issues with analog synthesizer build - Page 1
Forum section: electronics / projects

author: jonny_w
author_rank: Contributor
timestamp: August 03, 2014, 06:59:51 am
content: I'm a little bit frazzled with an issue here with my analog synthesizer project, and could really use some additional insights. The schematic is below, with the problem area marked in red:My issues are as follows: the voltage present on pin 7 on U1-B is 0.505V, when it *should* be 55mV. Additionally, Q1 is not turning on (VBE = 0.52V, VCE = 0.44V).Some notes & measurements:- R4 is set at 100K, and R12 is set at 100 ohms.- Voltage from the D1 cathode to U1-B pin 7 is 0.303V- Voltage drop across D1 is 0.447V- Voltage drop across R14 is 0.08V- Voltage drop from the D1 anode to ground is 115mV- Power rails of U1-B measure +12V and -12V respectivelyI've played around with the values of R4 & R12 pots, but could not get the transistor working/see a drop in voltage at pin 7 to 55mV.I would *REALLY* appreciate any wisdom/guidance from you guys. I've been troubleshooting this for a couple weekends now (with both my multimeter & Rigol oscilloscope)- Jonny

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 03, 2014, 10:54:32 am
content: Well, this thing is...Wait, what IS this thing...It looks like it's trying to be an exponential V-I converter, but it doesn't work. At least, I can't convince myself that it was supposed to, as shown.1. I don't know what the D1 loop is doing. It feeds op-amp output back into the input / Q1 B, which is a positive feedback path. The label "High Freq Comp Trim" seems to suggest it's to do with compensation, but with C1 loading the thing down, it's not going anywhere very fast. Seems like a very janky way to fix a problem that shouldn't have been there in the first place (say, by proper choice of C1, or an appropriate network -- possibly a nonlinear one -- instead of a dumb capacitor; or by fixing the rest of the circuit to begin with).The analog equivalent of patching a bug in software, which introduces even more bugs, and patching those, one by one, by ever-more-roundabout means.2. Q1-Q2 seem to be arranged as a matched pair (needless to say, the transistors must be matched, both in Vbe, Is and temperature), but they aren't quite being used as such.The wiper of R12 sets one base voltage (a small voltage above ground -- little over a volt I think, and probably more like <200mV), and the other is tied to ground. The emitters are common, so the (Vbe1 - Vbe2) voltage is set here. The restriction is, Q1 collector current must be constant (presuming the +12V supply is regulated, and R7 doesn't drift). Also, Q1 collector is held at 0V, which is perhaps a good idea (it puts Q1 near the "diode mode" range, i.e., Vce ~= Vbe, which if Q2 is held in the same condition, will maintain matched circuit conditions for both).Since Q1's Ic = 10uA, its Vbe will be fairly low, and your observation of 0.52V seems consistent with that. It's turning on. (But more on that later.)Since Ic is constant, Vbe should be constant as well, which means the voltage at R12 should appear on Q2's Vbe (upside down). So Q2's Ic should be exponential in the (negative) voltage at R12. (Presumably, the inputs are 0-5 or something like that, positive control voltages?)But here's the crap. Q2 is pulling collector current. Q3/Q4 are hanging from GND. So their base voltage has to be negative a little. But Q2 is already negative only a little (about a Vbe), so that node's voltage is going to be really tight. This also blows the assumption that Q1 and Q2 are under matched conditions (Q2's Vce is low, and may even saturate, in which case the current error will bewayworse than just Early effect).The worst part is, Q3-Q4 seem to be a current mirror, but in hFE mode, not actually a current mirror. So the matching of collector currents (and thus slew rates of the two OTAs) will be terrible, extremely drifty (as if all these discrete BJTs weren't bad enough), and unreproducible from build to build. There will also be gain, which might partly explain why the bias is so low (i.e., 10uA at 0V).What should be done, is Q3 and Q4 emitters tied to a +Vbe reference (which might be another op-amp half, following another 2N3904 biased at 10uA or so), so that their bases are closer to 0V, and add another transistor in parallel with Q3/Q4, with C tied to B, so that it actually becomes a current mirror as such.As for your measurements... have you accounted for instrument error? A 10M probe isn't much normally, but out of 10 microampere signals, well... Your numbers don't appear to be consistent, which tend to suggest something like that (voltages are "squishy" and being pulled up/down by the meter), or it's oscillating and things are all over the place.Tim

author: jonny_w
author_rank: Contributor
timestamp: August 03, 2014, 07:37:24 pm
content: Thank you for your reply, Tim. I'll include the circuit explanation below (taken fromhttp://www.musicfromouterspace.com/index.php?MAINTAB=SYNTHDIY&PROJARG=SOUNDLABMINIMARKII/page1.php&VPW=1270&VPH=523)The 12db/octave state variable filter works as follows. Control voltages are applied to the control voltage summer U1-A via the 100K resistors feeding it's inverting input. The fine COF pot feeds the control voltage summer via a 4.7M resistor (R35) so that the fine pot gives a much smaller range of adjustment to the cut off frequency. The filter goes from sub audible to about 18kHZ at the top end. The VCF modulation sources: LFO and AD1 (Attack Decay Envelope Generator 1) are fed into the control voltage summer via linear 100K pots R9 and R13 (wired as variable voltage dividers) respectively to allow for adjustment of the modulation levels of those sources. Switch S20 lets you add keyboard voltage as a modulation source for the VCF. The output of the control voltage summer feeds the voltage to exponential current convertor (U1-B and associated components). It is important to use a matched pair of NPN transistors for Q1 and Q2. The critical parameter for the transistors is that their VBE be within 2mV of one another. I also recommend that Q3 and Q4 be a matched pair of PNP transistors with VBE within 2mV of one another. I sell matched pairs both discrete and the SSM2210 integrated matched pair. Here is some information on matching transistors. The expo convertor's whole job in life is to convert the voltage presented to it via the control voltage mixer into current. But the conversion factor is very critical if you want the VCF to track 1V/octave. The transistor in the feedback loop of U1-B (Q1 2N3904) is the magic component. It doubles it's current output for every approximate 18mV change in voltage presented to it's base. So the 100 ohm trimmer R12's job is to adjust the voltage fed to the base of Q1 so that you apply the exact voltage change necessary to get the current doubling response. Current through Q1 is mirrored in Q2 and Q2's current is applied to the bases of Q3 and Q4 who also mirror (but invert) the current flowing into the LM13700 bias inputs from ground. The potential at the bias inputs is low so current can flow from ground into the bias inputs. The 10K resistors Q3 and Q4 act to keep the current into the bias inputs limited and balanced. The high frequency compensation adjustment trimmer is necessary because without it the filter's cutoff frequency would go flat at higher frequencies due to the bulk emitter resistance of transistor Q2. (For an excellent article on Log and Exponential (Antilog) Circuits see Bernie Hutchins' ELECTRONOTES S-019).1. Q1 & Q2 were supposed to be a matched pair (I purchased the pair from the kit provider). R18 is a 2k tempco resistor thermally coupled (with heatsink silicone grease) between the bodies of both Q1 & Q2.2. Your design modifications make sense, however this project is on a printed PCB project & I'm stuck with what the maker designed. There are examples of this synthesizer all over the net, so I know the circuit works (even if the design is not optimal)3. Looking at the data sheet for the 2N3904 (and remembering a little from the transistor course I took awhile back), shouldn't Vbe be at least 0.65V to turn on? Also, I figured Vce would be 0.1-0.2V in this application?4. All my sanity checks measuring pin 1 on U1-A come out perfect, so it appears the circuit is functioning up to that point.5. I just re-checked my regulated power supply, and I'm measuring +11.82V off the +12V feed & -12.12V off the -12V feed.6. I'll re-measure every point with both my meter & scope and post the results.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 03, 2014, 08:50:38 pm
content: Quote from: jonny_w on August 03, 2014, 07:37:24 pmOk, all well and good so far...Quote...But this is where his knowledge reaches a limit. In fact, Q1's collector current is forced to be constant, because R7 is supplying current into that node and nothing else is drawing it out. Which forces Vbe to be constant as well. Q1 is providing a constant Vbe offset, which compensates the voltage from R12 so it can be applied directly to Q2. The output current (assuming it doesn't saturate) will be 10uA when V(R12) = 0V, since the pair is balanced. For V(R12) < 0 (or positive voltage at the control inputs), the current goes up exponentially, an octave every Vth * ln(2), or 18mV (assuming room temperature). This is proportional to absolute temperature, so V(R12) should also vary with absolute temperature; if R12 were a silicon PTC, that should about do it (you wrote R18, which I think was a typo?).The other part is:QuoteQ3 and Q4 act to reflect the current, but they aren't doing it in direct proportion as a truecurrent mirror. There needs to be the diode-strapped transistor included.Also, no mention of how he expects Q2 to stay out of saturation. He might just be lucky and always manages to run >50mV there so it keeps working. But man that's marginal.QuoteIf Q3 or Q4 are unsaturated, the resistors havenoeffect on the current flow (well, except for Early effect, but that's a 10% kind of thing, and can provide no balancing action, regardless). Theyarea good idea, because with "suicide biased" transistors like these, you want to allow for the case when theydosaturate.They aren't really necessary if Q3-Q4-Q(new) were a proper current mirror, but you can leave them in with no harm.QuoteThat's interesting. The diode and resistance act in parallel with the emitter plus output resistance, but return to the circuit input, not the op-amp. Acting as positive feedback, which acts to oppose the effects of a degenerating resistance.Note, by the way, that Q1's emitter resistance is a big fat "do not care" -- because its collector current is constant, and small. Q2's matters at large currents, because Q2's current does vary.Alternatives might include low-R_e types, like high hFE (I think?) e.g. 2N5088, or power transistors (bigger scale effectively swamps the effect, though hFE reduction at low current, and the higher leakage current and junction capacitances, probably makes things worse).It probably wouldn't hurt to use a diode-strapped transistor in place of D1, also. 1N914 has a much larger junction (= smaller voltage drop) than 2N3904, which might put a little kink in the transfer function, or something.QuoteWell that's no excuse! Every pin or pad on the board presents an opportunity: to lay a wire across it, solder that on, and make an ugly rat's-nest of modifications. Traces can be cut, soldermask can be scraped away, and pins can be cut or just bent straight up and away! Once you've built a few circuits like this,and made as many modifications to existing (printed) circuits, you'll realize that nothing is holy.QuoteVce = Vbe - V(R12) for Q1; for Q2, Vce might be quite small (how much exactly depends on the voltage drop at Q3/Q4, which is poorly defined because they aren't a correct current mirror).Vbe is 0.6-0.7V for Ic in the 1-10mA range. But by the accompanying model (assuming Vbe is constant and Ic = hFE * Ib), this circuit is utter nonsense, and none of these functions could ever be possible.The Ebers-Moll equation, however, describes much more exactly, and more richly, how transistors work. Since Ic is an exponential function of Vbe (with scaling constants), it will still be true that Vbe doesn't vary much under limited signal-level current ranges (like 1-10mA), but for huge changes (like 1uA to 1mA), it will be noticeably different. What's more, any two transistors with matched coefficients will produce exactly the same current, from the same applied B-E voltage (yes, a fixed, constant, base voltage -- exactly what they tell you not to do -- okay as long as very specific conditions are met, though!).Or, 18mV/octave is 60mV/decade, so if Vbe = 700mV @ Ic = 10mA, three decades brings that down to Vbe = 520mV @ Ic = 10uA.QuoteMore importantly, measure one node (a low-sensitivity one, like op-amp output) while probing with another. If you see the first measurement change when the other probe touches things, you're skewing the circuit -- observer effect.Tim

author: jonny_w
author_rank: Contributor
timestamp: August 03, 2014, 09:14:45 pm
content: Measurement update (confirmed with both BK & Agilent multimeters)U1-B pin 8 = +11.8VU1-B pin 4 = -12.13VVR16 = 112mVVR12 = 34mV (trimpot set @ 100ohms)VR14 = 78mVVD1 = 326mVVR7 = 11.81V (complete +11.81V supply drop, verifying pin 6 is virtual ground)VR4 = 288mV (trimpot set @ 100k)VR5 = 13mVVR8 = 99mVVC1 = 513mVQ1:Vbe = 530mVVbc = 114mVVce = 414mVQ2:Vbe = 415mVVbc = 346mVVce = 110-120mV (unstable)Q3:Vbe = 240-260mV (not stable)Vbc = 10.7VVce = 11.14VQ4:Vbe = 260-270mV (not stable)Vbc = 10.7VVce = 11.14VVR17 = 0VVR18 = 0V

author: David Hess
author_rank: Super Contributor
timestamp: August 04, 2014, 06:05:24 am
content: Could it be relying on the matched Vbe of Q3 and Q4 being much less than the matched Vbe of Q1 and Q2?I think if I had designed this, I would have included a cascode to control the collector voltage of Q2 so Q3 and Q4 could be biased from the positive supply as a real current mirror.As far as the frequency compensation, that is always a problem with translinear circuits. I remember seeing one that used a varactor diode in place of the feedback capacitor.