Question about signal clipping in the following circuit

[boylesg]

It may be simpler and better to just put a resistive divider at the opamp's output. e.g. 10K and 2K will reduce 12V to 2V.

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.

It wont adress the clipping problem if the signal voltage exceeds the supply voltage though.

 

[jasonc2]

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**

 

[boylesg]

Thanks Jasonc2. I understand that explanation fairly well and now see how it relates to the water analogy. I wasn't 'getting' the relative biasing of the diode due to voltage differentials. I was really only considering the absolute biasing of the diode, as in which end is connected to Vcc and which end is connected to GND.

Yet another small knowledge gap is thus filled in and I have another useful tool to use in future with circuit modifications.

So you are saying that my diode protection on the buffer output is ultimately more robust than a resistor voltage divider and I should probably stick with it?

 

[jasonc2]

So you are saying that my diode protection on the buffer output is ultimately more robust than a resistor voltage divider and I should probably stick with it?

The diode output protection is generally more robust but the design of your circuit is such that your output will probably not exceed the op-amps output swing.

The diodes will protect against transient spikes from unknown sources, which the resistors will not do - to me that is important. They will also give you some flexibility to increase the power supply or substitute a different op amp without having to care about its output range / modify the output resistors. Additionally, they will be much less sensitive to input impedance of your sound card (which behaves like a resistor in parallel to the ground resistor of the output divider).

The resistors on the other hand are a bit more straightforward. They will let you utilize the full output range of the op amp which is good because it gives you a better SNR wrt. op amp noise and output offset (but again, sound card ADC resolution and your own design requirements may make this not matter). They will also generally not distort your output beyond any distortion from the op amp's clipping, whereas the diodes start to "compress" the signal a bit near the clipping threshold.

Personally, I guess I'd probably use both the 1/6 divider (to take advantage of full op amp range) as well as TVS diodes maybe at 2.5-4V or so to protect against transients without distorting an in-range signal. Perhaps a small trim pot on the divider that you can use to calibrate to the specific sound card (you wouldn't use this to adjust signal range, it'd be more a set-and-forget kind of thing). I see an output divider and output protection diodes as addressing two different, but related, issues.

Something like:

Input -> Input range divider (w/ adjustable range) -> Op amp input diode clamp -> Op amp w/ 10x gain switch -> Output divider (w/ calibration pot) to scale to soundcard -> Output TVS for soundcard transient protection

It's up to you though. Trade off of complexity vs protection, plus there's more than one way to skin a cat. For me the price of a new sound card greatly outweighs the extra complexity and cost of a few diodes, so I would still feel good about it even if the diodes never spent a microsecond of their life conducting.

Although honestly I'd kind of expect the sound card to have its own input transient protection but absolutely do not quote me on that.

 

[boylesg]

The diode output protection is generally more robust but the design of your circuit is such that your output will probably not exceed the op-amps output swing.

The diodes will protect against transient spikes from unknown sources, which the resistors will not do - to me that is important. They will also give you some flexibility to increase the power supply or substitute a different op amp without having to care about its output range / modify the output resistors. Additionally, they will be much less sensitive to input impedance of your sound card (which behaves like a resistor in parallel to the ground resistor of the output divider).

The resistors on the other hand are a bit more straightforward. They will let you utilize the full output range of the op amp which is good because it gives you a better SNR wrt. op amp noise and output offset (but again, sound card ADC resolution and your own design requirements may make this not matter). They will also generally not distort your output beyond any distortion from the op amp's clipping, whereas the diodes start to "compress" the signal a bit near the clipping threshold.

Personally, I guess I'd probably use both the 1/6 divider (to take advantage of full op amp range) as well as TVS diodes maybe at 2.5-4V or so to protect against transients without distorting an in-range signal. Perhaps a small trim pot on the divider that you can use to calibrate to the specific sound card (you wouldn't use this to adjust signal range, it'd be more a set-and-forget kind of thing). I see an output divider and output protection diodes as addressing two different, but related, issues.

Something like:

Input -> Input range divider (w/ adjustable range) -> Op amp input diode clamp -> Op amp w/ 10x gain switch -> Output divider (w/ calibration pot) to scale to soundcard -> Output TVS for soundcard transient protection

It's up to you though. Trade off of complexity vs protection, plus there's more than one way to skin a cat. For me the price of a new sound card greatly outweighs the extra complexity and cost of a few diodes, so I would still feel good about it even if the diodes never spent a microsecond of their life conducting.

Although honestly I'd kind of expect the sound card to have its own input transient protection but absolutely do not quote me on that.

From what I have read about the subject there is probably very little consistency among different sound cards with respect to input protection and no easy way of knowing with any particular sound card.

But I like your idea of bumping up the diode protection a little and then also adding the voltage divider anyway.

I have also been reading about 555 based boost converters with a view to boosting the 12V input to the 18V max of the opamp, thus alleviating any possible input signal clipping. There is no shortage of inductors on tv circuit boards so I am bound to find one that is about right for this purpose.

Now I understand to some extent what all the jargon about buck and boost is all about.

No shortage of trim pots either. What would you recommend basing the voltage divider on? A fixed 1kR resistor or a fixed 10kR resistor and appropriate trim pot? I suspect the former so as not to increase sound card imput impedance significantly above what the buffer circuit already provides.

 

[godfreyl]

...with a view to boosting the 12V input to the 18V max of the opamp, thus alleviating any possible input signal clipping.

Not worth the hassle, imo. Whatever the supply voltage is, there's a possibility that the input voltage will be higher, so you'll need the protection diodes at the input anyway.

What would you recommend basing the voltage divider on? A fixed 1kR resistor or a fixed 10kR resistor and appropriate trim pot? I suspect the former so as not to increase sound card imput impedance significantly above what the buffer circuit already provides.

I would use something like 10K and 1K or maybe half those values. The first should be fairly high so it's an easy load for the opamp. The second should be fairly low to give a low source impedance for the sound card.

That ratio of 10 to 1 means the maximum input to the sound card is only about +-1V,so you lose some resolution (maybe 1 or 2 bits). I don't think that's a problem though. For high quality sound recording you'd want the best resolution you can get, but for an oscilloscope I don't see any need for more than 10 or 11 bits.

 

[jasonc2]

Out of curiosity, what kind of sound card do you have?

 

[boylesg]

Out of curiosity, what kind of sound card do you have?

On my main PC I have a creative audigy 2 but I am not intending to use this device on it until I know it works properly and wont damage it.

I have put a crappy old Creative Esonique, that I pulled out of a discarded PC years ago, in my other PC and intend to use the buffer circuit on it for a while until I feel it is proven.