Questions about functions of certain components

[rhaggart1]

Question time!

I'm moving into some slightly more interesting designs now, though I am still a beginner.
I'm trying to understand some existing schematics so I know what's important and what isn't.

Right now I'm looking at the MXR Distortion Plus schematic:
Image [http://www.generalguitargadgets.com/pdf/ggg_dist_plus_sc.pdf]


At the moment I'm focussing on the input & power.
The way I understand it is as follows:
R1 is a pull down resistor which prevents the capacitor leakage from causing a switch 'pop' when you un-bypass the circuit.
C2 is a coupling capacitor to block the DC from power from driving the input so a proper DC bias of the input can be achieved
R2 & R3 act as a voltage divider to produce a 4V5 level which is used to bias the input

Now here come the questions:
What is the purpose of C3?
And R5?
    I read something about it maybe protecting the op-amp from ESD, though I didn't fully understand this...
What about C1 and R4?
    I assume R4 is there to prevent the non-inverting input to the op-amp from being just 4V5? But doesn't the high resistance mean
    we won't have the full 4V5 to bias with? Is it something to do with the high op-amp input impedance meaning that we have a
    voltage divider here and the op-amp input is much higher than 1M? (Stumped on C1)

[GibsonGM]

Hi Rhaggart, welcome!

C3 is kind of a 'snubber cap'...it shunts very high freq's that we don't want to ground before the input. Think RF, junk higher than 20Hz that we don't need - just get rid of them.  The value is so low (.001u) that it can only affect what we can't hear.

C2 also is a filter...lower values here will cut bass.

R5 limits the current on the opamp input (yes, protects it) and works with the opamp input impedance to form a lowpass filter (similar to C3 but more for tone-shaping...prevents shrill harmonics from being developed when the opamp clips). 

C1 filters the power supply, makes it a 'clean source'. If it were to fluctuate a little, it can supply the 'missing voltage' briefly.  Good practice - a battery may not really 'require' it, but it's a general rule of thumb to include it.  Wall-wart supplies, YES, you need it.  Think of it as smoothing the power supply to limit hum.

R4 provides a load, if you will, for the opamp input. Its value is so high, and the opamp input is so VERY high R, that it does very little to drop the bias voltage (ideal opamps draw no current, so there is no voltage drop...they sort of just 'feel the voltage' there).  So your assumption on opamp impedance is correct! 

Hope that helps!

[rhaggart1]

Thanks very much for all of this! Definitely helping me find more questions to ask!

C3 is kind of a 'snubber cap'...it shunts very high freq's that we don't want to ground before the input. Think RF, junk higher than 20Hz that we don't need - just get rid of them.  The value is so low (.001u) that it can only affect what we can't hear.

Without a resistor in series with the input, does C3 actually do this job? I would have thought that a resistor would be needed to form a passive LPF? Surely as it stands, the voltage across C3 will always just be equal to the input voltage as there's nowhere else to drop any of it between the input & ground? OR does the output impedance of the guitar do this job? (my understanding is that the output impedance of the guitar is in series with the input of the pedal... right?)

R5 limits the current on the opamp input (yes, protects it) and works with the opamp input impedance to form a lowpass filter (similar to C3 but more for tone-shaping...prevents shrill harmonics from being developed when the opamp clips). 

I think I get this... Is there a general rule for a value of this resistor? Maybe it depends on the particular op-amp specs?

C1 filters the power supply, makes it a 'clean source'. If it were to fluctuate a little, it can supply the 'missing voltage' briefly.  Good practice - a battery may not really 'require' it, but it's a general rule of thumb to include it.  Wall-wart supplies, YES, you need it.  Think of it as smoothing the power supply to limit hum.

I thought this might be it. Thanks for clarifying :}

R4 provides a load, if you will, for the opamp input. Its value is so high, and the opamp input is so VERY high R, that it does very little to drop the bias voltage (ideal opamps draw no current, so there is no voltage drop...they sort of just 'feel the voltage' there).  So your assumption on opamp impedance is correct! 

This one still has me stumped. Why does the opamp input need a load? And would a lower resistance value not drop even less of the 4V5? Would it not be better to use a lower resistance here to preserve as much of the 4V5 as possible?

Thanks very much for your reply though! It's definitely helping me understand all of this and I really appreciate it :]

[GibsonGM]

C3 = yup.  Think of it as tacked on to a circuit 'in motion'...the signal is passing by it, and the cap sucks out those high frequencies. Remember, there is a source impedance to whatever is feeding it (guitar etc), even if no R is there...the response of C3 will change with different inputs, but it's so small it's just pulling those ultra highs, we don't notice this.   That's how I understand it

The 'feed' resistor for the opamp from the bias network DOES have some effect on how it runs.  The tech talk on this is on page 7 https://www.renesas.com/sg/zh/www/doc/application-note/r13an0003eu0100-biasing-op-amps.pdf   
I'm no pro (at all!) but when I put something together, I usually just select a common value....if 10k in the bias network (extremely common), maybe 33k to 68k, seems to work without a lot of math    I'm not big on eqn's unless I HAVE to do them, and of course in some cases you do.


When I said it's a load (R4), I was sort of implying stability - the OA will work without it of course....your first impression was correct, it's more about protection.  That R also works with the input capacitance of the OA, and is part of a filter.    You could lower it, sure, but I think that may affect the tonal response of the OA (not 100% on how much).    We could take this to a much higher realm (look at the PDF, lol); there is a LOT going on mathematically here.  It's good to try to understand it tho! 

[PRR]

READ:
http://www.geofex.com/circuits/what_are_all_those_parts_for.htm

[rhaggart1]

READ:
http://www.geofex.com/circuits/what_are_all_those_parts_for.htm

Thanks for this! A useful reference for the future. However, the R_bias still isn't making total sense to me...
"Rbias: provides the (hopefully tiny) DC bias current to the noninverting input of the opamp, and also sets the effective input impedance"
Why do we need this DC bias current? And how does a resistor provide current?

https://www.renesas.com/sg/zh/www/doc/application-note/r13an0003eu0100-biasing-op-amps.pdf   

Again, this .pdf is useful for all sorts, so thanks very much for linking :]
However,
'The DC biasing of AC-coupled single-supply amplifiers also requires an input resistor connecting the noninverting input to the reference potential, Vmid'
doesn't quite answer my questions :[ why does it require one?

Sorry to all involved! This likely isn't even such a significant talking point but I really want to *properly* understand exactly what it's doing and why it's important! I'm surely slowed down by my sub-par electronics knowledge too...

Thanks for all the help so far. I'm sure we'll get there eventually ;]

[GibsonGM]

Well, a voltage is just a voltage until it meets something with resistance, which causes a current to flow in the R...perhaps, because the opamp presents SUCH a high impedance to this bias voltage...there is some odd thing happening where the tiny current it WILL take needs to be 'handed to it', ha ha.  In layman's terms, it is such a high input impedance, it can't develop its own current and needs that R there to make it flow? (?)    This is where my reading stops, bud!  I've never worried too much about that.   Youtube probably has some great vids that will demonstrate the principle! 

Oh, and remember, it sets the INPUT IMPEDANCE, as R.G. said in the article.  In many circuits, they use 10k for the bias network ...to allow a much larger (relatively) current in there compared to what the opamp needs...this I believe is also for stability. The 1M you see in Dist+ is a very 'starved' network.   If the bias R wasn't there, the input impedance would drop dramatically, and result in a nasty loss of high frequencies at the input....

[amptramp]

If R4 wasn't there, the DC voltage at the op amp noninverting input could float around all over the place.  Usually there would be enough of a bias current requirement that it would be pulled to the high or low power supply voltage but with some CMOS input stages, it could wander around anywhere - with almost no bias current, it would pull the input voltage wherever it drifted to.  In the operation of the circuitry, the op amp attempts to drive the inverting input to within a few millivolts of the noninverting input.

[rhaggart1]

Well, a voltage is just a voltage until it meets something with resistance, which causes a current to flow in the R...perhaps, because the opamp presents SUCH a high impedance to this bias voltage...there is some odd thing happening where the tiny current it WILL take needs to be 'handed to it', ha ha.  In layman's terms, it is such a high input impedance, it can't develop its own current and needs that R there to make it flow? (?)

Ok so what about biasing applications like the one below? Surely we'd be achieving what we'd like without R4, and instead, just connecting the junction of the divider directly to the input stage and the IN+?
As a side note, I've been simulating the input/power stages and noticed that removing either C1 or R4 significantly affects IN+, but removing both results in the same IN+ as keeping both in?
Perhaps R4 is there to counter some effect of the smoothing cap.? Or that they're related in some way... This is all very curious

[antonis]

R4 can't be mooved or there shouldn't be a voltage divider anymore..
(setting IN+ voltage at a level of (+Ve X R3) / (R3 + R4)..)

Absence of R4 is only utilized for symmetrical bipolar supply, in which case only R3 is needed to create a DC path to GND (0V)..

As it is, it's just a non-inverting op-amp and the only difference with inverting one should be Input connection, in which C1 should be omited and C3 should end up at Input instead of GND..
(meaning that on both cases IN+ MUST exhibit a reference voltage set by R3/R4 voltage divider..)

The reason for the above mentioned reference voltage is for signal to have a level to swing about..

P.S.
I think I can get your original query for a voltage without current path (infinite resistance) is simply a useless nuisance but it's too late (local time) to proceed on it.. 

[PRR]

> just connecting the junction of the divider directly to the input stage and the IN+?

That injects half of the supply crap into a sensitive input.

R.G. Keen has an essay "Noiseless biasing" which explains this; however I am not finding it in his decades of pages. Connecting bias to an unfiltered divider "would" work, for a perfect power supply.

Why bias an input? We need to keep all action between the supply rails. Speech/music is symmetric on the average, so "half supply" is the logical point. How does the opamp "know" to sit at half-supply? We connect it to a half-supply point. But then we want Signal to swing both ways around that bias point. We have to mix AC signal and DC bias. Won't they load each other? Yes, but we can make it very small. The AC comes in through a cap which is ideally infinite impedance to DC. The DC can be brought in through a 1Meg resistor which is not a big deal to our Signal.

Voltage drop? Always! EVERYTHING leaks, some more than others. That's why you need math. For the archetypical LM741, the datasheet tells us: "Input bias current - 80 nA typ, 500 nA max @ 25C, MAX 0C-70C 1.5 μA". Oye, teeny currents! First, never trust a "Typical". We can have 500nA at room heat, 1.5μA when too hot to touch. 500nA is actually 0.5μA. So figure we may never have a full One μA when we are playing. 1μA times 1MegOhm is.... 1 Volt. So if we have 4.50V on one end of the 1Meg, and a '741 on the other end, we might end up with 3.5V or 5.5V. This seems sloppy, but there is no reason we need "exact half", and we still have plenty of clearance from the 0V and 9V rails. It works. (Although with '741/4558-class chips we do prefer less than 1Meg bias resistance. But in 1969 that's all we had, and much less than 1Meg sucks the treble out of guitar.)

[Rob Strand]

R.G. Keen has an essay "Noiseless biasing" which explains this; however I am not finding it in his decades of pages. Connecting bias to an unfiltered divider "would" work, for a perfect power supply.

Tricky to find,
http://www.geofex.com/Article_Folders/modmuamp/modmuamp.htm

[rhaggart1]

OKAY!

Here's where I'm at:

Tricky to find,
http://www.geofex.com/Article_Folders/modmuamp/modmuamp.htm

Thanks for this, and thanks to all of your for your help on the topic. I haven't found any wealth of info on 'noiseless biasing' yet... It all still seems a bit poorly covered online. However, it has helped me think through some things.

So, in a previous reply I linked to another biasing tactic:

I think I understand why this works - the DC can't flow through the signal source (coupling cap.), so at the junction of the divider, the source is either sinking or sourcing through the top & bottom resistors at opposite times (bad terminology, probably... sorry).
I.e. for a 1Vp-p signal, in the positive half cycle, the voltage across the upper resistor will peak at 4V5 - 0V5 = 4V, and in the bottom, it will be 5V (opposite true for negative cycle).

This technique is fine for a perfect supply. Great!

If we want to deal with a real-world supply, we want to filter that noisy supply with a cap. to ground from the divider junction (C1 in the original schematic). Cool!

But now, when our signal sees this capacitance (low impedance for AC) in parallel with the large resistance, the effective impedance is very low (maybe 100-200R for 1kHz at these component values, I'm ignoring R1, C3, R5 for now). The coupling cap. has a much higher impedance (~15K), and so most of our signal voltage is dropped across this cap, rather than the bottom half of our divider.

By introducing this high resistance R4, we introduce a much higher impedance for the signal to drop over.
With 4V5 present at the top of R4, the 05V from the source pushes current up through R4 in our positive half cycle, dropping -05V (measuring top leg as +ive) across the resistor, and so the bottom leg of R4 (IN+ of the opamp) sees 4V5-0V5=4V at the peak. Opposite true for the negative half cycle.

Ok, this was more of a stream of consciousness of me trying to explain this to myself. To me, this makes sense (mostly)... I may have mixed up some signs or some terminology here but I hope this paints the picture of my understanding.

Whilst this is obsiously a solution to the real-world components, I am still treating this all as ideal, and so this explanation doesn't cover the actual purpose of these components (noiseless biasing), rather just the behaviour through each.

Please let me know if I'm off here! Still trying to make sense of it....

[antonis]

Off definatelly not..!! 
(On not completely..) 

See how signal "leak" can be prevented with the addition of only one resistor (Rbias) of high enough value..
(it's actuall value is restricted by IN+ leakage current for a real world acceptable DC bias off-set but let it be for the moment..)
By isolating signal from bias point (due to Rbias) you can lower R3/R4 values for a less noisy configuration toghether with a stiffer Vbias source..
(from divider flowing current and voltage ripple viewpoints..)