Circuit that allows snap of fingers creates equal flash from LED?

[audioguru]

Do the 1M+100K resisters effectively act as enough of an "open" to prevent the 1M+100K resisters from completeing a circuit back to the battery, and draining the battery?

Ohm's Law is the simplest and most important law in electronics. It calculates the current in a resistance as the voltage divided by the resistance. So 4.5V/1.1M= only 4.5 micro-amps which is almost nothing.
The datasheet at www.energizer.com shows that a 186 button battery cell can provide 4.5uA for about 2 years.

what do you need that circuit path for?

Normally it turns on the transistor enough for it to be an amplifier but this circuit is designed so that the transistor is turned off so its current doesn't quickly kill the tiny battery.

In Caster's circuit (that worked but may drain battery?) that same area adds up to 43.7K...so even less...

Caster's circuit had the first transistor with more base voltage so it was more sensitive but it was still turned off. His resistor values were not calculated but were just a bad guess that drew 25.2 times as much current as Colin's resistors.

And, will the battery see the 10K being in parallel with the 1M, cause the 1M+100K path to be even less?

The 10k resistor is not across the battery, it is in series with the mic which is about 3.3k ohms. Then the battery is loaded with 13.3k plus 1.1M.

What is the resistance in the red loop, or can it even be looked at that way?

Don't look at the total resistance. Instead look at the total current.
The 1M and 100k resistors do not turn on the first transistor so it and the 2.2k resistor have no current. Then the second transistor is also turned off and it and the LED also have no current.

**broken link removed**

Why not take the 1M resister out completely?

The 1M and 100k resistors form a voltage divider so that 0.41V is applied to the base of the first transistor. It is turned on when its base voltage is 0.65V. So the mic signal must be 0.65V - 0.41V= 0.24V for the LED to light fully. The output from the electret mic is a peak voltage of only 0.007V when you talk close to it so the signal level to it must be 34 times higher for the LED to light fully. If the 1M resistor is removed then the signal from the mic must be extremely high to light the LED.

 

[colin55]

The circuit does not work very well as you need to whistle very loudly into the microphone to turn the circuit on.
You really can't get a microphone to illuminate a LED with 2 transistors - even when all variations of 2-transistor arrangements are tried.
You need 3 transistors and a super-alpha arrangement.

 

[arizonaguide]

4.5V/1.1M= only 4.5 micro-amps which is almost nothing.

Cool, audioguru. So, I did understand correctly that there will be a very tiny current there (negligable). Thank you for the explainations, and I'll take a day or two to attempt to fully understand the rest of the parts and how they interact.

The 10k resistor is not across the battery, it is in series with the mic which is about 3.3k ohms. Then the battery is loaded with 13.3k plus 1.1M.

But, isn't the blue loop in parallel with the red loop, and each (in effect) across the battery?
In other words, isn't the 10K+3.3K(mic) total (blue loop), in parallel with the 1.1M (red loop), which would result in the normal resting current being drained through a "Parallel Resistances" total (13.3K x 1.1M)/(13.3K + 1.1M)=13.1K ohmns resistance? (which if I did it correctly is .3mA total drain during the "off" time)?

RESTING STATE CURRENTS:
**broken link removed**
Wouldn't the battery see the RED loop resistance in parallel with the Blue loop resistance (when the transisters are both OFF)?
In effect, normal resting current being drained through a "Parallel Resistances" total (13.3K x 1.1M)/(13.3K + 1.1M)=13.1K ohmns resistance?

Or does that capacitor completely change that?
Or, is it the coil in the microphone that changes the formula?
My head hurts already.

This at least has been a learning exercise for me. The time you folks have contributed is MUCH appreciated!

Audioguru, did you say you located a Key Finder circuit that wasn't applicable?
Turn me on to it anyway, as I would like to see how it works?
Did you say you saw it on another site?

Also, what wattage would you guess you have on your iron when you desolder PCB's? (the desoldering iron I got is 40W).
I can adjust this with an external rheostat, correct?

 

[colin55]

Most what you say is correct but there are a few points to clear up.

When we say the 13.1k and 1M1, we are inferring a parallel combination as we are not specifying series or parallel. You need to refer to the diagram to work it out for yourself.

The microphone is not really a resistance. It has a voltage drop of about 200mV to 600mV when operating correctly. In this circuit the load resistance should be 47k.
In this circuit the voltage across the microphone is 900mV and you take this voltage away from 4v5 and work out the current flow.

But this is all a waste of time as the circuit does not work.
It is so poor that it is a waste of time trying it.
You cannot get the tiny amount of energy produced by the voltage fluctuations from the microphone and entering the 10u electrolytic, to produce a very high amplitude on the collector of the first transistor. The second transistor is then required to amplify this voltage and a lot amplify the current to get the LED to turn on. Two transistors are not capable of providing the enormous amount of gain required.
I tried a piezo diaphragm and it did not produce enough output to drive the circuit. You need a piezo or a PIC chip and an electric mic. if you want to take microscopic power as the PIC chip can turn on for a few milliseconds every 200mS and listen for audio.

The electret mic does not contain a coil, just a FET transistor and nothing more (except a cell consisting of a very fine mylar diaphragm that has been permanently charged with a static charge).

 

[audioguru]

Here is a schematic of the Keys finder project that doesn't work.
It uses a peizo transducer as its microphone and beeper.
It uses a Cmos logic inverters IC as a linear amplifier that is powered most of the time so it quickly kills a tiny battery. The piezo transducer resonates at about 4kHz which is much higher than people can whistle. It is very sensitive at its resonant frequency. The highest frequency that I can whistle is 2500Hz but the transducer has poor sensitivity at this "low" frequency.
When it picks up a sound it rectifies it which turns off the preamp and turns on a beeper oscillator to make a beep.

My daughter bought a Keys Finder at The Dollar Store and asked me to fix it because it beeped whenever she talked and when the TV was turned on. Inside it had a black blob custom IC so there was nothing I could do to fix it. I think it had very low current because its battery was tiny.

 

[arizonaguide]

Thanks Guys!

Even though we couldn't make it work, it has been a great exercise in learning for me.
From my questions I'm sure you guys can get an idea of where my knowlege sits right now, and your answers (and circuit attempts) have helped point me in the right direction.

Thanks for the efforts and the instruction!