How to use full swing of audio signal to drive resistive load?

Sep 20, 2008 12 Replies

Hello all,



I am working on a circuit to detect the presence of an audio signal (from a pro sound card) and, from it, drive a resistive load. This isn't really an amp; I just need to sense a signal, then drive a power transistor into saturation as long as the audio is present.



I have a working prototype that uses an op amp, open-loop configuration, driving a power MOSFET. My load is on the drain side, grounded source. Audio is decoupled through a cap into the non- inverting input; the inverting side is biased to about 0.1 volt to keep the output low when no signal is present.



This all works fine, except that I only get output when the audio signal is in positive polarity, kind of like a class B amp, I guess.



Is there any (simple) way I can get my op amp to output high when the audio swings low?



Thanks!


On a sunny day (Sat, 20 Sep 2008 06:51:52 -0700 (PDT)) it happened "Doug B." wrote in :

google 'opamp full wave rectifier'.

formatting link

View in Courier: +V | [LOAD] | AUDIO>--[C]--+--|+\\ D | | >--[1N4148>]---+----G 0.1V>--------|--|-/ | S | | | [R1] [R2] | | | | GND>---------+--------------------+------+ Even simpler; the MOSFET will have some gate capacitance, so select R2 to discharge it in, say, a few cycles of the lowest frequency out of the sound card. JF

Why not just use a resistor, crap and diode?

The diode converts to dc as you don't care about the actual integrity of the signal, it charges a cap. The cap is what "drives" the fet. When enough cycles pass to charge up the cap then it will turn on the mosfet independent of any local variations in the audio signal.

The only problem is that the cap will stay charged(excluding leaking) but you can discharge through a large resistor. (it has to be large enough so the cap won't discharge too quickly)

This is a quick way but isn't perfect(it isn't accurate but you don't need accuracy)

That\'s a really shitty way of doing it, Jon. ;) JF

(from memory, in the 70's), just d/l the lm3914/5/6 datasheets,from national, full of circuits. They have what you need for signal conditioning, you might only need 2 op amps

martin

=A0 =A0 +V =A0

=A0 =A0 =A0|

=A0 =A0[LOAD]

=A0 =A0 =A0|

=A0 =A0|

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=A0 =A0|

Thanks for the quick reply, but there's something I don't get: I don't see why adding the diode will prop up the op-amp's output voltage when the input swings low. Is it basically intended to force the MOSFET to discharge its gate Q into R2 slowly enough to carry through a couple of input cycles? I tried modeling this in LTSpice and it didn't seem to do anything different from my circuit, sans 4148. Am I missing something here?

B."

Right, thanks. A few more parts than I was hoping to get away with, but I'll give it a try.

hehe, why? because it doesn't involve an op amp?

You are missing the play on words.

hehe... I didn't see it. lol.

It's not only a s***ty way but a nasty way too ;/

What you want is the old analog modem "energy detect". Unfortunately, I can't get any good google hits, so you may need a trip to the library unless you have access to the IEEE online. The energy detect circuit was basically a full wave rectifier (of the audio), a comparator to set the level, and a deglitch (maybe one-shot is a better word) to make sure the signal was present enough so that what was detected was audio and not some random event.

It is slightly more complicated than what you are doing. That is, my descriptions sounds worse than it really is.

No. When the input goes lower than 0.1V, the op-amp\'s bottom output transistor will turn on, draining the charge on the gate. The diode prevents that from happening and forces the gate to discharge through R2. If the gate capacitance is too low for a reasonable value of R2 to work, then add some capacitance in parallel with R2. JF

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