OK so it looks as though my hunch was correct. Since the 1N914 break down voltage is way in excess of any likely over voltage to the buffer circuit it must be leakage current through the reverse biased diode that 'carries away' any excess voltage on the input. Correct?
Not quite. It is current through the diode when it is
forward biased that "carries away" excess voltage. During normal operation when the signal voltage is below the +rail voltage, the diode is reverse biased and does not conduct. However, if the signal voltage
exceeds the +rail voltage (plus the diode's forward voltage), the diode is now forward biased, conducts, and pulls the signal back down to the rail.
The key concept is that the diode is forward biased when the signal voltage exceeds the rail voltage (plus the forward voltage) and reverse biased during normal conditions.
And since the resistance to reverse current flow through the diode is considerable to any likely over voltage, the diode acts as a very high value resistor.
Half right. The resistance to reverse current flow is what allows the circuit to function in
normal conditions (i.e. prevents the signal from being pulled to the rail when it is in normal range), just as a low water level in the container will not spontaneously flow up and over the wall.
But it only works relative to a supply voltage.
It works relative to whatever. Simplified: If a diode is forward biased it conducts. If a diode is reverse biased it does not. When the anode voltage is higher than the cathode voltage (plus the required forward voltage) the diode is forward biased, otherwise it is reverse biased. So... keep those basic rules in mind then you can start to see what interesting things diodes can be used for.
It's like a check valve in a pipe, or on an air compressor -- probably a much better analogy than the water container. If the pressure is higher on the entrance side than the exit side (plus any force needed to open the actual valve, analogous to forward voltage drop), then the valve opens and water/air flows through it.
The reverse breakdown voltage spec is an operating limit for the diode. If the relative pressure is exceedingly high on the exit side of a check valve, the valve breaks and material rushes through. Same with a diode. The diode reverse breakdown voltage in your circuit wouldn't come into play unless the supply voltage was so much higher than the signal voltage that the diode could no longer hold up to it.
If a tried the same method on the buffer output I would be no better off than not having the clipping diodes at all and the output is just less than 12V as specified by the original designer.
Your series diodes on the output buffer work. The concept is the same as your supply voltage clamp and follows the same rules as above. In this case you are clamping relative to 0V instead of 12V and are relying on the diode's forward voltage to provide the threshold (just as you're actually clamping your input to 12 + forward voltage, not 12). If the forward voltage is 0.7V then the anode must be at least 0.7V higher than the cathode for the diode to conduct.
If 3 diodes are in series and the cathode is at ground, then when the anode reaches 2.1V the 3 diodes conduct and "carry away" the excess voltage.
Roughly the same thing could be accomplished by connecting the buffer output to a single diode's anode and connecting the cathode to a (2.1V - 0.7V) = 1.4V reference.
Similarly, for diodes with Vf=0.7V the input signal could be clamped at 12.7V by connecting a single diode from signal to +12V, or by connecting ~18 diodes in series from signal to ground:
https://www.falstad.com/circuit/#$+...+64+0+34+37.41444191567111+9.765625E-55+1+-1
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Actually you have a point there. I already know the ouput is always less than 12V from the original designer. So a 1/6 voltage divider would ensure that it is always less than 2V.
I think I might remove the diodes off and do it your way.
If you do this, you may wish to update your op amp gain calculations to make sure you use the full range of your sound card's ADC input as much as possible for the signal input ranges you are designing for, so that you don't unnecessarily lose ADC resolution (if that matters).
E.g. if you want the highest possible resolution for a 5V input signal, and your op-amp's max output is 12V, and you have a 1/6 divider at the output, you'd want your op amp gain to be 12/5=2.4x so that your 0-5 in maps to 0-12 out maps to the full range of 0-2 at the divider.
You can test your soundcard's ADC range (and a lower bound for the max voltage as well) by inputting 0V (I'd go through a 600Ω resistor as discussed in your other thread) and
slowly ramping up until you see the values clip on the PC side.
For the signal input protection, you will want to use diodes. It is the simplest, safest, and most effective way of handling that situation. Clamping is one of the primary applications of diodes.
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BTW I sense that this tool may come in handy for you soon:
**broken link removed**