Common Emitter Config with Negative FB

IMHO, the input impedance to the circuit is between the source resistance and the input terminal of the circuit, that is, at the junction of the 600 ohm resistor and the 47uF capacitor.

John S

If you use waveform arithmetic V(n003)/I(Rs) you don't need to change the source to 1V. Of course, you may if you wish.

John S

Il Fri, 29 Jul 2011 12:31:23 -0500, John S ha scritto:

I was referring to the first circuit:

which don't have the source resistance yet, so that we are taking the same point :-D

Anyways, I would tend to "ignore" the bypass cap as its function is to allow for proper DC decoupling but otherwise should be "transparent". That is, if I were to talk about the input impedance of such circuit, I would do it mid-band, where the cap is a short (at least, in theory).

Still better ;-D I would *first* evaluate the input impedance *after* the cap and consider it the somewhat "intrinsic" impedance. Now, this will be resistive at the beginning and - hopefully - greater than the impedance of the source we intend to drive the circuit with. From this value we *then* decide which bypass cap is suitable to obtain the lower corner frequency of the circuit :-D

Yes, that's what I meant by writing

It's just a shortcut I got used to. Comes handy when you're manipulating more complex expressions (like extracting equivalent C,L,R from a model). M

When SPICE does AC analysis it linearizes the circuit at the quiescent operating point, then doesn't check that it's linearizing assumptions are valid. You could use 10^12V for your generator, and SPICE will give you the same answers for gain, phase, impedance, etc.

So you want to use 1V or 1A, etc.

Yes.

Well, he emphasizes 60Hz on the schematic. Otherwise, it can be done any way one wishes.

Of course. It is up to the designer to decide what data is most useful to him/her.

John S

That impedance is 282.7 ohms. Phase shift is insignificant.

John S

Il Fri, 29 Jul 2011 13:45:41 -0500, in sci.electronics.design hai scritto:

So if I got you right, you say that the input impedance is dominated by what I called "intrinsic" impedance, and if I remember correctly we're stimulating @60KHz, which means the 47uF looks like a 1/(2*Pi*6e4*47e-6) which is on the neiborhood of 56mOhm...Yes, I'd say is negligible ;-)

Now, my Buglar way to calculate the input resistance gave me

Rin = R2//rb//(R3/K_v) with rb = Vt/Ib = 26e-3*300/0.841e-3 = 9.2k

Rin = 6.8k//9.2k//309 = 286 Ohm

Not bad ;-D

Il Fri, 29 Jul 2011 21:12:44 +0200, Michele Ancis ha scritto:

^^^^^^^^^^^

I beg your pardon, Sirs ;-D

Michele Ancis expounded in news:15wvrbvmvytc9$.zwqtsz96pq08$. snipped-for-privacy@40tude.net:

Actually 60 Hertz, just to be clear. ;-)

Warren

Tim Wescott expounded in news:IJudndiNQv35aK_TnZ2dnUVZ snipped-for-privacy@web-ster.com:

..

Ok but when I tried it again, LTspice produced the same screwy results.

Warren

John S expounded in news:j0uqch$i4a$ snipped-for-privacy@dont-email.me:

I'll have to chew on this over the long weekend.

Thanks, Warren

Actually, the stimulus is 60Hz as Warren pointed out elsewhere. Also, the 282.7 ohms with negligible phase shift is measured at the junction of the capacitor and the base. So, the 47uF is out of the picture completely. There is actually a phase shift in the millidegrees which is probably attributable to the transistor's intrinsic capacitance and can be ignored.

But, you're right; not bad at all.

John S

Il Fri, 29 Jul 2011 19:24:50 +0000 (UTC), Charmed Snark ha scritto:

Oooops, lucky me I said "if I remember..." LOL :-D

Yeah ok then just multiply by 1e3 and get 56 Ohms, not /that/ negligible The phase difference between voltage and current in the source being atan(Xc*R) = atan(56/280) ~ 11 deg

No need to apologize. You are doing much better than some natives here.

"Ballpark" is easier to spell.

John

BTW, Warren, .TF is the DC transfer function. It doesn't seem to work well with AC.

John S

If you want .TF :

First, do a .OP (DC operating point) and get Vb.

Change your source to zero AC and put in Vb for the offset.

Short the 47UF.

Then run your .TF

Do you then get what you want?

Cheers, John S

John S expounded in news:j0vcpn$geu$ snipped-for-privacy@dont-email.me:

..

Ok, I got V(vbe): 0.649086 voltage

after I labeled the node "Vbe").

The results came much out better (thanks). I did have to short out the source resistance Rs=600, to get the correct results:

--- Transfer Function ---

Transfer_function: -222.635 transfer v2#Input_impedance: 282.74 impedance output_impedance_at_V(out): 6920.69 impedance

So gain is pretty close to the 220 that was computed.

The input impedance also came close to the 286 ohms.

OK, LET ME SUMMARIZE, TO SEE IF I HAVE THIS RIGHT:

AC Gain Kv : Kv = -gm * ( Rc || ro || R3 ) VA Early Voltage : VA = -99 (in this case)

So: gm = Ic / VT = 0.032

So: ro = VA / Ic = 118k

So: Kv = -0.032 * ( 8.2k || 118k || 68k ) = 220

Input Impedance:

Rin = R2 || Rb || R3 / Kv

Rb = Beta / gm = 9.4k

So Rin = 6.8k || 9.4k || 309 = 286 ohms

Output Impedance:

Rout = R1 || ro || R3 = 8.2k || 118k || 68k = 6.9k

Very nice. Thanks everyone! Warren