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Common base amplifier output impedance

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teachlit

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I have calculated (with help) the input impedance of this common base transistor stage as 6.8 ohms. Now I want to know the output impedance of this stage.
I calculated the reactance of the inductor attached to the collector pin (assuming a value of 100 uH and a frequency of 5 mHz) as 3141 ohms. But I know from several sites that the output impedance of this type of amplifier is very high - in the megohms. So my question is: how can I calculate the output impedance for this circuit?
impedance.PNG
 
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how can I calculate the output impedance for this circuit?
In an attempt to answer my own question...
I read on one site that ZOUT = VCE/IC.

So here are my calculations, may the force be with me. (Not that I believe in the force anyway!)
Voltage feed for this circuit is 9 v. Current for this circuit is 3.8 mA.
Vce = Vcc - Vc - Ve
Vce = 9 - IcRc - IeRe
Vce = 9 - (0.0038) - (0.0038 x 2200)
Vce = 9 - 0.0038 - 8.36 = 0.63

So Vce/Ic (Output Impedance) = 0.63/.0038 = 165 ohms (This is not what I am expecting, I have missed something here).
I know there is an AC reactance of 314 ohms at 5 mHz but how do I integrate this reactance with the DC resistance?

Ah, perhaps I need Z = sq rt of DC R Squared plus AC R Squared
This would come out at 354 ohms so this is still not right.

Okay, everyone. I did my best... please can someone help...
 
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I'm no expert on this configuration, but I think that in order to understand the output impedance of the stage, you have to know the impedance of the collector load. In this case, the load is not a simple inductor, it is a transformer. So, it will be necessary to know what the impedance of the load connected to the transformer secondary (to the right of the transformer) is and then transform that impedance through the turns ratio of the transformer (recall that for an ideal transformer the impedance ratio is the square of the turns ratio). The output impedance of the amplifier will be roughly the collector load impedance that you calculate. Usually this value is somewhat high, like more than 1K ohms.

I reviewed this tutorial to understand:
**broken link removed**
 
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