Dr. Boogie power shorted somewhere...

[jsandlin0803]

I have built a Dr Boogie. But, I am only reading 0.9 volts between the + and - on the board input, and the battery gets hot.

Tells me there is a short somewhere, but I am not sure where.

Here is a pic of my circuit board...



According to the parts layout page, I needed to bend the pins on the trimpots in order to get them to the correct orientation. Not sure if that would be the issue.

I am getting an ohm reading from the input jack all the way through the output jack, but I cannot find the issue here.


Thanks in advance,

Jason

[R.G.]

In the forum there is a list of sticky posts at the top that you have to scroll past to read the others. One of those is titled "Debugging: what to do when it doesn't work". It has some really good, time tested things to do to get help.

[jsandlin0803]

1.What does it do, not do, and sound like? No Power. Battery gets hot.
2.Name of the circuit = Dr. Boogie
3.Source of the circuit (URL of schematic or project) = http://gaussmarkov.net/wordpress/circuits/dr-boogey/
4.Any modifications to the circuit? Y - I put in and extra footswitch and gain pot to switch between 2 different gain levels.
5.Any parts substitutions? C5 (20pf) replaced with 22pf
6.Positive ground to negative ground conversion? N
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? => 8.3v
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 = 0.88v
Voltage at the circuit board end of the black battery lead = 0.0v

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 - J201 - 0.75v
Q2 - J201 - 0.75v
Q3 - J201 - 0.75v
Q4 - J201 - 0.75v
Q5 - J201 - 0.75v



Thanks for the help guys.


Jason

[R.G.]

Voltage at the circuit board end of the red battery lead = 0.88v

Most likely: D1 is inserted backwards. Check that first.

Q1 - J201 - 0.75v
Q2 - J201 - 0.75v
Q3 - J201 - 0.75v
Q4 - J201 - 0.75v
Q5 - J201 - 0.75v

Are all three pins of all five FETs at the same voltage?

[jsandlin0803]

Only the drain leg on the FET's have 0.75 volts. Other 2 legs on each are 0v.

[jsandlin0803]

On D1, I have the  end with the White stripe going to the ground side of the board.

[slacker]

That's the wrong way round, the stripe should be towards 9 volts.

[R.G.]

On D1, I have the  end with the White stripe going to the ground side of the board.

That is backwards. The non-white-stripe end goes to the ground side. Reverse it.

[jsandlin0803]

Cool. Got D1 switched, now I am getting proper voltage at the board and my LED's are coming on for my footswitches!

You are the man!

I will now need to bias my FET's and see if it produces sound.


Thanks again.


Jason

[twabelljr]

This is a perfect example of how effective the "Debugging" sticky process is.
Enjoy your build!