1) If the voltage divider measures only 3.8V then your 9V battery is almost dead at only 3.8V x 2= 7.6V and its voltage will continue dropping. Maybe the opamp is defective and is drawing a high battery current?
2) If pin 3 reduces the voltage at the end of the 1M resistor to 0V then either the resistor is bad or the pin 3 of the opamp is bad.
Be certain that the multimeter is set to measure voltage, not current or resistance.
Also be certain that the marked pin 1 on the opamp is at the pin 1 on the circuit.
Hi Bertus: Yes I am. I have both kinds, but on this build, I am using the Turned Pin type.
I just built the new voltage divider part of ther circuit. Before the 1M, my volt reading is 4.7. After the 1M, the volt reading is 4.2. I am using a new battery now. I think those readings are good- right?
When your multimeter has an input impedance of 10 MOhms, a voltage divider will be made.
You will measure about 10/11 X 4.7 Volts = 4.27 Volts.
It can be a little lower due to the input impedance of the opamp in parallel with the multimeter.
Thanks Bertus. Yes, I remember that. But the original schematic had a 470nF capacitor- whihc, as AG said, was way too big. So. given that it started with a 470nF, just wondering if it would work with a 15nF even though it may take a little more time that a 10nF. Far better than 470-nF I would assume.
I bought some 1uf capacitors. They only had 50volt ones, so I bought those. I didn’t realize that their cap size is bigger. Aside from the size inconvenience, there is no reason they wouldn’t worth in this circuit right?