debugging rat

[sir ladderhole]

Hi, I'm trying to debug my tonepad rat using the voltages from this thread http://www.diystompboxes.com/smfforum/index.php?topic=38828.0
I'm having a problem where the power connects to the board through a 47ohm resistor. With a 9.1v battery, the spot where the battery connects to the resistor is 9.1v but on the other side of the 47ohm resistor its down to about 0.7v. I know this is wrong, because this then connects to pin 7 of the IC which is supposed to have the same voltage as the battery. I've checked and double checked the resistor to make sure its the right value. Can anyone offer any help?

[Processaurus]

The resistor isn't soldered in right (cold solder joint maybe) or is much larger than 47ohms.   Did you measure it (without a battery in there) with the ohm meter?  I'd hook it up at the battery clip + and the other end up at the + supply pin for the opamp, that would tell the story if its really 47 tiny ohms, or if something is open somewhere on that path.

Nothing sounds like its shorted out, because the battery would both get hot, and also not be able to keep up its 9.1v.

What are you going to call yours when its done?  Love those rats.

[sir ladderhole]

I did check with the ohm meter and no battery to make sure it was 47ohms. I will resolder the resistor tomorrow; thank you for the suggestion. As for the shorting out, Im not sure what you mean by the battery getting hot, but I did accidentally touch the bottom of the pcb while testing and it was REALLY hot. I thought it was going to leave a scar. I just figured this was normal, though. Could this be another indication of a problem?

And the name? I havent decided yet. The first decal I made simply says "Rat". Not too exciting, so the name is the next problem I need to work out. 

[Processaurus]

Hmmm, your pcb shouldn't be hot at all, that means something is sucking a lot of power.  Nothing in a battery powered stompbox should do that.  Batteries get hot when they have to work really hard.  I've had them get super hot when I've stuck them in my pocket with a bunch of change that shorted it out.

Measure with the ohmeter from the + to ground on the circuit, to tell if there is an unintentional short from + to ground.  I'd guess it should be a few Kohms normally, once the power supply cap has charged up and the reading stops changing.

[R O Tiree]

The clue is in the fact that you have 0.7V from the resistor to ground. Have a look at the 1N4001 (diode) - My guess is you have it back to front, because 0.7V is a Silicon diode's Vf (give or take).  It's supposed to be reversed biased.

To check this, measure the voltage drop across the 47 ohm resistor. If it's 8.3V or so, then switch that diode around.

One thought occurs... You have got the battery the right way around?  If not, leave that diode alone, as it's done its job.

Next, go back and check the orientaion of every other polarised part before you switch it back on again, ie the IC, the electrolytic caps and the transistors.

I would seriously think about replacing the 47 ohm resistor and the 1N4001 diode.  It's a poor-man's polarity protection.  You did buy a 47R rated for 2W, didn't you?  If not, then you have just stuffed 1.47W through a resistor rated for 0.25W  The diode has also got very hot, which will degrade its performance and ability to withstand future polarity %^&*-ups.

[sir ladderhole]

Ah-ha! The diode is the wrong way around. Since Im going to be making another order at small bear to replace the diode and resistor, should I replace anything else? Could I have blown the transistor or the IC?

And am I right in thinking this diode is the cause of the low power problem AND the heat problem? I measured the resistance from ground to + with my meter, and its a whopping zero. Whoops. Is this the fault of the diode, or should I expect another problem elsewhere?

[R O Tiree]

The way you had it, you were dropping 8.3V across the 47R and 0.7 V across the diode. Therefore, there was only 0.7V across the whole of the rest of the circuit. That's not enough to break anything else. Or even power much.

This is the purpose of this bit of the circuit. With the diode the "right" way round as per the schem (reverse biased) no current can flow through the diode to ground. If memory serves, you only drop about 50mV across the 47R and all the rest goes to power the pedal. If you are silly enough to try to connect the battery the wrong way round, the diode is now forward biased and conducts. This is enough to protect the circuit against inadvertently touching the wrong contacts on the battery for a split second. It is not enough to cope with prolonged reverse polarity. The diode gets hot and so does the resistor. Both end up damaged.  Keep doing it and the damage gets worse. Ultimately one or the other will fail. If the resistor fails to open circuit, for example, all current stops flowing, so no worries. If the diode fails to a short, then you develop the full 9V across the 47R. Again, no worries. The bad one is if the diode fails to open circuit. Then you have 8.95V going backwards across the whole circuit. All the electrolytic caps will be damaged and the opamp will badly overheat.