replies, I almost got it!
V1 is an AC source, so it isn't zero AC impedance to ground. But, following your tip, I shorted it (see attach) and now I get a Zo = 508.7 ohms, which is fairly close to the theorical 495 ohms !
Originally Posted by crutschow 
Yes, it's a little bit below 1/2 of supply voltage. Originally Posted by gilmanli
I'm still missing something.
If I use a test current of 1A at output, I measure 508,7 V (from where I get Zo = 508.7V / 1A = 508.7 ohm).
BUT... I think that 508.7 V from collector to emitter is something crazy, unless maybe you are trying to destroy the transistor. At 508V output, this amplifier just isn't working. Am I wrong?
If I use a test current of 1mA, I measure 508 mV (this seems reasonable), which leads to the same Zo = 508 ohms.
Is it OK to use a 1A test current to find Zo in any circuit, even though that would make an impossible output?0
- 7th February 2012 12:18 PM #6audioguru
You should not feed 1000mA into the output of your circuit that idles at 1.3mA and has a peak output current of 2.6mA.
0Uncle $crooge
- 7th February 2012 03:19 PM #7crutschow It may work in an AC simulation, but it is not realiistic and would never work in the actual circuit. Originally Posted by patroclus 0
Last edited by crutschow; 7th February 2012 at 03:19 PM.
Zapper
Curmudgeon Elektroniker
--Inside every little problem is a big problem trying to get out.--
- 7th February 2012 03:23 PM #8crutschow An ideal AC voltage source does have zero AC impedance to ground. An ideal current-source has infinite impedance. Originally Posted by patroclus 0Zapper
Curmudgeon Elektroniker
--Inside every little problem is a big problem trying to get out.--
- 7th February 2012 03:56 PM #9patroclus
I see. In simulation software it is OK to use a 1A current to get Zo. Right now, I'm only interested in simulation.
Thank you guys.
True. Originally Posted by crutschow
But, if I don't short the AC source to ground, I get Zo = 3.6 kohms, which just isn't right (I can attach the LTSpice project if you want to quickly try it).
To get the right Zo I need to set AC source to 0V or short it.
According to "output impedance" definition I read in my book, it is the (output voltage) / (output current) with signal source = 0V. I understand that to be AC source short to ground.0
- 7th February 2012 05:51 PM #10crutschow
Certainly you have to set the AC voltage to 0 when you do the simulation. Otherwise the input signal interferes with the signal you are injecting at the collector.
Even though the simulation may work at 1A it's still better to simulate under real conditions rather than impossible conditions.0Zapper
Curmudgeon Elektroniker
--Inside every little problem is a big problem trying to get out.--
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