Please excuse the crude drawing. One, I don't know if this can even work in theory. If it can work then I'm having trouble figuring out what values for the components. Any help would be appreciated.

B2 is a momentary switch the normally closed.

What I want this circuit to do is briefly behave like a short when the button is pressed but then return to being able to use the microphone. When the button is released I want it to again act like a short and then, again, return to being able to use the mic. Each time it acts like a short I want it to be in this state for about .1 second or so. Perhaps .2 second to return to normal.

To describe that I want it's probably easier to describe what I hope this circuit does....

1. When the circuit is connected to the 1.8V power, nothing will happen with the transistor and all power goes through the MIC.

2. When the button is pressed the, current will pass though C1 until is charged and then C1 will make that part act like a open circuit an all current will pass though the MIC. This will act like a short for a very brief period. I want this to happen and I want it to take about .2 seconds to go from closed circuit to open.

3. When the button is released the capacitor creates a voltable drop across the base of the transistor (switching it on). Power then flow though the transistor until the voltable across the base drops below .7 volts at which point the transistor turns off. I want the transistor to stay about about .2 seconds.

I'm I out of my mind with this design?

 

The base of the transistor is floating when the switch is open. (as it is drawn now)
The transistor is inverted, and if it conducted it would not only short the microphone but also the power supply. Doesn`t this microphone need a resistor in series? You would then only short directly the microphone and not the power supply...

Is the purppose just to disable the signal coming from the microphone? There are other means to do this. How does the rest of the circuit look like?

 

I think you're looking for something like this:

The circle with the "X" in it is supposed to be your electret mic.

The 100 Ohm resistor is so that you don't burn the switch contacts up when you push the button.

The 1k/1MEG resistors on the base help to both control the ON time, and to help ensure that it switches off fairly decisively. Decreasing the 1MEG resistor will make the mic turn off more quickly.

 

I was wondering if I would ever have to use a Darlington pair for something. I guess I should go read up on how they work. Thanks for the information. I'll try to mock it up and see how it does.

 

The only reason I used a Darlington configuration was to increase the Vbe (voltage at the base using the emitter as the ground reference) at which the transistor would go into cutoff.

 

So let me describe how I think your circuit will work to make sure I understand it.

1. Before the button is pressed there is no voltage drop across the Darlington and the circuit draws no current through the microphone.
2. When the button is pressed (and held down) the lowest resistance path is through the capacitor, after it charges the Darlington pairs are activated and current is now flowing through the microphone.
3. When the button is released the capacitor discharges and keeps the transistors 'on' for a short time. Then they switch 'off' and the microphone no longer draws current.

If this is correct then there are a couple things I wanted to accomplish that I was hoping my crude drawing accomplished.

1. I want the microphone to be always on. Both when the button is up and when it is down.
2. I want the micophone to be very briefly shorted upon the action of the button being pressed and again when the button is released.

I'll see if I can use your circuit and make some changes. These Darlington things are pretty cool.

 

1. Before the button is pressed there is no voltage drop across the Darlington and the circuit draws no current through the microphone.

The first part is wrong, but the 2nd is OK.
When the transistor is off, ALL of the voltage will be dropped across it, because for all practical purposes, it is an open circuit.

2. When the button is pressed (and held down) the lowest resistance path is through the capacitor, after it charges the Darlington pairs are activated and current is now flowing through the microphone.

That's sort of correct.
The 100 Ohm resistor and cap in series are indeed the lowest path of resistance when the switch is first depressed. Once the voltage across the cap and resistor exceeds the cutoff voltage of the Darlington (somewhere between 0.1v and 1.2v), the Darlington's collector will begin sinking current from the microphone. C1 charges quite rapidly, as it is fairly small and so is R1. R1 limits the maximum current flow to Vcc/100 = 9/100 = 90mA when the switch is first pressed.

3-. When the button is released the capacitor discharges and keeps the transistors 'on' for a short time. Then they switch 'off' and the microphone no longer draws current.

That is correct.

If this is correct then there are a couple things I wanted to accomplish that I was hoping my crude drawing accomplished.

1. I want the microphone to be always on. Both when the button is up and when it is down.

2. I want the microphone to be very briefly shorted upon the action of the button being pressed and again when the button is released.

If you short across the microphone, that will remove most of the power from it, and it will basically turn off. It also means that the horizontal 1k resistor on top would have 9v less the Vce of the Darlington dropped across it; 9mA current.

I'll see if I can use your circuit and make some changes. These Darlington things are pretty cool.

Darlingtons are kind of neat because the gains of the transistors are multiplied. So, if the upper transistor has a gain of 80 and the lower has a gain of 30, you have a net gain of 80 * 30 = 2,400.

The disadvantage is the rather high saturation voltage, or Vce(sat), which is the voltage on the collector measured from the emitter with a collector current, when increasing the base current will not appreciably lower the Vce.

Depending on the collector current, the Vce(sat) could be anywhere from ~0.7v to ~1.8v or so; even higher. That represents a significant power loss.

Enhancement mode MOSFETs have largely replaced Darlington transistors for low voltage circuits. Rather than current amplifiers, they are more of a voltage controlled switch.

 

Interesting. It seems that every time I think I have enough information more stuff comes along! This is indeed a low voltage/ low power circuit so I guess I have some reading to do for "Enhancement mode MOSFET's".

 

Are you simply wanting to turn off the microphones' output when activated?

If so, then instead of using transistors to cut the power to the mic, you might use something like a CMOS CD4066 IC to turn it on and off - with a couple resistors, caps and a switch. That would still be inexpensive, but require much less current than the transistor method.
TI's information page for the CD4066:
http://www.ti.com/product/cd4066b

[eta]
There are also smaller packages than the CD4066; here is a dual CMOS bilateral switch in an 8-pin SMT package:
http://www.toshiba.com/taec/components2/Datasheet_Sync//165/7440.pdf
Available here: http://parts.digikey.com/1/parts/530980-ic-bilateral-switch-dual-sm8-tc7w66fu-te12l-f.html

 

I'll look the CMOS CD4066 IC. This device I'm trying to build is for a cell phone. The headset on cell phone has a Play/Pause button. All that button does is short the microphone. When this happens a phone fires a "button down" event which an app can act upon. When the button is released it recognized that the short is no longer across the microphone and the phone fires a "button up" event. I want to take advantage of this event handling capability of the phone but in addition I want the microphone to be active even while the button is held down.

To accomplish this I'm trying to come up with a circuit which will briefly short the microphone when the button is pressed, and then do it again when the button is released. I'll have to determine what is "brief" means by experimenting with what the phone recognizes.

 

OK, in that case - you might look at the 2N7000/2N7002 N-ch enhancement mode MOSFET. The 1st is in TO-92 package, the 2nd is surface mount. They have limited sink current, but depending on what the resistor is on the high side of the mic, it could work.

But, you will need to find out whether the microphone's output is on the high or the low side, how much current does it take to short it out, and which power rail is closer when it's shorted.

 

So I've tried the Darlington pair and some other slight variations. I've tried several different resister and capacitor values. I can almost get it to do what I want if I use a huge capacitor. The problem is that I think I need the microphone to be in a shorted state for a considerable amount time (both when the button is pressed, and upon it's release) I'm guessing about .1 seconds. Now that I've been playing with this I'm pretty sure I need to go in another direction.

Thanks everyone for help on this.

 

Why don't you post a schematic of where you are now?

 

The signal output from an electret mic is 5mV to 20mV. But shorting and unshorting it gives an output BANG! of a few volts. You don't want that.

Maybe you could make a circuit that ramps into and away from a short.