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Post by flateric on Oct 31, 2008 3:59:00 GMT -5
OK heres the voltages, depending on position of the 5k gain pot [r5] C ranges from 0.08 - 1.4v B ranges from 0.68 - 1.97v E ranges from zer to 1.29v
You mentioned earlier that I should be seeing 7.8v on the collector?
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Post by ChrisK on Oct 31, 2008 19:38:39 GMT -5
That was a guess as I am trying to determine the DC bias point for the transistor.
With no AC input signal, turning R5 should have ABSOLUTELY NO EFFECT since the wiper is capacitively coupled to the other pot and hence only presents a varying load to AC (capacitively coupled) signals. The DC effect on bias from the rotation of R5 has to be zero.
You either have an AC signal applied (with your readings being a multi-meter filtered DC average, or you have circuit oscillation or extraneous signal pickup with "no" signal applied, or R5 and the associated other pot/capacitors are miswired.
I know that a cable needs to be connected to turn the battery on, but the input point to the effect circuit must be grounded to effect NO AC signal or pickup.
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Post by flateric on Nov 1, 2008 11:37:04 GMT -5
I'm going to desolder all the components and start again with a new bit of stripboard......
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Post by JohnH on Nov 1, 2008 15:49:24 GMT -5
As noted above, the transistor is definitely reversed in the layout diagram, the curved side should be on the left looking down on it from above. The gain pot is showing an influence on the bias voltages, and it should make no difference at all as Chris says. In particular, the last set of reading shows that the 5k gain pot is able to take the emitter right down to ground zero. That suggests either that the pot lugs are wrongly wired, with an inner and an outer lug between emitter and ground instead of two outer lugs, or, the centre wiper connection on the 5k pot has a direct route to ground through or past C2 when it shouldn't. For example, a tiny solder blob between point B on the diagram and the adjacent grounded strip would explain that. Does he 1k pot affect the voltages at all? it shouldn't.
I typed the circuit into 5Spice, and got the following voltages, with a 9.5v battery (ie a fresh one)
Base 3.04V Emitter 2.41V Collector 6.25V
Real components will give different values but should be in that ballpark
Another thing to check is your resistor values, are you sure the colour band codes are correct?
120k brown red yellow 6.8k blue grey red 62k unusual value, blue red orange. 56k or 68k are more common.
After those , there'll be another band to show tolerance class.
Measure them on a meter to be sure. Variations of 10 or 20% probably won't stop the circuit from working.
John EDIT another possibility - your voltages are very close to those that you would expect if the 1k pot had a direct route to the wiper of the 5k, instead of through the capacitor.
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Post by flateric on Nov 1, 2008 18:24:28 GMT -5
Appreciate all your input on this guys! rechecked the R values with digital multimeter. All ok. Will also recheck for any tiny shorts between perfboard strips. Just want to clarify on Cap values, 474 means 47pF with 4 noughts, right? ie: 470000pF = 470nF = 0.47uF, have i got that right? 103 = 10,000pF = 10nF = 0.01uF, correct?
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Post by JohnH on Nov 1, 2008 18:31:55 GMT -5
I agree with the cap interpretations good luck
John
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Post by ChrisK on Nov 1, 2008 22:16:00 GMT -5
JohnH,
What beta value did you use for the 2N3904?
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Post by JohnH on Nov 1, 2008 23:36:29 GMT -5
Chris - the 2n3904 was an included selection in the 5Spice program, so I did not have to consider how to model it. But here are the modelling parameters that it uses:
.MODEL Q2N3904 NPN(IS=1.4E-14 BF=300 VAF=100 IKF=0.025 ISE=3E-13 + BR=7.5 RC=2.4 CJE=4.5E-12 TF=4E-10 CJC=3.5E-12 TR=2.1E-8 XTB=1.5 KF=9E-16 )
And I'll admit to understanding almost none of them.
From the model, it seems that about 4.7uA going into the base is resulting in 0.477mA at the collector, so current gain = 101?
John
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Post by flateric on Nov 2, 2008 3:35:02 GMT -5
What's the difference between PNP and NPN semiconductors, in terms of operation and how they need wiring? Do they both achieve similar functions but reversed polarities or something like that?
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Post by newey on Nov 2, 2008 7:31:48 GMT -5
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Post by flateric on Nov 4, 2008 14:55:16 GMT -5
I got hold of a couple of vintage AC128 Germanium transistors and a couple of the more modern equivalents to try in a fuzz circuit. Now I believe these are PNP type, there's a dot by one of the legs on the vintage AC128, but no markings on the more modern ones - it has a top hat with a small protruding tab. Which way round (E,B,C) should I wire these?
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Post by ChrisK on Nov 4, 2008 15:12:58 GMT -5
Search me. (Well, go search the web.)
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Post by flateric on Nov 4, 2008 15:31:00 GMT -5
Spent 40 minutes searching the web with no joy - trying to find which leg next to the dot on the case was C, B or E. So my web search again turns to Guitar Nutz for some definitive information. I can understand you're getting tired of all my questions by now but I live in hope of a helpful answer.
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Post by JohnH on Nov 4, 2008 16:13:25 GMT -5
That is an elusive one, I couldnt find it either (in about 10 minutes). I found one with some numbers but no reference to case style. Usually dozens of copies of datasheets pop up just by entering the part number in google. John EDIT - try this one: sound.westhost.com/trans.htmit says its in a TO-1 case. Also, if you have a cheapo yellow multimeter, or better, you can plug a transistor in and test it with different b, c, e arrangements until you find one that indicates some gain
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Post by flateric on Nov 5, 2008 8:02:15 GMT -5
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Post by ChrisK on Nov 5, 2008 12:41:53 GMT -5
One of the things that can be done to determine two of the leads on a transistor is to measure the diode drop using a multi-meter that has a diode detector (this is usually a position on the selector switch with a diode symbol).
Most silicon diodes will have a drop of about 0.65 VDC and most germanium diodes will have a drop of about 0.35 VDC. The operative word is "about".
You need to know the voltage polarity coming out of your meter when using this function. Use another meter set to 20 VDC (if your meter has a 9 VDC battery) and connect it to your meter while it is set for diode detect.
Measure the polarity of the emitted voltage. Keep track of which test lead is positive.
Once you know this, you can detect diodes such as the base to emitter junction inherent in every bipolar transistor by measuring from lead to lead to lead both ways (this is 6 tests).
Now, you know for relative certain that the third lead (the one that doesn't belong to a diode) is the collector. You also know that the other two comprise the base to emitter junction.
If it's an NPN, the base will be the lead connected to the positive voltage lead from your meter and the emitter will be the lead connected to the negative voltage lead from your meter.
If it's an PNP, the base will be the lead connected to the negative voltage lead from your meter and the emitter will be the lead connected to the positive voltage lead from your meter.
This doesn't tell you everything, however;
if you know the base material polarity (NPN/PNP) it tells you the pinout, and
if you know the pinout (CBE) it tells you the base material polarity.
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Post by flateric on Nov 5, 2008 16:59:20 GMT -5
Thanks for your guidance - glad to say the fuzz pedal is finally working tonight, was about to trash it and throw it in the fire. Now I have to go back to picking apart the treble booster again to try and get this working.
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Post by ChrisK on Nov 5, 2008 19:11:48 GMT -5
My approach is to measure every component before I solder it in.
Ya never know......
The other thing to be aware of are anti-static precautions..
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