input impedance

[djp8djp]

I've got a little opamp buffer. Works fine. But I was wondering about something. At the input there's a ferrite then a1Meg pulldown to ground, then a series 33nF cap. Then there's the biasing resistor, 510k to REF (half-supply), and from there into the non-inverting input. So my question is - does the biasing resistor contribute to the input impedance? If so, I've got 338k, and if not, it's 1Meg. Yes? No? Maybe?

[R.G.]

Everything the input signal touches contributes to the input impedance. The only question is at what frequencies.

The ferrite (bead??) has an impedance that's essentially zero at audio frequencies. It's there to hold radio signal outside.

The 1M to ground pulldown would make the input impedance 1M for audio if it was the only thing there.

Adding a series 33nF cap adds an impedance of Xc = 1/ (2*pi*F*C)  between the 1M and the bias resistors, so at high audio (call that 10kHz) it's a series 1/(2*pi*10,000* 33E-9) = 483 ohms. At low guitar (82Hz) it's 58.8K.

Connected to that is the bias resistor of 510K to Vref and the input of your opamp. Both the opamp input and the bias resistor suck some signal current, loading the signal down. Vref itself has an impedance of some amount, but we can assume it's very low compared to the 510K, so we'll ignore it.

The opamp itself has a "high" input impedance at its non-inverting input. JFET and MOSFET opamps have impedances so high that you may as well ignore them as an open circuit. The very popular NE5532 has an impedance there of about 100K. 

If the opamp is a FET version, the load can be simplified to just the bias resistor.

So your signal is loaded by 1M to ground, then a 33nF cap in series with a 510K resistor. The frequency you're worried about is the high end, where a guitar's pickup is inductive, and may be as high as 100K to 200K. So you're interested in the frequencies where the cap is about 500 ohms.

That leaves 1M to ground in parallel with 500 ohms in series with 510K. Ignore the 500 ohms, and the result will be close to 1M||510K, or the 338K you got.

Yes. 338K, plus or minus the not-clear effects of the capacitor, which is trivial at high frequencies, not so at low frequencies, the Vref, which may also be frequency dependent, and the opamp, which may be both lower or high and frequency dependent.

[notnews32]

Everything the input signal touches contributes to the input impedance. The only question is at what frequencies.

The ferrite (bead??) has an impedance that's essentially zero at audio frequencies. It's there to hold radio signal outside.

The 1M to ground pulldown would make the input impedance 1M for audio if it was the only thing there.

Adding a series 33nF cap adds an impedance of Xc = 1/ (2*pi*F*C)  between the 1M and the bias resistors, so at high audio (call that 10kHz) it's a series 1/(2*pi*10,000* 33E-9) = 483 ohms. At low guitar (82Hz) it's 58.8K.

Connected to that is the bias resistor of 510K to Vref and the input of your opamp. Both the opamp input and the bias resistor suck some signal current, loading the signal down. Vref itself has an impedance of some amount, but we can assume it's very low compared to the 510K, so we'll ignore it.

The opamp itself has a "high" input impedance at its non-inverting input. JFET and MOSFET opamps have impedances so high that you may as well ignore them as an open circuit. The very popular NE5532 has an impedance there of about 100K. 

If the opamp is a FET version, the load can be simplified to just the bias resistor.

So your signal is loaded by 1M to ground, then a 33nF cap in series with a 510K resistor. The frequency you're worried about is the high end, where a guitar's pickup is inductive, and may be as high as 100K to 200K. So you're interested in the frequencies where the cap is about 500 ohms.

That leaves 1M to ground in parallel with 500 ohms in series with 510K. Ignore the 500 ohms, and the result will be close to 1M||510K, or the 338K you got.

Yes. 338K, plus or minus the not-clear effects of the capacitor, which is trivial at high frequencies, not so at low frequencies, the Vref, which may also be frequency dependent, and the opamp, which may be both lower or high and frequency dependent.

RG - seriously? You are great. Laying things out plainly like this, with math when there needs to be and detailed descriptions when there needs to be... I would love to read a book about electronics laid out in this way. Thank you for explaining input imp like this.

[Transmogrifox]

The very popular NE5532 has an impedance there of about 100K. 

Don't forget the effect of negative feedback in a buffer.  The input impedance usually is stated for an open loop:
http://www.ti.com.cn/cn/lit/ds/symlink/ne5532.pdf , specifically Page 5 where ri is stated.

(1)  All characteristics are measured under open-loop conditions, with zero common-mode input voltage, unless otherwise specified.

Adding negative feedback (such as in a unity gain buffer) dramatically increases effective input impedance to the point where for this application you may as well consider it non-existent next to 1 Meg.  IIRC input impedance is multiplied by the loop gain, but even if that isn't exactly correct, I can accurately say input impedance is proportional to loop gain. 

To make intuitive sense of this, imagine placing something small like a 1k resistor between + and - inputs.  You would expect input impedance is still quite high just like you would see if you were driving a 1k resistor into a voltage source in phase with the driving source, only multiplied by a gain of 0.999999999999...The effective V/I relationship would be a very large ratio.  Now imagine that resistor increased to ~100k, then the ratio gets even larger.

That input impedance number isn't very meaningful unless you're using the op amp in a high gain configuration (where %feedback is much much smaller) or as a comparator.

In something like a Tubescreamer, the op amp input impedance changes with the "drive" knob since you are changing the amount of feedback.  More interestingly this input impedance changes as you transition in and out of clipping -- maybe one of the reasons people swear different IC's sound different in this circuit.  Still it's a stretch since these often have a gain of more than 100,000.

[GGBB]

Everything the input signal touches contributes to the input impedance. The only question is at what frequencies.

The ferrite (bead??) has an impedance that's essentially zero at audio frequencies. It's there to hold radio signal outside.

The 1M to ground pulldown would make the input impedance 1M for audio if it was the only thing there.

Adding a series 33nF cap adds an impedance of Xc = 1/ (2*pi*F*C)  between the 1M and the bias resistors, so at high audio (call that 10kHz) it's a series 1/(2*pi*10,000* 33E-9) = 483 ohms. At low guitar (82Hz) it's 58.8K.

Connected to that is the bias resistor of 510K to Vref and the input of your opamp. Both the opamp input and the bias resistor suck some signal current, loading the signal down. Vref itself has an impedance of some amount, but we can assume it's very low compared to the 510K, so we'll ignore it.

The opamp itself has a "high" input impedance at its non-inverting input. JFET and MOSFET opamps have impedances so high that you may as well ignore them as an open circuit. The very popular NE5532 has an impedance there of about 100K. 

If the opamp is a FET version, the load can be simplified to just the bias resistor.

So your signal is loaded by 1M to ground, then a 33nF cap in series with a 510K resistor. The frequency you're worried about is the high end, where a guitar's pickup is inductive, and may be as high as 100K to 200K. So you're interested in the frequencies where the cap is about 500 ohms.

That leaves 1M to ground in parallel with 500 ohms in series with 510K. Ignore the 500 ohms, and the result will be close to 1M||510K, or the 338K you got.

Yes. 338K, plus or minus the not-clear effects of the capacitor, which is trivial at high frequencies, not so at low frequencies, the Vref, which may also be frequency dependent, and the opamp, which may be both lower or high and frequency dependent.

Thanks R.G. That's the first time I've been able to understand input impedance.

Can we call this "The Technology of Input Impedance" and make it a sticky or something? 

[R.G.]

Adding negative feedback (such as in a unity gain buffer) dramatically increases effective input impedance to the point where for this application you may as well consider it non-existent next to 1 Meg.  IIRC input impedance is multiplied by the loop gain, but even if that isn't exactly correct, I can accurately say input impedance is proportional to loop gain. 

To make intuitive sense of this, imagine placing something small like a 1k resistor between + and - inputs.  You would expect input impedance is still quite high just like you would see if you were driving a 1k resistor into a voltage source in phase with the driving source, only multiplied by a gain of 0.999999999999...The effective V/I relationship would be a very large ratio.  Now imagine that resistor increased to ~100k, then the ratio gets even larger.

That input impedance number isn't very meaningful unless you're using the op amp in a high gain configuration (where %feedback is much much smaller) or as a comparator.

It depends on the kind of feedback. In a series feedback setup, like an emitter follower, the feedback voltage is in series with the input voltage and does raise the input impedance. In an opamp circuit with shunt feedback going only to the inverting input, the feedback can't raise the impedance of the non-inverting pin, so it remains just the impedance of the pin itself.

There is a third arrangement that can increase the effective input impedance at the noninverting pin of a shunt feedback circuit. That's bootstrapping, where a portion of the output is fed back not only to the inverting input, but also to the non-invering input. This is tricky, as you must ensure that the portion fed back can never make the loop gain to the noninverting pin be equal to one at any frequency, or it will oscillate. But as long as the bootstrap voltage is smaller than the input voltage, this amounts to having the output of the amplifier supply a portion of the input current, and the effective impedance on the input goes up.

In a plain-vanilla gain circuit with the noninverting input connected only to a bias resistor and the input signal through a cap, you see the same impedance at the input pin, unchanged by the feedback. It makes for some strange results with the 5532, as it has a quite low impedance at the noninverting input, anomalously so for an opamp.

[djp8djp]

Let me take a moment to echo notnews32, and say something that's perhaps overdue (at least from me).

I'm still pretty much new to all this, but have learned a lot. Some of that has come from my old high school electronics course, and some from miscellaneous reading. But mostly from the internet. And most of that from a small number of sources, which includes this (great!) site. It's been a wealth of knowledge for me, and while I haven't posted much, I've eagerly consumed as much as I can from many threads.

And from the topics here, and elsewhere, I have to say that the best, the most correct, the most helpful, and by far and away the greatest amount, has come from R.G. Keen (here and Geofx), Jack Orman (AMZ and other places), and a few others I know only by username, and yes, PRR & Mr. Hammer, I mean you. I've gotten as far as I have by many archived posts from these particular individuals, and they've also brought me back on course when confused or led astray. And I'm sure that many others would agree that these people deserve our heartfelt thanks. Your willingness to teach and share provides all of us with an invaluable resource.

So, thanks - it's much appreciated!!

ps - of course, all other contributors here have also contributed to my learning, and all of your selfless help and generosity is also greatly appreciated. This is really a great community. Cheers to one and all!

[djp8djp]

R.G., that was a great reply, thanks. I wasn't sure about whether I should calculate in the impedance from the biasing, as it wasn't going to ground, but to Vref. But of course, Vref goes (eventually) to ground too. You mentioned that Vref itself has an impedance of some amount, and that we can assume it's very low compared to the 510k. And yes, mine is, 20k.

For the use of others, I'm going to deviate from input impedance a little, and provide what I found for Vref. I'd wondered how to settle on the "correct" values for my voltage divider, and had gone hunting for guidance. A few sources I found were
- intro level general info
https://learn.sparkfun.com/tutorials/voltage-dividers
- some great general info (from a great source of much info)
http://sound.westhost.com/vda.htm
- a discussion
http://electronics.stackexchange.com/questions/28897/how-to-choose-value-of-resistor-in-voltage-divider
- two generic voltage divider calculators
http://www.raltron.com/cust/tools/voltage_divider.asp
http://www.ohmslawcalculator.com/voltage-divider-calculator
-  a calculator more suited/specific to voltage dividers used for applications like Vref
http://howardtechnical.com/voltage-divider-calculator/
- detailed info applicable to comparators
http://www.ti.com/lit/an/slva450a/slva450a.pdf

There's probably more great info out there, but those are a few I found. There was one other that showed the calculations for resistor size, that uses the expected current draw and the input voltage, giving a maximum value for the divider resistor(s). But darned if I can find that source now. But I'd put together a quick spreadsheet for it, and here's the process:
1) Vcc (voltage available from supply) = 17V
2) desired Vref (for me, half, so..) 17V / 2 = 8.5V
3) Rref (value of reference resistors to be used) = 500k
4) Iref (current expected from each application of Vref) = Vref / Rref (x  1,000,000 for uA) = 17uA
5) # Vrefs expected = 2
6) total Iref = Iref x #Vrefs = 17uA * 2 = 34uA
7) max Iref variance expected/allowed = 10%  (personal choice)
8 ) total Iref capacity required = total Iref / variance = 34uA/0.1 = 340uA
9) total required resistance (from both divider resistors) = Vtot / tot Iref (x 1,000,000 to remove u-scale) = 50,000 = 50k
10) value of each divider resistor = total / 2 = 25k

So, the resistors can be no larger than this, to "guarantee" the provision of an adequate Vref for the given application. And BTW, the results are independent of Vcc, and actually depend only on Rref, #Vref, and margin. But it's nice to keep it all in there to illustrate how it all works.

In my case, I'd had planned 500k reference resistors, a 17V supply, and 2 expected Vrefs, so those are the values I used (as above) and got 25k. But I want my buffer to have an input impedance of 1Meg, and as the Vref resistor is dominating the input impedance, I'll now be using 1Meg. And in this application I'll now have 3 Vrefs, so plugging all that in, I get 33k "recommended" max values. I'd used 20k resistors before, a safe-ish amount below my previous calc of 25k. And I think that's still pretty good for my new requirement, so I'm leaving 'em be.

Anyhoo, I hope somebody finds all that useful. And any necessary corrections are welcome !!

[djp8djp]

As for my buffer's input 33nF series capacitor, it's for DC blocking of course, and I chose that value to keep the frequency range plenty wide enough for full audio range. I calculate the -3dB corner at about 9.5Hz. Maybe (? definitely!) overdoing it a bit, but at least it's not stopping anything. But now that I'll be doubling Rref to 1Meg (or more, and upping the preceding resistor too), if I want to keep the same response, then I'll drop the cap by half to 17nF. Leaving it at 33nF would put the corner at 4.8Hz, which is really low but there's probably nothing wrong with that either. Is there?

But I'd like to ask about the impedance on the output too, so long as we're on a roll here?

At the opamp output I have a series 2k resistor. This helps to isolate the opamp output pin from any capacitive load. It also combines with a subsequent capacitor-to-ground, to form a low-pass filter. I feel that anything from 1nF to 2.7nF will be fine for my "full-range" requirement, as those values (with the 2k resistor) will have -3dB corners at 36kHz and 87kHz, respectively. Yup, way high. Dog-hearing high. No loss there. But those are down -1dB at 18kHz and 46kHz respectively, so even the higher 1nF value will keep me flat right up through any meaningful harmonics. Way safe.

The resistor-capacitor values of 2k/2nF could be changed to 1k/4.7nF, or 500/7.5nF, or 100/20nF, which'd all provide about the same low-pass performance. I wouldn't want to drop lower than 100R, to keep the isolation effect. The only salient difference between those combos is in their contribution to output impedance.

Now, after that I have a series 4.7uF electrolytic capacitor, for DC blocking. Most circuits have a similar cap at their input, so I can probably "expect" that from anything that'll follow this buffer. If that following circuit has, as mine does, a first component of a large-value "pulldown" resistor to ground, then that'll combine with my output cap to form a high-pass filter. Assuming a 1Meg value, the -3dB corner would be vanishingly low (<1Hz). A 500k resistor would put it at ~0.1Hz, still way low. So all good. And even if going into some minimally-acceptable input resistance such as 10k, it's still at 3.4Hz. Still good. I could even use a 1uF cap, and with a following 10k input impedance, it'd be -3dB at 16Hz. So all is well at 1uF and higher. I'll keep it higher, as insurance against a following unit's having a small-ish one.

OK, so this is where my question really begins. I've run across instances and suggestions regarding a final component of another resistor to ground. It could be handy to throttle back a larger-than-unity output. Some suggest it to help forestall switch-popping. Others say it prevents the buffer's output impedance from being "too high". I think that's in reference to a FRFR opamp output often being interpreted/heard as "cold", and that this is due not just to the frequency content, which is all accepted at the high-impedance input, but also because of the extremely low output impedance, which preserves that content. And yeah, I know that this low output impedance is often the exactly POINT to using a buffer, because it can drive without loss through long or sketchy cables, and that it's relatively immune to loading by whatever follows.

So what about that sometime-suggested final pulldown resistor? Use 2k-10k there to bring up the total (added to the other, series resistor) output impedance? If the following unit were to have a low-ish input impedance, and I use a low-ish output impedance, they'd interact in parallel to produce an even lower input impedance for that unit. Would I start losing low frequencies? Is there an output impedance value which would be of any benefit in particular situations, or is it just always best to keep it low? (and let the "cold" sound be handled elsewhere).


That topic can sort of be applied to the other end too, the input. Guitar output is typically pretty high impedance. And some (old) amps and other audio equipment may have some relatively low input impedance. Plug 'em together, and you have an interactive conglomeration, that is never wideband lossless. So if we want a sequence of guitar-buffer-amp to sound exactly the same as guitar-amp, then do we need to "recreate" the impedance/coupling of that bare-bones rig? I could swear that I've seen recent commercial pedals with an input impedance adjustment, so they can do just that.

With the buffer's usual high-input-impedance, we eliminate any of the effect like the loading between guitar (pickups and pots) and amp. And therefore it also eliminates any differential interaction with the guitar when its pots are adjusted. Putting the direct-from-guitar setup aside for a sec, there's a different situation, where we'd want the buffer between other pedals, in which case "transparent" high-in and low-out impedances are great. Can we have our cake and eat it too? Is there a middle ground, or are these two situations so far apart that there's no practical compromise?

[PRR]

> series feedback setup, like an emitter follower, the feedback voltage is in series with the input voltage and does raise the input impedance. In an opamp circuit

Exact same effect. The feedback voltage is applied to a buffer (the other-side of the input pair) to the emitter of the + input transistor.

The "input resistance" (poorly defined; they should say differential input resistance *) is just like a resistor from base to base. By NFB theory, the differential input voltage tends to zero, leading to infinite closed-loop impedance even with 100K input to input.

This is all brain-game (also long experience with even weepier chips like the '709). I am too sore this month to poke a breadboard.

To check my math, I asked the idiot. SPICE with 100K across the inputs of a near-ideal opamp, unity gain follower, plus 100K from source. If the effective input were truly 100K, the 100K:100K would cause a 2:1 loss. Initially there was "no" loss. I dropped ideal opamp gain to just 100. Now 1V in gives 0.990,2V at + input pin, apparent Rin of 10,204K, which is ~~100X what the parasitic pin-pin resistance R3 is.



(*) There is also a common-mode input resistance. By the layout of the '5532, I would expect many Megs; the long-tail collector impedance times hFE of the input transistor.

[PRR]

> that sometime-suggested final pulldown resistor?

You are over-thinking.

BS aside, it is there because your 4.7uFd (electro) cap leaks DC. If it leaks into an open, then you plug/switch it to a load, that stray charge will POP as it bleeds off.

You used a 4.7u electro because it is cheaper/smaller than the 0.5u Film which otherwise would be fine for all guitar-cord interfaces.

You want to pick the C and this final bleeder R for not-many seconds. Few hundred K perhaps. 10K (0.3Hz) to be very sure of a quick bleed, or some bleed even when that cap gets old and leakier.

As you can figure, 2K series and 100K bleed wastes-off 1dB-2dB signal (not ideal) and reduces the 2K node to 1.6K (who cares?).

> anything from 1nF to

1nFd = 1,000pFd = 30 feet (10m) of cable. Unless you only keep short cords, you may stick in an extra 1nFd without knowing. Also some (even many) boxes have ~~1,000pFd caps at input to kill radio. I'd pick a deliberate cap larger than that.

2K is very conservative build-out. The opamp can drive that perfectly. We are mainly trying to prevent supersonic hassles which then will ne negligible amplitude. With '072 and our low voltages, 1K or 500r is a fine build-out. If you must drive loooong cables; but that is another arena.

[R.G.]

To check my math, I asked the idiot. SPICE with 100K across the inputs of a near-ideal opamp, unity gain follower, plus 100K from source. If the effective input were truly 100K, the 100K:100K would cause a 2:1 loss. Initially there was "no" loss.

The 100K between inputs i

[Chugs]

As a pulldown resistor affects the input impedance does it also affect the frequency cutoff?

If we have 1M bias resistor and a 0.022uf cap the input frequency cutoff is 7.2hz. Adding a 1M pulldown resistor changes the overall input impedance to 500K, so does the frequency cutoff change to 14.5hz?

[merlinb]

If we have 1M bias resistor and a 0.022uf cap the input frequency cutoff is 7.2hz. Adding a 1M pulldown resistor changes the overall input impedance to 500K, so does the frequency cutoff change to 14.5hz?

No, because the pull-down comes 'before' the input cap. You can imagine the pull-down as being 'absorbed' or 'lumped' into the source impedance of the guitar, as far as the filter is concerned.

However, if you have an output cap in one effect, plugged into another effect that has a pulldown, then it will affect the cut-off of the output filter.

[djp8djp]

Hi PRR! Glad to see you here. See reply#6 this thread.

I'm overthinking it? Yeah, probably. But when approaching a situation that isn't fully understood, a person will likely do one of two things ... consider unnecessary things, or fail to see the necessary things. That being said, I admit that I went a bit off the deep end and rambled away from the topic.

But as to the core issue of configuration and values, what and why, your reply really helped a lot, thanks. The final pulldown stops pops. Got it, makes sense! I could use a few 10k's worth this time, as I actually could lose a little dB just to get to exact unity gain overall. That'd put my corner around 0.5 to 3.5 Hz, so all good.

But the next unit likely has a pulldown at its input, let's call it 1Meg. If my output were to have a pulldown of say, 100k, the parallel combination is about 90k. But with 10k instead, that'd go to 9.9k, and that's probably too low.

merlinb, this connects to your answer to Chugs, which was great, thanks. But here I'm thinking not of my unit as the second of two, but as the first - and what's getting "lumped together" is my final output pulldown, and the input pulldown of the next.

So am I grasping the situation correctly?... should 100k maybe be my minimum "to be safe" if I'll be connecting to unknown units?

And PRR, thanks also for the guidance on cap values, that makes a lot of sense. But could you please clarify something for me? You called 2k at my opamp output "very conservative", but then 1k or 500 as "fine". Is more conservative = more resistance, or less? And what would be the max/min? (And is " '072" a reference to TL072?)

[djp8djp]

anybody ?

[Transmogrifox]

But the next unit likely has a pulldown at its input, let's call it 1Meg. If my output were to have a pulldown of say, 100k, the parallel combination is about 90k. But with 10k instead, that'd go to 9.9k, and that's probably too low.

No, that wouldn't be too low for something that can drive 10k.  If you use a 10k pull-down on a stompbox with a low impedance output (like an op amp) then the op amp will drive that just fine.  Adding another 100k in parallel doesn't hurt anything.

We make high impedance inputs for the worst case input source:  the instrument.

The guitar pickup really wants to see more than 100k and most pedals aim for somewhere between 500k to 1 Meg.  If you can drive the input reasonably with a guitar including the 250k volume pot then you can drive it with any guitar FX unit.

[antonis]

What Transmogrifox is trying to tell you is that your gess is wrong because you have to consinder your output's impedance IN SERIES with your output.
(like circuit equivalent of any power source's internal resistance..)

This impedance forms a voltage divider with the (possible) pull down resistor of the next unit in paralel with anything comes next.

To be accurate, you have to calculate capacitive reactances and phase shiftings for every branch of series/parallel combination but, IMHO, it's just an overkill..

Also, as many guys have told, capacitors are much more involved in "tone loss" rather than in significally altering impedance values...
(at least for "regular" frequencies..)

So, you can safely consider capacitors as short circuits in impedance calculation but not in any kind of filtering..

[ashcat_lt]

I know this goes back a little, but...

R.G., that was a great reply, thanks. I wasn't sure about whether I should calculate in the impedance from the biasing, as it wasn't going to ground, but to Vref. But of course, Vref goes (eventually) to ground too. You mentioned that Vref itself has an impedance of some amount, and that we can assume it's very low compared to the 510k. And yes, mine is, 20k.

If you did it "correctly", there's a big cap to ground from the VRef point, so for AC the impedance at that point is zero.  Those resistors contribute (close enough to) nothing for these calculations.

[djp8djp]

Thanks guys, one and all. Big help.