"Overdrive" effect and transistor mixing issues... - Page 1
Forum section: electronics / beginners

author: iampoor
author_rank: Frequent Contributor
timestamp: December 15, 2015, 11:22:03 am
content: Hi allCurrently prototyping a transistor based "overdrive" pedal for guitar. All 3 of the circuit blocks work fine, but when they are connected together...I start running into issues.The primary issue is that the mixer pot's dont work correctly. They sweep pretty cleanly through the first 80% of there range, but at the last 20% the channel will cutout completely, At 15% (If Im turning U$5 The "Overdrive volume" pot) the clean signal will start reappearing, even if the clean signal pot (U$1) is at 0. The fundamental issue seems to be with the mixer. I have never done a mixer using transistors, and am baffled. I got the mixer schematic from Rod Elliots articles on mixers.http://sound.westhost.com/articles/audio-mixing.htmCan anyne point me in the right direction?Thanks!

author: krivx
author_rank: Frequent Contributor
timestamp: December 15, 2015, 11:26:45 am
content: No emitter or collector resistors on Q1?edit: also can you say what you mean by "last 20%" - 20% from maximum volume? 20% from minimum? Are you using linear or audio (log) pots?

author: krivx
author_rank: Frequent Contributor
timestamp: December 15, 2015, 11:33:29 am
content: Is Q3 drawn correctly? Input applied to emitter, output at collector, base biased from the collector?

author: iampoor
author_rank: Frequent Contributor
timestamp: December 15, 2015, 11:41:50 am
content: Quote from: krivx on December 15, 2015, 11:26:45 amWhoops! Forgot to put that in the schematic. 10k resistor on the emitter going to ground, I will reupload the picture with that correction.20% from minimum. I am using linear taper pots.Quote from: krivx on December 15, 2015, 11:33:29 amYes

author: krivx
author_rank: Frequent Contributor
timestamp: December 15, 2015, 11:49:38 am
content: Quote from: iampoor on December 15, 2015, 11:41:50 amThen it works as expected. Your hearing is logarithmic so a linear pot will have an abrupt jump-to-silence at one end, many commercial pedals that use linear pots also behave like this. Use a log pot if you want gradual control over the whole range. See herehttp://www.diystompboxes.com/pedals/FAQ.htmland herehttp://www.geofex.com/Article_Folders/potsecrets/potscret.htmif you are new to this.Is Q3 set up like this for a low input impedance? it's unusual to see a common base configuration in a guitar pedal.

author: iampoor
author_rank: Frequent Contributor
timestamp: December 15, 2015, 12:03:52 pm
content: Quote from: krivx on December 15, 2015, 11:49:38 amThe issue is that the "clean" signal will reappear once the pot is past 15% from minimum. From 100% to 20% it acts quite normally (And yes it does fade out rather abruptly like expected O0). Sorry I havent been explaining the issue super clearly.Yep, you got it!I am thinking my issue might just be simple control interaction through the R7 R8 mixing network?

author: Andy Watson
author_rank: Super Contributor
timestamp: December 15, 2015, 12:22:29 pm
content: One potential problem is that at low signal levels, i.e. when Q2 is behaving linearly, there is an inversion in the signal such that at certain settings of the mixer controls, the signal will cancel.Edit: Minor points, either C1 or C5 is superfluous. C7 is back-to-front.

author: krivx
author_rank: Frequent Contributor
timestamp: December 15, 2015, 12:23:01 pm
content: Quote from: iampoor on December 15, 2015, 12:03:52 pmPossibly, you could try disconnecting the clean side and see if the problem persists.Also - just my opinion - if you are mixing two signals then a signal "mix" pot is more useful than individual volume controls.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 15, 2015, 08:44:40 pm
content: The resistor values are all over the place.I don't know what VB is, but I'll assume it's around VCC/2. Then,- R4 should be more like 100k (and then C3 can be smaller, and R2 probably won't matter)- C5, C6 are definitely impractical (1F??)- R5 is just straight up wrong (1m, should be, 100k maybe, but Q2 is suicide biased as shown)- R7, R8 need to be larger than the pots they're drawing from (and the pots need to be log taper)- Q2 inverts, so the mixture of U$1 and U$2's signals will oppose, mostly leaving clipping/harmonics. It won't simply be an additive mixture of undistorted versus heavily clipped, it will be subtractive.- Q3 is simply... wired wrong. I mean, it will work, sort of, but a common base amplifier has an extremely low input impedance, which makes it the worst possible choice after necessarily raising the signal's impedance, first with the pots (even if they were 1k pots, it would be a lot), and second with the series resistors. You are getting less than unity gain with that stage, and it would literally work better without it entirely.I'd suggest an emitter follower, just to buffer the output as is (good for driving long cables?), or a normal common-emitter amplifier (with emitter degeneration this time).I understand that there are a number of awful things done intentionally here, to generate harmonics and such, but these differences are very basic: values wrong by six or more orders of magnitude, a stage that actually reduces output, inconsistent bias that will be drifty and unstable (and therefore the timbre will change), and ultimately more frustrating to use.Tim

author: techricky
author_rank: Contributor
timestamp: December 15, 2015, 09:08:41 pm
content: The Q2 distortion stage inverts the signal, so with the right proportion of mixing it will be cancelling out the direct signal

author: iampoor
author_rank: Frequent Contributor
timestamp: December 15, 2015, 10:28:42 pm
content: Quote from: techricky on December 15, 2015, 09:08:41 pmQuote from: Andy Watson on December 15, 2015, 12:22:29 pmAhhh!! Of courseThanks for pointing that out! Q2 will invert in saturation too, right? Not just the linear region?Quote from: krivx on December 15, 2015, 12:23:01 pmAt this point, I think Tim has convinced me some major circuit changes are needed so I am probablly going to rebuilt it now...Long term thats the plan! I am just using separate controls to test the concept and see if its even good or not.Quote from: T3sl4co1l on December 15, 2015, 08:44:40 pmHi TimThank you for your thoughtful analysis as always.Yes, VB is VCC/2-I am trying to keep the input impedance of the circuit over 1m (100k will load down pickups pretty heavily)....of course with opamps this is easy, just make the damn VB resistor higher.I am assuming a voltage divider arrangement would be much more appropriate(See the attachment)? Q1 should be biased at 4.5v and be nice and linear.-C5 and C6 are 1uf, I used an arbitrarily high value because I have a whole bag laying around, Gut instincts tell this should be closer to 100nf in the finalized design. Are there any cons with using a larger coupling capacitor (within reason)? I am not aware of any.-Is there a rule of thumb for this? EG Mixing resistors should be 2x or 3x the value of the pots?-Thank you for pointing this out, thats a huge issue to say the least...!-Regarding Q3, I had another AH-HAH moment today The mixer circuit is simply apassivemixer stage followed by a buffer. I didnt really understand anything going on with the circuit initially, so I thought it was the transistor equivalent of a virtual ground summing mixer. If I want to maintain unity gain, I should use mixing resistors and then a common-emitter amplifier at the end to make up the gain that was lost in the passive mix stage, correct?Yes, All of the associated components around Q2 should produce distortion, Making the Bias of Q2 variable might produce some interesting results too.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 16, 2015, 04:17:41 am
content: Quote from: iampoor on December 15, 2015, 10:28:42 pmUhh...100k >> 1m, and 1m will load the pickup as a short circuit.Did you mean 1Meg???Case counts!QuoteAttachment?If you aren't using VB anywhere else, you can reduce the series resistor + voltage divider to a single voltage divider of slightly different values.If this is running right off the pickup, you probably want some preamplification as well. A very strong pickup will only be "hot" enough to push the diodes into clipping on the attack, and it will be very "mellow" otherwise, with poor sustain.QuoteYes, if the stage itself is driven into clipping. Base rectification shifts the bias point, and you get time dependent compression or clipping. Time constants (coupling C * sum of nearby R's) in the seconds will be irritating, for example rapidly strumming and stopping notes will push it into saturation on the leading edge, then leave it quiet inbetween. If you wanted to follow a loud note by a quiet note, for example, that would be difficult or impossible. It still sounds like something that might be useful, but I suspect a more sophisticated 'squelch' system would be in order, like a proper noise gate, that behaves itself no matter what tempo you're playing at.QuoteYes, 3-10x is good. Too small and the resistors "load down" the pot too much, making its curve (volume vs. rotation) distorted. (To a small extent, this can be used to make a linear pot look somewhat logarithmic -- but it's pretty awful for any useful log range, like >20dB. A good pot should be log-ish over >40dB.)Accordingly, your final amp needs to have a pretty high Zin to 'read' that high impedance. So, 10k pots say, 100k resistors, 100k to 1Meg feedback resistor maybe.QuoteIt's not entirely wrong -- the purpose of a virtual ground is to maintain isolation between the inputs, to limit cross-talk. The emitter input is a very low impedance, so although it doesn't respond the way an op-amp's virtual ground does (i.e., there isn't a feedback resistor virtually nulling it to ground), it still has the same benefit (it's a darned low impedance, so prevents crosstalk), and as it turns out, has the same behavior, but inverted (namely: the change in collector current equals the change in emitter current, so the collector resistor sets gain, just as the feedback resistor does with an op-amp).But again, the impedance is too mismatched to work; 100k resistors at the input, and ~1k resistor at the output, means your gain is more like 1/100.Now,QuoteYou may find it interesting to use Q2 itself as the clipping element. Simply use a common emitter stage (with emitter degeneration, bypassed), with a variable base voltage divider (i.e., you'd use a pot to set "VB", and supply that to the base with a resistor -- you can't put the pot right on the base, because the pot's Thevenin resistance varies with position!). Then amp up the input signal, so it's more than enough to drive it into saturation and cutoff. Varying input level and base bias will shift which parts of the waveform get clipped.The output level will be quite "hot" (i.e., basically VCC peak to peak), so you'd probably want a voltage divider off of it, then a buffer or follower for final output.I use such a circuit and adjustment on my Theremin (which has the big difference that its signal level, at this point in the signal chain, is constant, not varying with intensity), and find the range of character sounds pretty good.Other options include a differential amplifier (two transistors, tied at the emitter) with variable input level and bias (the clipping will be symmetrical, rather than soft on one half), or an op-amp with diodes in the feedback loop (so it has clipping behavior, but it's not absolute, still letting much of the signal through; or, it has clipping behavior around zero, adding crossover distortion).Tim

author: krivx
author_rank: Frequent Contributor
timestamp: December 16, 2015, 10:19:59 am
content: Quote from: T3sl4co1l on December 16, 2015, 04:17:41 amHiya Tim, do you mind going into a little more detail on this? "Base rectification" doesn't seem to bring up much on Google.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 16, 2015, 10:24:32 pm
content: Ah, not really a specific subject, but a combination of effects...The BJT base-emitter is a diode. Simple as that. Under small signal conditions (namely, on the order of 10mV), it looks like a more-or-less constant resistance, and we can simply apply a signal and be done with it. But for increasing signals, it looks more and more like a diode, and the current drawn by the base will depart from a resistive response. The result is that more current is drawn during positive peaks than is drawn (or removed) during negative peaks, which changes the voltage drop in the base voltage divider. This shifts the base DC voltage negative. Which charges the input coupling capacitor (so it exhibits a time constant), and shifts the bias point.The effect is that, with an optimally biased common emitter amplifier: as you increase signal level, the signal will begin clipping on both peaks symmetrically (it reaches saturation and cutoff at the same level), then as you advance, it becomes asymmetrical: the saturation peak grows less quickly than the cutoff peak.If the signal then decreases, at a rate faster than the coupling capacitor will discharge at, it will "stick to one side", namely, staying towards the cutoff side while the saturation side stops clipping entirely. After a few time constants, the DC value returns to its quiescent value.You also get this effect with JFET amplifiers, because the gate to channel (D and S) is a diode junction as well. Normally it functions entirely outside of diode conduction (unlike the BJT which functions entirely because of B-E conduction), but when driven hard, it will still do this.A MOSFET will not, because it is entirely insulated. (Okay, there are MOSFETs with internal protection diodes, which will eventually conduct, and cause clipping and bias shifting for very large signals, like 30V, but these are well beyond the realm of normal signal levels!) You can also use a much larger gate bias resistor, because there's no base current to supply, and no rectification to deal with.Tim

author: krivx
author_rank: Frequent Contributor
timestamp: December 17, 2015, 08:14:23 am
content: Thanks again for another great post. I had never thought of the input capacitor reacting with changes in bias in this way, though it does go some way in explaining how some guitarists describe "bloom" on note attack on some fuzz and distortion pedals. I'm going to look into this and see if I can watch this change