Impedance at a Glance...

[sonolink]

Hi all,

I didn't really know where to post this question so I apologize if it's the wrong place.

I'm trying to understand and figure out Input/Output Impedance and for the life of me although I understand the CONCEPT (resistance in a circuit where voltage is applied, that changes depending on frequency) I am unable to reliable figure it out for a circuit. DIY Electronics is just a hobby for me but learning is another hobby I love and the resulting feeling of being blind and unable to see is very frustrating...

I see people answering posts like "since the output impedance of your schem is blablabla..." in no time, but then I try to calculate that same impedance and I end up needing hours if not days and the result is far from reliable. The only reliable thing that happens is a headache and brain confusion, haha.
I've never been too mathematical but there MUST be a simpler way.

I'm a sound engineer and I don't make maths when I have to compress a signal, or apply gain or EQ, but I know where the ballpark figures are and I'm generally roughly pretty right. This, of course is the product of experience more than theory itself.

Is there a simple way to calculating input/output impedance or being able to answer roughly those questions just by taking a glance at a schematic is product of experience and having watched and sweated through quite a few schems?

Cheers
Sono 

 

[ruffrecords]

It is relatively easy for op amps, less so for discrete  ones.

For op amps it is generally safe to assume the output impedance of the op amp itself is zero. The final output impedance is then largely determined by any series resistor in the output circuit - it is often between 50 and 200 ohms.

Similarly for op amp inputs it is usually safe to assume the input impedance of the op amp is effectively infinite. Once again, the actual input impedance is then determined by the combination of resistors at the input.

For 90%+ cases this is all you need to know.

Cheers

Ian

 

[sonolink]

Wow, thanks a lot for that tip Ian! That is great info indeed! I was going crazy summing up and dividing by 1 resistances, reactances, etc!!

So if I understand correctly, for instance in your "Single Supply Fet Opamp Make Up Gain Stage" circuit,

Output Impedance would be 47R (the final resistor)?
Input Impedance would be infinite since there is no resistor between the circuit input and the opamp + input or would it be 480K? (10k+470k on the 24v power rail?)

Thanks again for your help
Cheers
Sono

 

[Bo Deadly]

The concept of impedance is everything in electronics. Electronics ultimately boils down to 4 components: resistors, capacitors, magnetics (inductors and transformers) and semiconductors (diodes, transistors, triacs, etc). If you want to be able to glance at a schematic and "see" how it works, you need to understand the characteristics of those 4 components which is to say what their impedance is at each net (wire) connecting them together. Impedance does not depend on frequency. Impedance is to AC as resistance is to DC. But the impedance of all components ARE dependent on frequency. More so capacitors and magnetics but technically resistors and semiconductors have frequency dependencies (a transistor will only amplify up to a certain frequency for example). But at this point, we'll focus on impedances of components around the audio frequency range.

So let's use your op amp input / output impedance as an example. The output impedance of an op amp is effectively 0 at audio frequencies. So with a 47R resistor in series, yes, the output impedance of the circuit is 47 ohms. The input impedance of an op amp is effectively infinite (if biased within it's operating range) at audio frequencies so again, yes, the input impedance is 470K.

But let's pick this apart a little more. In the second circuit, there's a 330n capacitor in series. Capacitors have an impedance. But at high frequencies the impedance is low. At low frequencies the impedance is high. The frequency at which the impedances are the same depends on the size of the capacitor. This is because small capacitors will basically charge or discharge completely if the alternating current (AC) is moving slowly. But as the AC increases in frequency, the capacitor never completely charges / discharges and so there is a voltage drop across it and thus you have frequency dependent attenuation. This is why they make high pass and low pass filters. So in your example, the 330n capacitor and 470K resistor makes a high-pass filter. At really low frequencies, the voltage across it is moving so slowly that the capacitor is discharged completely through the 470K resistor at any particular moment in time. At higher frequencies, the resistor cannot source / sink current fast enough to discharge the capacitor and thus the audio signal get's through. You can calculate the frequency using math but I never do that. Instead, I cheat and use an online calculator like this one:

  http://sim.okawa-denshi.jp/en/CRhikeisan.htm

So we punch in 470K and 330n and we get a cutoff frequency of 1Hz. Meaning all audio is effectively passed through. Basically it just blocks DC so that the circuit can be biased to half of the supply because it's single supply circuit.

 

[sonolink]

I really appreciate you taking the time to write such a comprehensive reply Squarewave

I'm also glad I got it right!!  8)

So, after reading you reply, at the output of the circuit, is it right to assume that C4/R9 make another HP filter? If so the cut-off frequency would be 34mHz (mili) and then all audio spectrum would be allowed through. This would make sense since that cap is meant to block DC and protect against i.e. phantom and R9 is there to discharge C4 and allow a DC path to GND. Or so I think...

If that's right, why did Ian need to put R5 (the 47R resistor) after that filter? Is that to make sure that the circuit's output impedance is higher than 0 and in the 50-200 Ohms range because that will match any other circuit's input connected to that output?

Thanks for your help
Cheers
Sono

 

[JohnRoberts]

With properly designed gear on sending and receiving ends you do not need to match impedances, or even think about it.

Some old legacy gear is a throw back to earlier times when interfaces matched impedance for optimal power transfer.  This old power transfer strategy has been replaced by voltage transfer. For optimal voltage transfer we aim for "bridging " interfaces, namely source impedance 1/10 the load impedance.

Good luck, do you have an actual problem interfacing gear?

JR

 

[sonolink]

Thanks for chiming in John

My biggest problem in here is that I am a complete noob and Electricity and Electronics are not like learning how to skate hehe

But seriously, my main doubt/problem is that I built a tube preamp in a stompbox for bass that sounds awesome through a power amp. I then added a DI out to be able to connect directly to a mixer and/or a soundcard for recording. This Di Out was based on an Impedance Balanced circuit by Elliott Sound, and corrected kindly by Volker, Ian Thompson and ClintRubber.
Alles gut so far...

Since then I tried the pedal with a Strat into a clean guitar power amp. The sound is absolutely awesome. So I thought it would be great to be able to carry this little box instead of an amp and plug it directly into a mixer/soundcard, but of course a guitar is not a bass and a guitar without a guitar speaker neither

So I found a nice a simple cab simulator that is said to behave great.

The main doubt I have is: can I build a PCB for this so that if I'm using it with a bass I can leave the cabsim out of the circuit, either by means of a jumper or switch, or by means of leaving a part of it unpopulated, or would the fact of inserting the cabsim between the tube preamp and the balancing output circuit need further changes like preventing impedance problems, DC blocking, etc?

Since I never got a direct answer to that specific question ( i did get a lot of help with the rest), I was trying to answer it myself....but of course it's a bit of a maze hehe.

In any case, if you would like to answer it these are the circuits in question:

The tube Preamp + Di impedance balanced output:


The cabsim schem:


The question itself: would the cabsim inserted like this cause any problem? The switching would be done while powered off, of course. Maybe a jumper would be a better idea...



Thanks a lot for your time and help
Cheers
Sono

 

[Bo Deadly]

So, after reading you reply, at the output of the circuit, is it right to assume that C4/R9 make another HP filter?

Yup. But ...

If so the cut-off frequency would be 34mHz (mili) and then all audio spectrum would be allowed through. This would make sense since that cap is meant to block DC and protect against i.e. phantom and R9 is there to discharge C4 and allow a DC path to GND.

Not exactly. You have to consider what the output is connected to. Most line inputs have an impedance of 10K (although some older gear like passive lab filters and stuff like early revisions of the Urei 1176 have 600 ohm inputs). So you have to imagine a 10K from the downstream side of the 47R to ground. That means the impedance that the output of the op amp "sees", which might also be referred to as the "load impedance", is actually the 100K in parallel with 47R in series with the 10K (the shorthand expression for this is 100K || (47+10K) = 9.13K). But because the 10K is much smaller than the 100K, the "load" is effectively 10K which makes the filter frequency 340mHz. But whatev.

So now you might ask, "why is the 100K needed at all"? The reason is because electrolytic capacitors leak a tiny current. It's not much but it's enough that if there is even a moment where nothing is plugged in, such as when disconnecting the cable or even when just flipping a switch or relay, a charge can (will) build up on the downstream side of the capacitor such that when the switch closes, it can cause a loud pop. So that 100K drains the cap and keeps it at ground potential. That resistor is commonly referred to as a "drain resistor".

If that's right, why did Ian need to put R5 (the 47R resistor) after that filter? Is that to make sure that the circuit's output impedance is higher than 0 and in the 50-200 Ohms range because that will match any other circuit's input connected to that output?

The 47R is called a "build out" resistor and it's for current limiting and stability. When driving capacitive loads like really long cables, the output can oscillate. So pretty much all line outputs on modern gear have that build out resistor.

 

[Bo Deadly]

A graphic equalizer makes a much better cabinet simulator. And they're super cheap. Just get a DBX 131 or 215 for $40 on Ebay. It's noise will be 90dB and will blow away that cab sim. Just google the frequency response of whatever cabinet / speaker. Then put white noise through your graphic EQ, look at the response in the FFT window of your DAW and match it as closely as you can. That's one of the great uses for a graphic EQ. It can be very convincing.

 

[user 37518]

Input Impedance would be infinite since there is no resistor between the circuit input and the opamp + input or would it be 480K? (10k+470k on the 24v power rail?)

Thanks again for your help
Cheers
Sono

The input impedance is actually only 470K because the 47uF capacitor essentially acts as a short to ground, so one end of the 470K resistor is at AC ground.

 

[ruffrecords]

To summarise the other replies, when trying to work out input or output impedance 'by inspection' you can assume that in a well designed circuit all the capacitors are effectively zero ohms at audio frequencies ( unless it is an EQ).

Of course there are lots of other considerations a designer has to take into account. For instance, all capacitors need to be able to charge up at switch on and discharge at power off. That's what the 'bleed' resistors are for. You don't really want a capacitor charging up when you plug your circuit into something else because at least you get a click at worst you blow up a speaker. Equally when you turn off a tube amp you don't want capacitors sitting at a few hundred volts for long periods of time.

Cheers

Ian

 

[sonolink]

So now you might ask, "why is the 100K needed at all"?

Indeed I did, Squarewave, as I wrote:

R9 is there to discharge C4 and allow a DC path to GND. Or so I think...

I am acquainted with drain resistors or bleeders. I use those when I build tube amps to discharge the PSU caps 8)
I had once the very uncomfortable (to call it something) experience of manipulating a tube amp WITHOUT bleeders...I can still feel the snap in my hand...and I am a VERY careful guy (total pussy), but at the time I took for granted something you should never take for granted: your life haha!!
Anyway, bleeders are a must, and knowing how to locate them to check that the big lytics are empty is definitely the first thing one should do when opening an amp

The 47R is called a "build out" resistor and it's for current limiting and stability. When driving capacitive loads like really long cables, the output can oscillate. So pretty much all line outputs on modern gear have that build out resistor.

That's very interesting. I will look into that

That's one of the great uses for a graphic EQ. It can be very convincing.

That's a great idea, but the idea behind this project is to fit everything inside a Hammond 1590BB enclosure, so a Graphic is out of the question. This cabsim is a well known circuit and is supposed to sound good (I say "is supposed" because I haven't heard it yet, but I do trust other people's word for it).

The input impedance is actually only 470K because the 47uF capacitor essentially acts as a short to ground, so one end of the 470K resistor is at AC ground.

Sorry but I don't understand that user 37518. The 47uF cap is not part of the voltage divider? Could you please explain?

To summarise the other replies, when trying to work out input or output impedance 'by inspection' you can assume that in a well designed circuit all the capacitors are effectively zero ohms at audio frequencies ( unless it is an EQ).

Thanks for chiming in Ian!
So, when designing, modding or adapting a circuit one should make sure that those rules prevail, right?

Please, pretty pretty please, could someone PLEASE answer my other question about the cabsim? I would be VERY grateful

The question itself: would the cabsim inserted like this cause any problem? The switching would be done while powered off, of course. Maybe a jumper would be a better idea...

Thanks again to all for your help 
Cheers
Sono

 

[ruffrecords]

I do not see any problem inserting the cab sim as shown in your schematic. Not sure you need R40 though.

Cheers

Ian

 

[sonolink]

Thank you VERY much Ian for taking the time to check the schem

Looking at the it now, and after this great thread  8), my guess is that R40 is probably a "build out resistor" of the original circuit? If so, adding the balanced impedance output would make it redundant I guess and since the balanced impedance output will always be present I suppose I can just leave it out, right?

Thanks again for your time, help and to everybody else for the learning experience!

Cheers
Sono

 

[user 37518]

Sorry but I don't understand that user 37518. The 47uF cap is not part of the voltage divider? Could you please explain?

Remember that capacitors exhibit capacitive reactance, if the cap is big enough the reactance will be low, thus, if the capacitor is connected to ground on one end and to the 470K resistor at the other end if the reactance is very low, it is as if the capacitor acts like a short to ground, with a 47uF capacitor the capacitive reactance at 20Hz is 169 ohms, so it is basically 470K in series with 169 Ohms, that is very close to 470K, at higher frequencies, say at 1KHz the reactance is just 3.4 Ohms, which for all practical purposes can be considered a short to ground, so the input impedance is just the 470K resistor.

 

[sonolink]

which for all practical purposes can be considered a short to ground, so the input impedance is just the 470K resistor.

Mmmm ok. I understand the reactance bit, but I don't understand what you said earlier about AC ground. Are there different GNDs for DC and AC?
Also, why would you want a short to GND there? Sorry if I'm a bit slow here

Cheers
Sono

 

[ruffrecords]

Mmmm ok. I understand the reactance bit, but I don't understand what you said earlier about AC ground. Are there different GNDs for DC and AC?
Also, why would you want a short to GND there? Sorry if I'm a bit slow here

Cheers
Sono

OK, for the purposes of this discussion there is basically only one ground which we will call analog 0V. In the single supply op amp example, the + input needs to be dc biased to about half the supply voltage, which is what the two resistors do. However, for audio you want it to be referenced to analog 0V which is one reason the bottom resistor has a chunky electrolytic across it. The other reason is that any noise on the power supply is fed through the pot divider straight into the + input of the op amp. The same capacitor shunts any such noise to analog 0V and stops it reaching the op amp + input.

Cheers

Ian

 

[user 37518]

Mmmm ok. I understand the reactance bit, but I don't understand what you said earlier about AC ground. Are there different GNDs for DC and AC?
Also, why would you want a short to GND there? Sorry if I'm a bit slow here

Cheers
Sono

A capacitor acts as an open circuit at DC and as a sort of variable resistor in AC, with the difference that it also affects the phase of the current-voltage relationship, the value of that variable resistor is called reactance. At AC the capacitor acts as a low value resistor from ground to the end of the 470K resistor, you want one side of that 470K resistor going to ground for AC frequencies since it determines the input impedance.

 

[sonolink]

However, for audio you want it to be referenced to analog 0V which is one reason the bottom resistor has a chunky electrolytic across it.

Of course!!  :
Thanks for opening my blind eyes Ian

At AC the capacitor acts as a low value resistor from ground to the end of the 470K resistor, you want one side of that 470K resistor going to ground for AC frequencies since it determines the input impedance.

Oh I see what you mean now!

Thanks again to all for your help

 

[Rob Storms]

With properly designed gear on sending and receiving ends you do not need to match impedances, or even think about it.

Some old legacy gear is a throw back to earlier times when interfaces matched impedance for optimal power transfer. This old power transfer strategy has been replaced by voltage transfer. For optimal voltage transfer we aim for "bridging " interfaces, namely source impedance 1/10 the load impedance.

Good luck, do you have an actual problem interfacing gear?

JR

Hmmm.... In "principal" wouldn't you want to transfer as much "power" (as opposed to voltage or current) as possible from the source to the next stage to preserve as much of the s/n ratio as possible at the final output? I realize that there are lots of other reasons not to do this. I am thinking of a giant tone arm/stylus playing a huge record with an output z of 8 ohms. Wouldn't you get a better s/n ratio by hooking it directly to an 8 ohm speaker than any kind of amp with a higher input impedance? ( Ignoring stylus loading and lots of other stuff).