I found the circuit below on the web and thought it might be useful in a project I'm researching and "MIGHT" attempt to create one day.
The circuit, for my purpose, is routing an audio signal to one of 8 different paths. I had a look at the datasheet for the CD4051B Multiplexer and it states "DC Input Voltage -0.5/Vdd+0.5" and "DC Input Current -10/10".
One possible audio source (guitar pedal) operates on 9 volts. So, in theory (my theory anyway), in a bad situation, 9 volts could possibly be transferred through the AUDIO IN and damage the 4051. And, under minimal operating conditions, the source audio voltage can peak above the max 0.5 volts. I measured the voltage coming through my guitar cable and saw 2 peaks of 0.54 volts when strumming hard. Not something I generally do. But it's a concern, nonetheless.
With that in mind, what would be the best way to insure the audio voltage stays in range for this circuit? I have done some research on the subject and have read about 3 different methods. But what I have read thus far, has not given me enough information to say for certain which is best in this case.
The 3 methods are a linear regulator, a switching regulator, and a shunt regulator. Is one of those "best" for this case? Or have I missed a method that would be better?
I'm looking for direction here. Not "Here's this circuit. Fix it for me".
Vdd is the supply voltage for the chip. Inputs cannot exceed 0.5V above that. Usually not a problem.
Inputs also cannot go more than 0.5V below Vss. Audio is an AC signal which usually has equal positive and negative voltage. So your input signal should be measured with the AC scale on your meter. +/-1V is common for "line level" audio. Guitars and microphones use lower levels.
But those ratings on the multiplexer are "absolute maximum". It will have no output for any negative input. So the sound will be destroyed. You need to shift the input up so there is no negative in the signal. Conversely, you can use a negative voltage for Vss, which is why that pin is not labelled "ground".
Voltage regulators are for DC power, not for audio signals or data lines.
These are the "absolute maximum" values and you might get distortion on the negative side of the AC audio signal. Your 0.54V reading is RMS, which is about 0.75V peak (with positive & negative peaks of ~ 0.75V or ~1.5V peak-to-peak).
I seriously doubt you could hurt the chip with a guitar pickup because a pickup can't deliver much current. An active pedal could be more dangerous, but it probably wouldn't kill the chip.
In any case you should bias the input at 2.5V with two equal-value resistors and a capacitor. [u]This page[/u] shows an audio bias circuit. With a guitar you want a higher impedance so use 1M or 2M resistors.* And, with higher value resistors you can use a lower value capacitor. (You can leave out the 47nF cap.)
With the input biased at 2.5V you can handle a signal from +2.5V to -2.5V (5V peak-to-peak). 5V peak-to-peak is about 1.75VRMS.
The guitar signal "sees" these two series resistors in parallel so two 2M resistors makes a 1M load on the guitar (or pedal).
MorganS:
Vdd is the supply voltage for the chip. Inputs cannot exceed 0.5V above that. Usually not a problem.
Inputs also cannot go more than 0.5V below Vss. Audio is an AC signal which usually has equal positive and negative voltage. So your input signal should be measured with the AC scale on your meter. +/-1V is common for "line level" audio. Guitars and microphones use lower levels.
But those ratings on the multiplexer are "absolute maximum". It will have no output for any negative input. So the sound will be destroyed. You need to shift the input up so there is no negative in the signal. Conversely, you can use a negative voltage for Vss, which is why that pin is not labelled "ground".
Wow! I have known what Vdd meant since I started dabbling with electronics nearly 2 years ago. And somehow, my brain just skipped over it and saw +0.5. Just wow. I think I need a beer.
Voltage regulators are for DC power, not for audio signals or data lines.
These are the "absolute maximum" values and you might get distortion on the negative side of the AC audio signal. Your 0.54V reading is RMS, which is about 0.75V peak (with positive & negative peaks of ~ 0.75V or ~1.5V peak-to-peak).
I seriously doubt you could hurt the chip with a guitar pickup because a pickup can't deliver much current. An active pedal could be more dangerous, but it probably wouldn't kill the chip.
In any case you should bias the input at 2.5V with two equal-value resistors and a capacitor. [u]This page[/u] shows an audio bias circuit. With a guitar you want a higher impedance so use 1M or 2M resistors.* And, with higher value resistors you can use a lower value capacitor. (You can leave out the 47nF cap.)
With the input biased at 2.5V you can handle a signal from +2.5V to -2.5V (5V peak-to-peak). 5V peak-to-peak is about 1.75VRMS.
The guitar signal "sees" these two series resistors in parallel so two 2M resistors makes a 1M load on the guitar (or pedal).
Thanks for the links! And as far as current from a guitar goes, I couldn't even get it to register on my multi meter. Made me think I was hooking it up wrong. So, I checked current on a 9v battery. Nope, wasn't hooking it up wrong.
And I will certainly look into biasing. Thanks again.
