Super fuzz de-buggery!!

[MikeH]

I'm using this layout: (Unicord superfuzz)

http://ustomp.com/?p=12

To make a univox superfuzz clone.  Schematic here:

http://members.fortunecity.com/uzzfay/superfuzz1/usf.html

I know the layout isn't exactly right; I used jumpers for the bias pot, the input cap and resistor are in different order, there are a couple  values that were different so I subbed in the Univox value in those cases.  For transistors I used all 2sc828 trannys like a univox superfuzz rather than the mix of 2sc828s and 2sc539s in the Unicord.  The pinout for the 2sc828 is BCE rather than CBE, which is what the layout calls for, so I insulated the legs of the trannys and bent around the collector and base pins.

Voltages (based on the layout):

Q1
C  9.94
B  0.00
E  0.00

Q2
C  9.01
B  8.94
E  0.01

Q3
C  9.00
B  3.41
E  0.00

Q4
C  9.00
B  1.61
E  0.00

Q5
C  9.00
B  1.61
E  0.00

Q6
C  9.00
B  1.16
E  0.00

Signal gets lost a Q1, and obviously that base voltage is suspicious.  I checked for continuity between Base and Emitter and they're not bridged or anything... so why the %&# is it at 0 volts?!?  Oh and I tried a different tranny with the same results.

One more thing- I didn't have enough 10uf caps so I used a few 6.8uf caps instead.

[Dan N]

The pinout for the 2sc828 is BCE rather than CBE

No, they're ECB rather than EBC.

[MikeH]

The pinout for the 2sc828 is BCE rather than CBE

No, they're ECB rather than EBC.

Well right, same difference.  Point is, I verified the layout and I know they are oriented correctly.  Base and collector pins are flipped.

[MikeH]

I was looking at what I posted before and I made a mistake: Q1 collector is 8.94v, not 9.94v.  And the battery was at 9.01V.

[R.G.]

Here's an interesting question - how does the base *ever* get pulled higher than 0V?

[MikeH]

Here's an interesting question - how does the base *ever* get pulled higher than 0V?

Damn.  I guess this means I've officially been around too long to just get an answer hand-fed to me.   

Ok... Let me think.  In this situation the transistor is set up as an amplifier (obviously).  Initially I thought since it was the first gain stage it need some kind of voltage swing on the input to conduct, but the reading is the same even when it's fed an input signal from a guitar.  I guess I don't know; I've never had this problem with base voltage before.  It's kind of strange that I have signal at the base even though it's at 0V.  Unfortunately I have to answer 'I don't know' 

Another hint?

[R.G.]

Damn.  I guess this means I've officially been around too long to just get an answer hand-fed to me.   

Sorry - that's what you get for being a long-timer! You have to learn to think. 

Another hint?

OK. Smaller steps are OK too.

There are three resistors connected to Q1 base. One is a 22K from the input cap, so no DC is coming in that way. One is a 100K to ground; that only helps pull the base to 0V. But the OTHER 100K is connected to the emitter of Q2. That's the only possible way to get Q1 base pulled up.

Hmm... what's the voltage on Q2 emitter? 0.01? AGH! No wonder base of Q1 is at 0V. Nothing pulling it up. And we know that Q2 base has to be one diode drop above Q2 emitter... doesn't it? Wait a minute! What's Q2 base doing up at 8.94V? That's impossible for a working NPN transistor. The base CAN'T be pulled up that high unless
- the PAD is pulled up and the real base lead is not connected to it
- the lead is soldered to the pad but the base is open internally - bad transistor (placed here for completeness, but the least likely of the possibilities)
- the pad on the emitter lead is at 0V, but the solder joint is open and the real emitter lead is up at 8.3V or so.
- the emitter lead is soldered to the pad but the emitter is open internally (same comments on transistors)

Maybe think about one of those.

[MikeH]

Ahh, I see.  I was so focused on the bad voltages on Q1 I didn't really think about Q2.  So the base- way too high.  I'm guessing I'd prefer to see something in the 3Vish range.

I checked for continuity between the socket (both empty and on the transistor leg) and the pad underneath for both Q1 and Q2 and everything is connected.  The solder joints appear to be solid; they do not appear to be floating in any way.  Tried another tranny and got the same voltages.

So if I'm understanding the basics here; Q2 is not conducting because the base is too high?  Which in turn causes Q1 to not conduct either?

I also double checked all of the resistor values to make sure I hadn't mistakenly used a 470R instead of a 470K or something but all values are correct.   The 10uf cap connected to Q2s emitter is a 6.8uf on my build, which shouldn't effect the base voltage in a negative way correct?

[bluesdevil]

Hey Mike - Sorry to hear the problem on your build. I'll be starting the same project as soon as I get a parts order in. I was told by a forum member that the 10uf's can be replaced by just about anything above 2.2uf to work, so the 6.8uf probably isn't the problem. Maybe try a .001uf instead of the .002 going to Q1 and Q2's collectors (if you went with the Univox values). Not sure if that really makes a difference, but worth a shot if R.G.'s suggestions didn't help. The guy at UStomp swears his layout works, so keep pluggin' away.
    Good luck!!

[MikeH]

Yeah, I've been over the layout about a thousand times, and it seems fine to me.  And yeah, I did use a .001 cap like in the univox.  I just can't seem to find the problem with Q2!! 

Is there any other potential problem that could be causing the base to be pulled so high?  I've checked the values of the resistors with my DMM and they are all ok (some sort of fluctuate when I meter them, I'm guessing it's because they're tied to a cap?).  I know the problem is right under my nose (literally), I just have no idea where to go next.

[R.G.]

So if I'm understanding the basics here; Q2 is not conducting because the base is too high?  Which in turn causes Q1 to not conduct either?

Close. Q2 is not conducting because the emitter is so low. With such little voltage across the emitter resistor, no current is flowing, according to Ohm's law.
If, for instance, the emitter resistor is 0.1 ohm instead of  10K, then that might be an honest voltage for conduction. But with a 10K there, a voltage of 0.01 means that the transistor is not conducting. Add to that the bit that the base is more than one diode drop above the emitter and you have a flashing red indicator that all is not well, and right here, right now.

This goes back to the very basics about how NPN transistors work. To electrons, they look like two diodes in series, a reverse biased one (the base-collector) and a forward biased one (the base-emitter). The base-collector keeps current from flowing 100% of the time. This prevention is "poisoned" by injecting current into the base. Each charge carrier injected into the base allows many charge carriers from the collector to sneak through.

That basic operation means that the base can never be many volts higher than the emitter. As the base voltage is raised, the base-emitter starts conducting at 0.5V and by 0.7V it's in heavy conduction. The base-collector is letting through from ten to several hundred times as much current as the base current. Sothe base simply *can't* get more than about 0.7V higher than the emitter in an NPN unless something is broken. In this case, The base being at over 8V when the emitter is at nearly 0V is a conundrum. If it's really hooked up the way you think it is, that is impossible. Back at the Sherlock Holmes School of Electrical Design, the maxim is "Eliminate the impossible, dear Watson, and whatever remains must be... the truth!" It's impossible for a properly biased NPN transistor to have the base over 8V higher than the emitter and still have the right external connections. So we have to get info to sort through the possibilities and find out which ones are impossible in this particular circuit.

I hit on a few of these. If the emitter lead were open from its pad by a bad soldering joint, then the real emitter could be up at the base voltage minus a diode drop, but not able to pull up the 10K resistor on the emitter. One elminates that by measuring the voltage on the lead and on the pad. If they're identical, they're connected. Same issue is possible on the base - if the base lead is separated from the pad which brings it the collector voltage for Q1, then the Q2-base pad will read nearly 9V from Q1 Collector and the Q2 base lead will be down at a little above the emitter voltage.  If Q2 collector was disconnected, it would be down near the emitter voltage, so it's probably OK.

So much for the connections story. If the device is the wrong one, with a different pinout. that could cause this. Likewise if the PCB does not connect where you think it does. Part orientation could do it. A bad Q2 with an open base or open emitter, or both, could cause it, although this is the least likely of the possibilities.

[MikeH]

So I need more juice on the emmitter of Q2- It's obviously improperly biased.  Since the collector is tied directly to V+ this would only happen because the resistance from the emitter to ground is.... what?  Too high?  Or is it too low?

Now, when I measured the resistance of the 470k resistor tied to the emitter of Q2 it was kind of floating upward (as, before; guessing because of it being connected to a cap?) but it didn't get moch higher than 150K, or so.  I assumed this was because it is creating a series resistance with the 47K resistor and 10uf cap.  I don't know what the resistance of a 10uf cap is, but it can't be that high.  Is this actually my problem and I just glossed right over it?

As far as pinout goes, I suppose I may have it wrong, but it seems obvious to me comparing the layout to the schematic that the layout calls for a CBE (from top to bottom) pinout on Q2.  The datasheet for the 2sc828 says the pinout is BCE.  Of course I realize that these trannys are really old, and I might not have the datasheet from the exact manufacturer, but I've oriented the other trannys in the same fashion, based on where their Collector is on the layout, and their voltages look good, right?

[R.G.]

Take your Q2 transistor out of the board. Use your multimeter to see if the following is true:

1. Base-emitter conducts when the base is positive, and not when the emitter is positive. Conducts both ways=shorted, conducts neither way is open, and in either case the device is dead.
2. Base-collector conducts when base is positive and not when the collector is positive. Same consequences.
3. With the transistor out, measure ohms from the base PCB pad to the collector of Q1 and to collector of Q2. It should show short circuit to Q1, open circuit to Q2C.
4. With the transistor out, measure ohms from the emitter to ground. Should be 10K, perhaps slightly less because of the 470K/100K/100K resistors in parallel with it to ground.
5. With the transistor out, measure ohms from the emitter to the base of Q1. This should measure about 100K, maybe a little less.
6. With the transistor out, measure ohms across the 100K from the base of Q1 towards the 470K to the emitter of Q2. This should be somewhat less than 100K.
7. With the transistor out, measure ohms across the 470K; should be about 140K.

[MikeH]

Take your Q2 transistor out of the board. Use your multimeter to see if the following is true:

1. Base-emitter conducts when the base is positive, and not when the emitter is positive. Conducts both ways=shorted, conducts neither way is open, and in either case the device is dead.
2. Base-collector conducts when base is positive and not when the collector is positive. Same consequences.

3-7, I'm all over.  1 & 2, not sure how to tackle.  Do I do this by just sticking it on my bread board and putting 9v on one pin and grounding the other (of the two that I'm testing conduction between)?  Or do I use the diode tester on my DMM?

[bluesdevil]

I would try it with the DMM.

[R.G.]

diode tester on DMM, or just low ohms range.

[MikeH]

Take your Q2 transistor out of the board. Use your multimeter to see if the following is true:

1. Base-emitter conducts when the base is positive, and not when the emitter is positive. Conducts both ways=shorted, conducts neither way is open, and in either case the device is dead.
2. Base-collector conducts when base is positive and not when the collector is positive. Same consequences.
3. With the transistor out, measure ohms from the base PCB pad to the collector of Q1 and to collector of Q2. It should show short circuit to Q1, open circuit to Q2C.
4. With the transistor out, measure ohms from the emitter to ground. Should be 10K, perhaps slightly less because of the 470K/100K/100K resistors in parallel with it to ground.
5. With the transistor out, measure ohms from the emitter to the base of Q1. This should measure about 100K, maybe a little less.
6. With the transistor out, measure ohms across the 100K from the base of Q1 towards the 470K to the emitter of Q2. This should be somewhat less than 100K.
7. With the transistor out, measure ohms across the 470K; should be about 140K.

#3 - Check
#4 - 9.65k - Check
#5 - 91.4k - Check
#6 - 84.4k - Check
#7 - 143.2k - Check

# 1 and 2- The only time I get a reading with the diode tester (and this is on all of the tranny's; I pulled tham all to check) is when the positive lead is on the emitter- and the negative lead is on the collector or on the base (or I guess I should say "what I believe to be the emitter, collector or base")  and the reading I get is always around 0.7ish (not sure on the units).  Is this right or do I have the wrong pinout going here?  And if so, why aren't my other voltages screwed up?  Or are they?

[MikeH]

How about these for Q1 and Q2 voltages?

Q1
c  8.93
b  5.56
e  4.95

Q2
c  5.56
b  0.54
e  0.02

Can't try it out now; my wife is literally dragging me out the door as I type, but I think I may have it.  If so, why were my other voltages so normal looking??

[R.G.]

#3 - Check
#4 - 9.65k - Check
#5 - 91.4k - Check
#6 - 84.4k - Check
#7 - 143.2k - Check

So the board and resistors are OK.

# 1 and 2- The only time I get a reading with the diode tester (and this is on all of the tranny's; I pulled tham all to check) is when the positive lead is on the emitter- and the negative lead is on the collector or on the base (or I guess I should say "what I believe to be the emitter, collector or base")  and the reading I get is always around 0.7ish (not sure on the units).  Is this right or do I have the wrong pinout going here?  And if so, why aren't my other voltages screwed up?  Or are they?

Here's the deal. A bipolar transistor is two diode junctions, the base-emitter and the base-collector, that happen to share one common electrode, the base. The collector base diode is run "backwards", or reverse biased. Shoving current into the base temporarily poisons the ability of the collector base junction to block current flow. So when you put a small current-limited voltage source across these junctions in isolation, they act like the diode junctions that they are. This means that you can not only crudely test an NPN transistor for good/no good with an ohmmeter, you can identify the base lead from the other two.

==>>- the base lead is the only one which shows diode action to both of the other two leads -<<==

[/b][/i]

That "emitter" you say conducts to the other two only when it's more positive, that's the base. The transistor is N...P...N, so when the base is more positive, it will conduct to either collector or emitter if they're more negative than it is.

[R.G.]

How about these for Q1 and Q2 voltages?

Q1
c  8.93
b  5.56
e  4.95

Q2
c  5.56
b  0.54
e  0.02

Can't try it out now; my wife is literally dragging me out the door as I type, but I think I may have it.  If so, why were my other voltages so normal looking??

Go with her. It could be important.   

When you come back, you can ponder that bipolars also work "inverted", with the nominal emitter more positive than the base and the nominal collector more negative.
Originally, there was no difference between collector and emitter junctions. Later we found out that if you make the emitter junction really thin and small, it give more gain and higher breakdown voltage, so you can make the collector junction more burly. But they still show transistor action at below the emitter-base breakdown voltage, which is usually 6-10 volts depending on the device. Yours both have a "collector-base" voltage of less than 6V, so it's possible they work this way, albeit more poorly .