HELP Debugging Boost.............

Started by Stu Diddly, December 21, 2007, 07:00:16 PM

Previous topic - Next topic

Stu Diddly

Hey Everyone, 

I just etched my first PCB and built the GUS NPN Boost.  I've run into an issue and I want to see if anyone can help.  I have researched my issue and can't find a solution.  Please let me know if you can help.  I have followed the "Debugging Procedure" and I hope that I've included enough info.

Here is the checklist to fill out:
1.When I plug the pedal in, I get sound in Bypass mode, but when I engaged the effect i get no sound except when I strum the strings really hard and then I get a jolt of distortion. I'm not getting any other sound except when I strum hard.

2.Gus NPN Boost Pedal (Begginners Project)

3.

4.Any modifications to the circuit? No

5.Any parts substitutions? No part subs. 

6.Positive ground to negative ground conversion? No

7.Turn your meter on, set it to the 10V or 20V scale. Remove the battery from the battery clip. Probe the battery terminals with the meter leads before putting it in the clip. What is the out of circuit battery voltage? Battery measures 8.1v
Now insert the battery into the clip. If your effect is wired so that a plug must be in the input or output jack to turn the battery power on, insert one end of a cord into that jack. Connect the negative/black meter lead to signal ground by clipping the negative/black lead to the outer sleeve of the input or output jack, whichever does not have a plug in it. With the negative lead on signal ground, measure the following:
Voltage at the circuit board end of the red battery lead = 8.1v


Now, using the original schematic as a reference for which part is which (that is, which transistor is Q1, Q2, etc. and which IC is IC1, IC2, C1, and so on) measure and list the voltage on each pin of every transistor and IC. Just keep the black lead on ground, and touch the pointed end of the red probe to each one in turn. Report the voltages as follows:

Q1 (2N 4401)- Transistor
C = 0.8v
B = 1.4v
E = 0.7v

Here is a few pics that I took.







Can someone help?

SISKO

Your transistor may be broken. Try replacing it. Also try to solder it quickly as it burns preety fast.

OR, your 10k res (from the collector) its damaged or wrong valued. Measure it

Luck

--Is there any body out there??--

PerroGrande

Yeah -- your transistor voltages are a bit off from what I would expect. 

On this circuit, I'd expect to see the collector close to the middle of the power supply voltage (4.5v or so from a fresh battery).

I'd expect to see:

C: ~4.6v
B: ~2.8v
E: ~2.2v

Your PCB looks good in terms of trace quality, etc (I didn't verify it to be correct in terms of the actual connections).  You don't appear to have solder bridges.

So I would double-check the resistor values (just in case), verify that the transistor is oriented correctly (correct pinout, etc), and then swap the transistor if those other items are correct.  Sisko is right -- transistor cannot stand long-term soldering heat, so work fairly quickly when connecting it.

Stu Diddly

Thanks Guys! I just de-solder the transistor I had in and I solder in a socket so that I could easily change out transistors if need be.  I just inserted a new 2N2222A transistor and measured the voltages and I came up with the same reading as the other transistor.

C: ~0.8v
B: ~1.4v
E: ~0.7v

I also double checked my resistor values and even replaced the 10k resistor from the collector and I still came up with the same readings.  This circuit is beginning to bum me out. I first built the circuit on perfboard and ran into this issues and I was hoping that when i switched it all to pcb, that I would fix this problem, but I guess not.

Do you guys have any more suggestions? I would almost like just to send my circuit to someone to see if they could troubleshoot it.

Thanks for all the help<

Shaun

PerroGrande

It kinda looks like there might be a solder bridge (upon closer inspection) between the "in" and "out".


PerroGrande

Oh -- and take voltages with the transistor removed from the circuit and post, please.

SISKO

Ok, this circuit tired me out.
Sadly i dont have my simulation progam here, but after the changes youve made, its seems obvious (for me, at least) that the transistor is saturated.

So if you want a quick, dirty and cheap solution. Replace de  10k base res whit a pot wired as a variable resistance (just one outer lug and the centre one) and start variyng the resistance untill you see 4v - 5v on the collector.

Again, its not even close to a "clean" soluttion, but that is wahat i would do if i were you.
--Is there any body out there??--

Gus

Like John posted

  Remove the transistor and take voltage readings at the collector base and emitter points referenced to ground

With no transistor and power to the circuit

The 10k collector resistor should have the battery voltage on both sides because of no current being drawn
If it is still .8VDC then there is another path to ground that should not be there.

The base should be a little less than 1/3 the battery voltage.

The emitter should be close to 0 VDC


PerroGrande

Hi Sisko,

I don't disagree with the saturation comment.  This circuit should not be anywhere close to saturation with no input, however.  The bias points are correct and stable, and values could move a fair amount without driving the thing to saturation.  I'm just not sure messing with bias voltages and the like is what will get this thing running.

The voltage divider is going to produce a base bias of 2.8 volts (give or take).  In turn, the emitter should be around 2.2v.  This gives us about .44mA of emitter current with the 5K resistance the pot shows at DC.  The 10K resistor sets the collector voltage of around 4.6v (.44mA * 10K creates a 4.4v drop).

So -- I guess my point is that this circuit isn't "on the edge" by any sense of the imagination.  Ergo, something other than minor biasing errors is causing the saturation condition - such as oscillation, wiring error, solder blob, major component value error -- wrong component installed, etc.





Stu Diddly

Quote from: PerroGrande on December 22, 2007, 12:26:07 PM
Oh -- and take voltages with the transistor removed from the circuit and post, please.

I just removed the transistor from the circuit and these are the voltages that I came up with.

C: ~8.2v
B: ~2.5v
E: ~0.0v

Quote from: PerroGrande on December 22, 2007, 12:20:14 PM
It kinda looks like there might be a solder bridge (upon closer inspection) between the "in" and "out".

There is NO solder bridge between the "In" and "Out." The picture looks as thought there is, but there's not.

PerroGrande

Good!  The voltage divider is working properly with nothing there to load it down.  And the lack of a solder bridge is also a good thing.

Apart from a bad transistor, there aren't a whole lot of other possibilities. Of course, verify the proper orientation of the polarized capacitors.  I'm going to look at your PCB design (fighting my mild dyslexia) and see if I can note any trace errors.

slacker

Your voltages seem to match what Gus posted. The layout looks ok to me.

I'm not 100% sure but from your pictures I think the transistor is the wrong way round.

If you're using a 2N4401 in this picture then the curved side should be facing towards you, with the flat side towards the wires. If you're in any doubt check the datasheet for your transistors.

http://www.alldatasheet.com/view.jsp?Searchword=2N4401


Stu Diddly

I've just recorded some audio of what this thing is doing.  I'm sorry for the out of tune guitar.  I first play in bypass mode and then I engaged the effect and then back to bypass mode.

http://www.juliandrive.com/diy/Gus%20Boost%20MP3.mp3

Check it out and let me know what you think.

Stu Diddly

Ok, so I changed transistors again and put a NEW 2N2222A in and I think that did the trick.  I measured my transistor voltages again and this is what I came up with.

C: ~4.6v
B: ~0.9v
E: ~0.3v

Does this look right?

PerroGrande

Well, the voltages look better -- especially the collector one.  The base/emitter voltages are lower than expected.

Did you have the input shorted (tied to ground) when you took this measurement?

Stu Diddly

Quote from: PerroGrande on December 22, 2007, 02:53:09 PM
Well, the voltages look better -- especially the collector one.  The base/emitter voltages are lower than expected.

Did you have the input shorted (tied to ground) when you took this measurement?

Input shorted?  I had a 1/4" cable plugged in the input and I touched the black lead of my meter to the ground on the input and then moved the red lead of my meter to test the voltages of the transistor leads.

PerroGrande

Generally, you want to take voltage readings with the circuit "quiet" (quiescent is the term used in Electronics).  This is best accomplished by taking the "input" line (to the pedal) and grounding it.  That way, the pedal isn't trying to amplify noise, etc -- which could throw off your readings a bit.