What is the input impedance of this CMOS inverter circuit?

[anotherjim]

I can't get my head around this.

Is it the source impedance + whatever the cap impedance is? The minimum value required for the input cap suggests it is.

Does the negative feedback make the inverter input a virtual ground?

But if the input signal is hitting a virtual ground, how come it doesn't load the treble out? It sounds ok.

Duh 

[R.G.]

Is it the source impedance + whatever the cap impedance is? The minimum value required for the input cap suggests it is.

Does the negative feedback make the inverter input a virtual ground?

To the extent that negative feedback can make it so, yes, the output drives the inverting (and only!) input to be a virtual ground.

But if the input signal is hitting a virtual ground, how come it doesn't load the treble out? It sounds ok.

Probably because the inverting opamp stage is actually a *current cancelling* stage.

The rationale goes liike this:
The input to the amplifier cannot absorb any current. This is massively true for a CMOS amplifier. So any current that goes in through the cap must be cancelled to as nearly zero charge as the amplifier output can make it by driving a voltage through the feedback resistor. The current through the feedback resistor is equal to the inverse of the current coming in through the input [whatever].

As to why it isn't loading, well, it is. I suspect that what happens is that it's better to think of this as a current input, where the declining impedance of the cap with frequency allows in more signal current at the same rate the loading decreases. That is not true of a resistor, which has a constant load at all frequencies.

It's a theory, anyway.

[anotherjim]

Hmmm, many thanks for the reply RG, I'm going to have to think it through a bit.

I just tried lower feedback at 100k and that noticeably sucked treble out. I'm going to have to think about this too.

[R.G.]

You might want to put this into a simulator where you can stick current sensing sources in series with the input cap and feedback resistor.

Also notice that for a circuit like this, the source impedance is a critical part of the amplifier "input resistor".

[anotherjim]

OK, I think I'm comfortable with this now.
The effect of lower feedback R seems to clinch it.
The neg feedback became strong enough to noticably defeat signal where impedance of C is low.
Weaker feedback allows enough treble, although it must still be cutting some, I'm just not hearing significant cut over the guitar bandwidth.

Is there a name for this I wonder? Auto impedance?

[PRR]

> the source impedance is a critical part

+1

Or as I say, "draw the WHOLE circuit".

> The effect of lower feedback R seems to clinch it.

Don't JUST look at the "feedback R". Look at the *ratio* of that resistor to the SOURCE impedance (which you have not revealed).

The impedance of 0.2uFd is 10K at the bottom of the guitar band, 100r at the top. If Source Impedance is >10K, this plan is "flat" across the guitar band. If Zsource is <100r, this is a Differentiator, rising all up the guitar band.

That's pretending the amplifier is infinite gain. But it is a 4069, which has "some" gain, a lot at 3V, not-much at 12V; you have not revealed your supply voltage. Anyway the quasi-linear gain is not on the data-sheet, except as implied by the digital performance, so it is all guess-work.

When you have uncertain source impedance and uncertain gain, the problem takes a lot of scratch-paper. It can be done with a pencil; and the effort may lead you to insights the computer will never have or tell. Still it is wise to have a second opinion (however stupid). And one advantage of the idiot-assistant is that you *must* show a source, though many sims will happily default to zero impedance (impossibly infinite grunt).

Here is what I get for high and low source impedances with high and low gain.

(Sorry about the Vio/Yel cross-up.)


For some conditions the input "is capacitive", as you guessed. But there is much more going on. In either the hi-Z source or the low-gain situation, capacitance is swamped by stray resistance (either a 20K source or the 500K/30= 17K feedback resistor).

There's too much going on here for simple hand-calcs. If you need results (not just puzzles), pick a different implementation.

[anotherjim]

Paul,
Sorry, input was a single coil passive guitar (Chinese Squire workshop test plank). A short cable - 1 metre. I was keeping it's vol on max to minimise it's source impedance.  Supply is 5volt.
I haven't tried a bunch of 4069's in this circuit, but  all the ones I've used up to now (including this one), have very modest open loop gain.

I've never used anything like Spice. A search showed there might not be an accurate 4069UB type model out there anyway.

This doc'
http://www.ti.com/lit/an/scha004/scha004.pdf
...quotes 26dB @ 5v for an unbuffered NOR gate. A gain of 10 to 20 seems about right for the inverters I've used.


Fig 15 in this doc'...
http://www.intersil.com/content/dam/Intersil/documents/cd40/cd4069ubms.pdf
...shows an inverter with nothing else but a 10M feedback resistor and calls it a "High input impedance amplifier". Sadly it provides no other facts.

I think this suggests that the feedback resistor is so high that it only succeeds in auto biasing the inverter into linear mode. It cannot maintain a low impedance virtual ground at the input.

In that case, the feedback R might also be defining input impedance seen at the inverter input in ratio with the open loop gain. 10M/10=1M.

Then my circuit could be 100k Zin at the inverter input.

[PRR]

> input was a single coil passive guitar

A pretty complicated and complex source.

Also with live-picking there is so much going on that it is hard to be sure what you are hearing. Test-tones are a useful side-check.

Go ahead however you want. However I *suspect* it will be somewhat source-critical (different guitars give different response aside from whatever difference in the guitars). And CMOS is notoriously hissy.

I'd also want a few Kohms between an input jack and a CMOS input pin, to retard static-zap.