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Vintage threads from the first ten years

Re: Ultimate DIY debugger award/IT IS OVER!

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5/28/1999 8:27 AM

Re: Ultimate DIY debugger award/IT IS OVER!

The input ballance (and other parts of the circuit) of the Motorola differs a lot from the Texas chip. I think the guilty part is Q18 (on the Motorola), which will conduct when the voltage on the ballance pin is greater than 0.7V, cutting Q12 and Q13, and this will cut the quiescent current on J2 (+ input). J1 (- input) will conduct by Q14 and the opamp will fire to +V.

Summary:

So you can't use a Motorola. The ECG replacement doesn't work either.

Aron used a Texas and it worked. The National worked for me.

Try any other brand at your own risk.

5/28/1999 10:46 AM

I took a look at a National data book. The National chip doesn't have a "protection" transistor like the Q18 in Motorola. The National is different from the Texas in that in the Texas the ballance pins connect to a ~1K resistor to ground and the emitter of a transistor, while in the National they connect to a 100R resistor to ground and a 2K resistor to the emitter of the transistor. So the National works, but maybe the original Texas works better.

5/26/1999 8:22 AM

Re: Ultimate DIY debugger award/No Go

From those voltages, I would replace the IC. If pin 3 is 2.55V, pins 2 and 6 should not be more than 10mv different in amplifier usage. The 0.2V difference is too big.

You see the same problem when the transistor is inserted - pin 3 remains at 2.55 volts and the - input and output jump. While this could be because of a massive imbalance current inserted into the balance/offset node, this looks very suspicious.

I will try to breadboard this thing over the next day or two as time allows.

5/26/1999 8:32 AM

RG I think the difference in voltage is due to the multimeter loading the circuit. The non-inverting pin is biased by a 1M resistor while the inverting pin is biased by a 100K resistor. If the multimeter has 10M impedance voltage at non-inverting pin will be read with a 10% error while voltage at inverting pin will be read with a 1% error and this would give ~0.2V. I've measured 2.16V and 2.37V on my prototype and it WORKS.

I think the problem is the use of a "replacement" IC instead of the original.

5/26/1999 9:53 AM
R.G.	... yeah, could be, maybe...

5/26/1999 11:44 AM
aron	R.G., GFR I tried 3 different TL071acp chips and 2 ECG857M chips. Uhhh, do you happen to know what acp means in this case? I will get a REAL honest to goodness TL081 chip next. BTW: I even decoupled the TONE circuit part and just left the 100K and 10K resistors in the feedback loop part. If anything was a candidate for "this is how you bias this", this circuit is it. Thanks, Aron

5/26/1999 12:03 PM

Hey, you know, this is starting to remind me of the times I'd get a board populated, then turn it over to do debug because you have access to all the wiring on the bottom and then waste an hour debugging because I hadn't put an IC in the socket.

That's clearly not the case here, but are you using a plug-in breadboard? If you are, I bet you plug the IC in the same holes every time when you swap chips, not wanting to move all those other plug in items. I've had occasion to have a breadboard have adjacent-hole shorts and opens where the contacts are slightly worn or oxidized. Is you *breadboard* maybe bad on a hole or two?

Of course if you hand wired your proto, this is clearly not the problem.

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