Controlling audio equipment

E

Elledan

[H]ard|DCer of the Month - April 2010
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Circuit properties:

- detects whether there is an active (audio) signal (amplitude/mV?)
- after a certain number of seconds/minutes of inactivity, it turns the equipment off
- after detecting activity, it turns the equipment on

The timing part can be done with a simple logic circuit and a counter, so all that is needed is a circuit that sends a (TTL-compatible) signal, e.g., activity = HIGH, inactivity = LOW.

What kind of voltages are we talking about with an amplified audio signal?
 
Use an opamp feeding a capacitor via a diode. The op amp amplifies the signal and isolates it from the rest of the circuit. The diode rectifies the signal so it can charge up a capacitor. The capacitor turns a transistor on pulling in a relay and turning on the rest of your stuff.

The signal could be around 100mV or less for a line level signal or a few volts for speaker level.
 
What would be the typical response time for such a circuit?
 
How about something like this?

audiodet.jpg


R5 is 39K, and the bottom trace wire is signal ground.

I have no idea what R1, R2 and C1 values should be. R1 will be much smaller than R2, in fact R1 could even be eliminated. Wire this up in PSPICE and see what you get...
 
First off, the use of the 2134 seems excessive--the things arent' cheap, and since you're looking for only a binary output anyhow, something like a 1458 (a dual 741) would be more suited, or perhaps a dual version of the LF356...

And beyond that, I don't think it will work as intended. The first op amp in the circuit has its noninverting input biased to half the positive supply voltage, and the diode in the feedback loop will insure that the inverting input will always be at least 0.7 V below the noninverting input... My guess is that PSPICE will show an output that is stuck at the high rail, or as close as the output amp can swing to it...

My suggestion would be to get rid of the resistor to V+ at the input amp, and take the feedback for the first amp straight from the first amp's output to make it a noninverting voltage follower. YOu may need to add a bit of gain if the input signal is small, by adding a pot between the input amp's output and ground, and taking the feedback for the negative input from the pot's wiper. The "time to turn off" for the circuit with a zero input voltage would be related to the product of R1 and C1, and the threshlond voltage determined by the setting of pot R7. BY the way, I"d just connect the two sides of that pot between V+ and GND, and take the signal to the nonverting input of the second amp from the wiper, after getting rid of the connection between wiper and V+.

ANyhow, that's my $.02, from my EE studies... Have fun-- I've wanted to build somethign like this myself BTW...=)
 
Indeed... the 2134 is excessive. I was designing another project in Eagle that uses a 2134, and I just cloned parts from my circuit into a bare area to make the above sketch :D Something like a MC33072 is a lot more appropriate.

The circuit does work. The diode's forward voltage isn't a problem since the diode is contained within the feedback loop of the op-amp; if you think of the diode being a 0.7V voltage source, the effect is that the output voltage of the op-amp will be 0.7V higher than usual, but the voltage at the other end of the diode will be the same as the input voltage. With a "real" diode in place, the behavior is exactly the same except the op-amp can't sink current and discharge the RC circuit on its own.

There's two "states" that happen here:

(1) Input voltage higher than RC circuit voltage - the op-amp will source current and force the RC circuit's voltage to be equal to the input voltage via feedback.

(2) Input voltage lower than RC circuit voltage - this wil force the op-amp's output to V-, but the diode doesn't conduct. The RC circuit discharges itself.

The second op-amp just acts as a comparator.

And the two resistors are in series with the pot, in order to make the adjustment range on the pot sensible. If you're using a 20 turn Bourns you can get away with it, but adjusting a 3/4 turn trimmer connected across a 12V supply to detect the presence of 0.1Vrms audio... that'd be painful.

How's EE going? I graduated just over a year ago.

Aristarchus said:
First off, the use of the 2134 seems excessive--the things arent' cheap, and since you're looking for only a binary output anyhow, something like a 1458 (a dual 741) would be more suited, or perhaps a dual version of the LF356...

And beyond that, I don't think it will work as intended. The first op amp in the circuit has its noninverting input biased to half the positive supply voltage, and the diode in the feedback loop will insure that the inverting input will always be at least 0.7 V below the noninverting input... My guess is that PSPICE will show an output that is stuck at the high rail, or as close as the output amp can swing to it...

My suggestion would be to get rid of the resistor to V+ at the input amp, and take the feedback for the first amp straight from the first amp's output to make it a noninverting voltage follower. YOu may need to add a bit of gain if the input signal is small, by adding a pot between the input amp's output and ground, and taking the feedback for the negative input from the pot's wiper. The "time to turn off" for the circuit with a zero input voltage would be related to the product of R1 and C1, and the threshlond voltage determined by the setting of pot R7. BY the way, I"d just connect the two sides of that pot between V+ and GND, and take the signal to the nonverting input of the second amp from the wiper, after getting rid of the connection between wiper and V+.

ANyhow, that's my $.02, from my EE studies... Have fun-- I've wanted to build somethign like this myself BTW...=)
Click to expand...
 
Okay, now for the big question: why would I want to use the more complex circuit gee suggested instead of chopsuwe's circuit?
 
Because his works. ;)

Mine's roughly the same thing, just the concept idea.
 
gee said:
Indeed... the 2134 is excessive. I was designing another project in Eagle that uses a 2134, and I just cloned parts from my circuit into a bare area to make the above sketch :D Something like a MC33072 is a lot more appropriate.

The circuit does work. The diode's forward voltage isn't a problem since the diode is contained within the feedback loop of the op-amp; if you think of the diode being a 0.7V voltage source, the effect is that the output voltage of the op-amp will be 0.7V higher than usual, but the voltage at the other end of the diode will be the same as the input voltage. With a "real" diode in place, the behavior is exactly the same except the op-amp can't sink current and discharge the RC circuit on its own.

There's two "states" that happen here:

(1) Input voltage higher than RC circuit voltage - the op-amp will source current and force the RC circuit's voltage to be equal to the input voltage via feedback.

(2) Input voltage lower than RC circuit voltage - this wil force the op-amp's output to V-, but the diode doesn't conduct. The RC circuit discharges itself.

The second op-amp just acts as a comparator.

And the two resistors are in series with the pot, in order to make the adjustment range on the pot sensible. If you're using a 20 turn Bourns you can get away with it, but adjusting a 3/4 turn trimmer connected across a 12V supply to detect the presence of 0.1Vrms audio... that'd be painful.

How's EE going? I graduated just over a year ago.
Click to expand...


Yeah, you're right about the input amp thing... Sometimes the critical thinking skills arent' all that swift when you're got an adult beverage in hand, and it's late;) But I'm still not fond of biasing the noninverting input to half V+... But now that you point out my stupid, I do believe that your circuit will work, although to maintain my pride, I still think that my design approach has merit, particularly in that adding gain in the input stage would precipitate a more consistent turnoff time, that's less dependant on input signal amplitude...

EE is going well. Am working on a MS, after graduating with a BS in ME a year or so ago. Am currently developing an ambulatory EEG system for use in brain-computer interface studies. =)
 
Aristarchus said:
Yeah, you're right about the input amp thing... Sometimes the critical thinking skills arent' all that swift when you're got an adult beverage in hand, and it's late;) But I'm still not fond of biasing the noninverting input to half V+... But now that you point out my stupid, I do believe that your circuit will work, although to maintain my pride, I still think that my design approach has merit, particularly in that adding gain in the input stage would precipitate a more consistent turnoff time, that's less dependant on input signal amplitude...

EE is going well. Am working on a MS, after graduating with a BS in ME a year or so ago. Am currently developing an ambulatory EEG system for use in brain-computer interface studies. =)
Click to expand...

The input is AC coupled to +V/2 for a reason; if you didn't do this, the circuit would need to share a ground with the audio chain, and the op-amp would require a negative supply rail. This circuit works off a single rail.

You're correct that the turn-off time is dependent on the signal amplitude. Probably the best thing to do run the first op-amp open loop - keep the diode and RC present, but disconnect the inverting input of the op-amp from the RC circuit and add a 2nd pot which divides down V+ to 'slightly above' V/2 - this is your "input sensitivity" adjustment.

This way the RC circuit is brought up to V+ every time there's an input signal preset, and the pot feeding the 2nd op-amp will adjust the turn-off delay.
 
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