Is this all impedance really is? That simple?

[Thecomedian]

From the way Z out is going, it seems to imply that Output impedance is actually related directly to only the resistor at the collector between Vcc and collector.

Is this right?

[greaser_au]

Yes... and no... it depends on the load imposed by the next stage. As you have the build shown,  the load is infinite, so from a service approach the Zout = Rc.

Tack a load resistor of say 2k,  5k or 10k between output & ground, and reduce c2 to 0.1u, and things will be somewhat different. The frequency response begins to take a hit, and the output begins to have a phase shift compared to the input as the frequency increases...  how is your imaginary numbers?

david

[Thecomedian]

From what I know, phase shift is mainly important because positive current will cancel out negative current in proportion to regular old addition and subtraction of positive and negative numbers, right? That's what causes attenuation of specific frequencies and is one of the main reasons to try to keep things "in phase" if you want full voltage?

as for imaginary numbers, they would be non-existent . I haven't gotten into the imaginary number side of working with circuits yet.

[greaser_au]

as for imaginary numbers, they would be non-existent . I haven't gotten into the imaginary number side of working with circuits yet.

hahaha yes... In simple terms,  what it means is that for RC circuits  (series R, parallel C)  with series capacitors the 'real' magnitude is reduced as the frequency increases (and the 'imaginary' part increases).  How's your trigonometry?  The 'real' magnitude at the output corresponds to the cosine (and the imaginary part is the sine).   For example, at the pole of an RC filter, the phase shift is 45 degrees, and the 'real magnitude is reduced to COS(45deg)=0.707,  the 'half-power' point (of course,  this reverses with seriesC'/parallel R)!

What we have (and what the simulator believes) is a circuit with a perfect voltage source, in series with R3, in series with  C2,  and a nonexistent/infinite  load impedance. i.e C2 is irrelevant. Not even imaginary real life

david

[R.G.]

Zout for a common emitter circuit is to a first approximation the collector resistor because the impedance of the transistor itself is so very high in this mode.

The collector impedance is in parallel with the collector resistor, but is so high that it changes the overall impedance almost nil.

[PRR]

That program gives some incorrect answers.



The results say Z_in is 100k. But base resistors 120k and 300k act in parallel as 85.7k.

Ah... it may be neglecting R1 R2, because R3 levered by transistor will be just about 100K alone.

However your guitar (or whatever) has to drive 85k||100k= 46k. That's the number we need.

I've also found an extreme case which claims gain of 1000.0, but realistic comps say 166.

> Output impedance is actually related directly to only the resistor at the collector between Vcc and collector.

For *that* circuit, most healthy transistors, this is essentially true. 10K at C will sum to about 9.9K output impedance.

(Vacuum tubes: kinda true for pentodes, way off for triodes.)

There are many good reasons to (and many products which) use a resistor collector to base. Z_out will be lower.

[Thecomedian]

I think you're supposed to select the Z_in and Z_out, and it will recalculate the resistors in relation to the desired impedance and the transistor and amp class used.

[PRR]

> I think you're supposed to

My stupidity is not involved. That image is on his website.

[PRR]

Here's my own attempt. I specified 100K input. It gave me 22K and 7.5K input resistors. Also the 18uF input cap for 10Hz response seems awful large, closer to 1K than 100K (so *some*thing in the code knows the true input impedance).



Also: setting supply voltage to 1V (or 1.3V) makes the program lock-up solid (incomplete input-checks). Hmmmm.... setting 120V brings up an error "800m..100", but it doesn't handle 800m, 1V, 1.3V.

[Thecomedian]

> I think you're supposed to

My stupidity is not involved. That image is on his website.

I was stating only as a fact.