Input selector using PS gating of opamps? - diyAudio
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Old 22nd February 2004, 06:46 PM   #1
tcpip is offline tcpip  India
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Default Input selector using PS gating of opamps?

I was chatting with one or two friends about this idea, and I thought I'd ask you what you think. Basically, can one build an "analogue switch" by using ordinary transistors (FETs or BJTs or CD4066 even) to switch on and switch off the power supply to opamps?

Essentially, imagine a circuit where there's one dual-opamp, set up as unity gain buffer, for each input. Dual opamps can handle left and right channel in one chip, hence dual. The power supply of each opamp is controlled by two analog switches. These switches can be built using transistors/CD4066 type things.

The purpose of doing all this is to get low-distortion analog switches. It is believed that even the old NE5532 is lower distortion than most FET or MOS analog switches. Hence this line of thinking. The aim here is to pass the audio signal through only good opamps, never through the analog switches themselves. For the moment, assume that the user will be willing to tolerate the sub-second audible glitch/noise when one opamp switches off and another switches on, when the input selector is changed.

Please forgive me if this idea is totally ludicrous. I'm too new to all this to know the difference.

Tarun
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Old 22nd February 2004, 07:00 PM   #2
sreten is offline sreten  United Kingdom
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AFAIK the simple answer is no.

Signals will still get through the feedback resistors of the op-amp,
I'm came across this in an effects pedal, the sound was poor,
I didn't realise the op-amp had gone west as some sound was
still getting through.

sreten.
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Old 22nd February 2004, 07:25 PM   #3
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To some extent, this would depend on if the topology is inverting or non-inverting. A standard inverting circuit would leak, but the non-inverting should be quite OK.

However, how a signal travels through a non-powered OP-amp is most likely undefined, so it's hard to be sure of a blanket statement. Either test it and work from there, or just regard it as unsupported.

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Old 22nd February 2004, 08:10 PM   #4
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Default TI and Intersil

among others have OP-AMPS with "Disable" pins -- I believe that you can search by this parameter in the mfr's database.

a signal will definitely fly through an unpowered opamp, but it won't look so pretty at the output.

I am using some of the OPA3684 current feedback opamp with disable in a DIY phase-meter.
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Old 23rd February 2004, 12:51 AM   #5
tcpip is offline tcpip  India
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Default Thanks, guys

Thanks, guys. What you say makes sense. And it never occurred to me to question what happens to the output of a powered-off opamp if there's signal at the input. I guess I'll have to try it out. Or else I guess I'm back to the standard SSM switches, right?

Tarun
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Old 23rd February 2004, 01:56 AM   #6
mirlo is online now mirlo  United States
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Even if the signal didn't make it through the powered-off op-amp, it would still be like connecting some unbiased PN junctions (ie nonlinear impedances) in parallel with the output. These would cause distortion on the signal that was driven by the one powered op-amp.

Use analog switches or op-amps with disable pins.
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Old 23rd February 2004, 02:07 AM   #7
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not a good design to feed an op amp unpowered sharing a signal
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Old 23rd February 2004, 06:20 AM   #8
azira is offline azira  United States
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So.. instead of gating the power pin, why not short the input to gnd?
For a non-inverting opamp, you can add a series resistance with the source. For an inverting opamp you have a built in to the configuration. You'd still need either an active or passive mixer circuit on the outputs.
I think you could use a tri-state buffer/tranceiver (74LS245 off hand) to short the signals... They should be able to sink the kind of input curren't you're talking about.
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Danny
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Old 23rd February 2004, 12:19 PM   #9
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well, you can't just nilly willy short the input to ground, you have to look at the circuitry ahead of the input opamp and make sure that by doing so you aren't inadvertently lowering the impedance of the next following circuit.

Op amps are so cheap that I am taken to buffering all my differential signals in instrumentation (or just using an instrumentation amp.)

Get some samples from TI - is my suggestion.
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Old 23rd February 2004, 03:51 PM   #10
azira is offline azira  United States
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Quote:
Originally posted by jackinnj
well, you can't just nilly willy short the input to ground, you have to look at the circuitry ahead of the input opamp and make sure that by doing so you aren't inadvertently lowering the impedance of the next following circuit.
Sure, but that's why I mentioned adding some input resistance.

If there isn't an EQ circuit (or even if there is, who cares if the source isn't being used) then a pretty normal configuration is to run the input straight into the opamp. Your input impedance for the preceding device is set by the input impedance to the opamp. For an inverting configuration R1 taps to a virtual ground already. For a non-inverting configuration it's good practice to add an input resistor anyway. You won't short your previous stages output by shunting the inputs to gnd.
If there is other stuff on the same signal path before the opamp, I agree then you should consider the impact.
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