Question about input and output impedances between connected devices

[GGBB]

I have a general understanding about input and output impedances but I'm having trouble finding clarification about one thing. When connecting two devices, e.g. two pedals or pedal into amp, are the respective input and output impedances combined? Here is an example of what I mean - a simple inverting BJT buffer - but it might be opamps or FETs:

So effectively B is two As connected together, and the active devices Q1 and Q2 combine their respective impedances. Q1 sees Q2's input impedance as exactly the same as Q1's output impedance seen by Q2. Seems like this should be simple, but something about it doesn't seem right. What am I missing? Note - I am thinking about the general concept, not the precise formulas.

[amptramp]

The output impedance in A is dominated by R4 in parallel with R6 since the transistor acts as a current source i.e. infinite impedance.  Below the audio frequencies, C2 has some effect.

The input impedance of A is dominated by R1 in parallel with R2 and R3 and the input impedance of the transistor.  If C3 is not there, R5 is multiplied by the hfe of Q1.  If there is no R5 or you are in a part of the audio where the reactance of C3 is much smaller than the resistance of R5, the transistor itself dominates with an effective emitter resistance of (26 ohms / emitter milliamps) for silicon transistors.  If you have 1 mA, the emitter resistance is 26 ohms, it you have 2 mA, the emitter resistance is 13 ohms etc.  This is in series with the paralleled reactance of C3 and R5 and this value is multiplied by hfe to give you the input impedance seen "looking into" the base.

In circuit B, there is no need for R6 and R7 or C4.  You can have C2 connected directly from the collector of the first transistor to the base of the second transistor.  You have a voltage divider between R4 (the output source impedance) and the paralleled R8, R9 and base input impedance of the second stage.  I will leave precise formulae to the more anal retentive to make any corrections to my verbal diarrhea.

[GGBB]

Thanks, but I'm not after the precise formula - I'm asking about the general concept of how the input and output impedances interact when connected. B is not a single circuit - it is the combination of two A circuits back to back as one might have with two pedals or a pedal before an amp. R1 and R7 are the typical input pulldown resistors, R6 and R12 are the typical output pulldowns (or perhaps output volume controls). The connection between R6 and R7 would be a cable. So the question is, in diagram B, is the input impedance of Q2 as seen by Q1, and the output impedance of Q1 as seen by Q2 one and the same?

[anotherjim]

The short answer is that with with high Zin of around 1M, Zout of most designs is not going to need much thinking about.

In the case of frequency response of the output cap, the total source and total load impedances are in series. However, if source Z is 10k and load is 1M, then source Z is almost negligible in the RC equation.

If the device is also expected to drive a low Z such as a 10k mixer line in, then Z out would have to be made low using an op-amp, say 100R and the output cap would be sized bigger to give needed low end with 10k load in the equation. This may be why some designs have apparently oversized coupling caps - maybe over 1uF - while 1M load means 100nF is more than big enough.

For the discrete designs with relatively high Zout, a 10k load will also pad down the signal amplitude quite a lot, but so long as Zout isn't much more than 100k, a 1M load will not noticeably load it down.

In both A & B, there are resistors (R6 & R12) across the output. If they are anti-pop resistors of very high value (>2.2M), they may be negligible. If they represent volume control pots set to full of moderate R (47k-250k), then they lower the load resistance seen by the output caps (C2 & C5) so R6 or R12¦¦Zload. To drive 10k inputs, it really looks more sensible to place any output volume control before an output buffer, so the output cap only has to handle the 10k load and our volume control taper is not changed by the low load resistance.

There are designs that place RC filters right at the output. Those are assuming a high Z load will be next and would need buffering before a line input.

If anyone wants to sell pedals to keyboard players, then I think they should ensure they can handle driving line inputs, because that's how most will use them.

[PRR]

Your straw-man may be easier to discuss or think-about with values. With values we may see that some parts "do not matter". Is, say, R6 1 Ohm or 1Meg?

Your straw-man is hairy. R6 R7 can obviously be combined. And on second thought, thrown-out, because they derive from original A's R1 R6 which are probably only there for switch/jack pop control. Then C2 C4 can probably be combined.

For that matter, the circuit drawn is probably "too wicked" for any sane audio purpose. (Guitar distortion may be an exception.) The emitters are fully bypassed. 20mV input to a BJT will be large distortion, 60mV pretty clipped. These are smaller than many audio sources. (Guitar 20mV-200mV). Q1 will usually be slammed. Then Q2 is over-slammed. This is a clipper, not an amplifier.

Your formula does not consider Q1's collector internal impedance (usually too high to care) or Q2's base input impedance (often quite low, dominating the whole network).

[GGBB]

I truly appreciate the answers, but I don't think I'm any closer to an answer (that I can understand) - I guess I need to rephrase my question. Forget the example circuit - let me ask it this way - when you connect two pedals together, does resistance on the output of the first pedal (say a volume pot - some series resistance and some resistance to ground) after the coupling cap from active output device (say a transistor or op-amp) affect the input impedance of the second pedal (assuming it has an active input section) when you consider the combination of the two pedals as one circuit? (Is that any better?)

[PRR]

> does resistance on the output of the first pedal ... after the coupling cap from active output device ... affect the input impedance of the second pedal ... when you consider the combination of the two pedals as one circuit?

Well, when combined as "one circuit", the impedances are combined.

Impedance is usually not the goal, but a tool for predicting. We choose in and out impedances for effective convenient coupling. We like to know what we will get. "Matched" impedance makes the most out of very expensive gain. "Bridging" impedance makes connection calculations much simpler. One non-pedal application: the power line to my house is 0.4 Ohms, while my house is down-to 5 Ohms. I can expect to get nearly full 250V (maybe 230V).

Or is your question about the *caps*? Yes, a naked transistor with 5K collector resistor, driving a 5K pot, you figure the coupling cap on 10K, not 5K.

[Rob Strand]

Forget the example circuit - let me ask it this way - when you connect two pedals together, does resistance on the output of the first pedal (say a volume pot - some series resistance and some resistance to ground) after the coupling cap from active output device (say a transistor or op-amp) affect the input impedance of the second pedal (assuming it has an active input section) when you consider the combination of the two pedals as one circuit? (Is that any better?)

The output impedance of source device and the input impedance of the driven device do not change.  That are what what they are connect and not connected to each other.

However, and this is where you might be getting caught-up, when you start calculating gains and cut-off frequencies the two circuits *do* interact and the response is *not* the same as what you would calculate for the two in isolation.

The 'p' and 's' means series and parallel.

Example 1:   Effect on gain.
Routs = 5kohm  (appears in series with the output)
Rinp = 10kohm  (appears from the input to ground)
When connected the two circuits form a voltage divider with voltage gain:
A = Rinp / (Rinp + Routs) = 10 / (5 + 10) = 0.67

Fairly easy understand that one loads the other.

Example 2:   Effect on high-pass filter
Routs = 5kohm
Rinp = 10kohm
Cins  = 160nF   (Cins is in series with the input and Rinp goes to ground)

Before connecting two the -3dB point for high-pass filter at the input is,
f3  = 1/(2*pi*Rinp * Cins)  = 1/(2*pi*10k*160nF) = 99.5Hz
The assumption here is the thing driving the input has low (ie. zero impedance).

However when you connect the two the capacitor "sees"  Rinp and Routs in series.
f3 = 1/(2*pi*(Routs+Rinp)*Cin) = 1/(2*pi*(10k+5k)*160nF) = 66.4Hz.

To understand what the capacitor sees you need to treat the source as an ideal source (zero ohms)
in series with Routs.   You the look into the circuit from the perspective of the capacitor terminals
and you will see Rinp and Routs appear in series.

Example 3:   Effect on a low-pass filter
This example is simplified as it has no DC input impedance on the input side.
Routs = 5kohm (in series with output)
Rins = 1k (from input terminal then to the cap Cinp)
Cinp  = 1.6nF  (this connects from Rins to ground, then to the next stage input).

Before the connection the low-pass cut-off frequency before the connection is,
f3 = 1/(2*pi*Rins *Cinp) = 1/(2*pi*1k*1.6n) = 99.5kHz
The assumption here is the thing driving the input has low (ie. zero impedance).

After the connection the capacitor sees Rins and Routs in series so you get
f3 = 1/(2*pi*(Rins+Routs) *Cin) = 1/(2*pi*(1k+5k)*1.6n) = 16.6kHz

Example 4:   Effect on a low-pass filter with an input impedance  (harder)
Routs = 5kohm (in series with output)
Rins = 1k (from input terminal then to the cap Cinp)
Cinp  = 1.6nF  (this connects from Rins to ground, then to the next stage input).
Rinp  = 10k      (this connects in parallel with Cinp)

Before connecting the two, we assume the input is connected to a zero impedance source,
the capacitor actually sees Rins and Rinp in parallel.
Rp  = 1/(1/Rins + 1/Rinp) = 1/(1/1k +1/10k) = 909 ohms
The -3db cutoff is then
f3 = 1/(2*pi*Rp*Cinp) = 1/(2*pi*909*1.6n) = 109kHz

After the connection the Routs and Rins appears in series.   The capacitor then sees,
Rinp in parallel with the series combination of Routs and Rins.

Routs in series with Rins = 5k + 1k = 6k
Then that in parallel with Rinp is,
Rp = 1/(10k + 1/6k) = 3.75k
f3 = 1/(2*pi*Rp*Cinp) = 1/(2*pi*3.75k*1.6n) = 26.5kHz

You will need to draw out the circuits and stare at them for a while.   There's plenty of engineers that don't get this stuff.   If you look-up "Thevenin Equivalent Circuit" it might help with the idea of separating the source and its impedance.  The idea of "what the capacitor sees"  is actually embedded in the analysis of circuits.  There are circuits with many caps where this idea does not work in an obvious way.  This is when you have have to analyse circuit with a lot of maths.

Most of the time we make approximations where we try to ignore the interactions but there are many instances where you can't do this and have to use the ideas above.

[GGBB]

Thanks guys. Much clearer now, but as Rob said, I'll need to "stare at it for a while" (particularly example 4).