At work, they are using this structure to protect an analog pin on a micro-controller from overvoltage.
I simulated the circuit and wired it up to test it (physically), but the output is linear in both (which I expected). One engineer at work assures me this works, and that once V5 = V6 + 0.6V, increasing V5 will not increase the output.

If it does work as overvoltage protection, can anyone explain or point to a resource that can help understand it?

 

I don’t get it. Why doesn’t D5 blow up from overcurrent? Why is V6 relevant at all? It won’t matter unless/until the 30V breakdown is passed.

 

It may work for over voltage protection but this method would not be my first choice. Are you aware to capacitance of each of the diodes? they will interact eith the source impedance of the signal source to make a low pass filter. D5 will limit voltage excursions below GND to whatever the forward drop of the diode is, which is a function of current. D6 will limit excursions above one diode drop. Your friend gave you the figure of 0.6V which might apply to a silicon diode, but I would expect a Schottky to be better than that

FWIW, I would use a cheap opamp follower and a series resistor to limit the current into the opamp input. The opamp will also help to overcome the capacitance of the diodes.

In your simulation - do an AC analysis to see what happens to fast signals on their way to the A/D converter.

 

I don’t get it. Why doesn’t D5 blow up from overcurrent? Why is V6 relevant at all? It won’t matter unless/until the 30V breakdown is passed.

Diodes don't blow up in simulation -- only in real life.

 

I don’t get it. Why doesn’t D5 blow up from overcurrent? Why is V6 relevant at all? It won’t matter unless/until the 30V breakdown is passed.

Thanks wayneh

As explained to me, V6 = 5V. When V5, which is a user input, increases to about 5.6V, the diode D6 will start conducting. According to this person at work, once the diode starts conducting, even if V6 is increased, the voltage will remain the same.

D6 doesn't blow up because there is no load or resistor pulling a large current.

 

Diodes don't blow up in simulation -- only in real life.

Lol, yes. I missed the polarity, for protection against negative voltage. I thought it was forward biased. My bad.

 

I use the dual diode clamps quite well, but there is always a series resistor from the input signal to limit the current.
The signals I measure are not fast so the diode capacitance is not a problem. In fact, there is always a cap to 0V for noise filtering anyway.
"V6" sets the +ve max volts to the micro input to V6+D6fwdV.
Likewise, the max -ve volts is 0V + D5fwdV.
This setup works quite well in the real word.

 

Thanks wayneh

As explained to me, V6 = 5V. When V5, which is a user input, increases to about 5.6V, the diode D6 will start conducting. According to this person at work, once the diode starts conducting, even if V6 is increased, the voltage will remain the same.

D6 doesn't blow up because there is no load or resistor pulling a large current.

Yes, I can buy that it works. If the voltage gets too high, the current in D6 will roast it.

 

It may work for over voltage protection but this method would not be my first choice. Are you aware to capacitance of each of the diodes? they will interact eith the source impedance of the signal source to make a low pass filter. D5 will limit voltage excursions below GND to whatever the forward drop of the diode is, which is a function of current. D6 will limit excursions above one diode drop. Your friend gave you the figure of 0.6V which might apply to a silicon diode, but I would expect a Schottky to be better than that

FWIW, I would use a cheap opamp follower and a series resistor to limit the current into the opamp input. The opamp will also help to overcome the capacitance of the diodes.

In your simulation - do an AC analysis to see what happens to fast signals on their way to the A/D converter.

@Papabravo thanks

The circuit doesn't really run at AC, it is only meant to read a voltage from 0 -> 5V, so I'm not too concerned about the frequency response, but I do agree there would be a low frequency filter.

 

@Papabravo thanks

The circuit doesn't really run at AC, it is only meant to read a voltage from 0 -> 5V, so I'm not too concerned about the frequency response, but I do agree there would be a low frequency filter.

The problem with that thinking is that if it can be any value from 0 to 5 then it is NOT DC. If it is allowed to change even at a sub 1 Hz. rate then it becomes AC. for an extreme example have a look at this article about an organ piece that takes over 600 years to play. THAT my dear fellow is really at the low end of the frequency spectrum.

https://en.wikipedia.org/wiki/As_Slow_as_Possible

 

I use the dual diode clamps quite well, but there is always a series resistor from the input signal to limit the current.
The signals I measure are not fast so the diode capacitance is not a problem. In fact, there is always a cap to 0V for noise filtering anyway.
"V6" sets the +ve max volts to the micro input to V6+D6fwdV.
Likewise, the max -ve volts is 0V + D5fwdV.
This setup works quite well in the real word.

@dendad Thanks for including the name "dual diode clamp." I was able to look up and I'm now able to find other resources that are helping paint a better picture.