# Voltage Clamping Concern: Zener Diode + Resistor

Hello,

I have an analog signal, from a sensor, between 0 and 5V - which I want to feed into an Arduino Due.

With the 3.3V tolerance, I was thinking of using a resistor-zener circuit to clamp the voltage at 3.3V (and a negative range of -Vf of diode).

I apologize for the crude drawing.

Vsig ----[Resistor]------ Vout
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[Zener, pointing up]
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I am not sure however, what value of the resistor to use. The zener I am interested in using is linked below. Notice the Vf is at 0.9V at 10 mA. Does this mean that in order to ensure proper switching of the diode I have to select the resistor to provide at least 0.9V?

I do prefer to use a passive approach, my space is limited.

Thank you.

The problem with a Zener on an analogue input is that the knee is not very sharp, so it distorts your signal when close to the rail.

Use either an op amp or clamping diodes to the supply.

You need to look into the datasheet and put a resistor which ensures that the current through the zener diode does not exceed its maximum current limits.

You could also try using three diodes in series each with a forward drop of around .5v. That would drop away the excess voltage.

Hi, if you use zener clamping in the way you have described, you will only be measuring input 0 to 3V3, from 3V3 to 5V the input to the analog pin will be clamped.
So with your input signal 0 to 3V3 == AtoD count 0 to 1023, then 3v3 to 5V = 1023
I suggest you put your 0 to 5V through a potential divider first, to convert it down to 0 to 3V3, then add the clamp, this way the clamp only comes into action if your 0 to 5V goes out of range, and the analog input goes higher than 3V3.
You then get input signal 0 to 5V === 0 to 1023
Tom.....

If the analog signal is between 0 and 5 volts, then why don't you make a voltage divider with two resistors with the ratio 1.7 to 3.3. Why the zener diode?

ejo60:
If the analog signal is between 0 and 5 volts, then why don't you make a voltage divider with two resistors with the ratio 1.7 to 3.3.

1.8k and 3.3k (1k8, 3k3 ) being standard values.

ok, so take anything close enough, like 4k7 and 2k2, or take a trimmable potmeter.

You'd need to check the source impedance of the sensor - likely its nice and low, but you
haven't said what the sensor is (if its already a divider of some sort the source impedance
will matter).

ejo60:
OK, so take anything close enough, like 4k7 and 2k2, or take a trimmable potmeter.

The values I suggested are particularly elegant, since they make the value for 5V ever so slightly less than the 3.3V.

Thank you all for your suggestions:

ejo60:
If the analog signal is between 0 and 5 volts, then why don't you make a voltage divider with two resistors with the ratio 1.7 to 3.3. Why the zener diode?

Yes, the analog signal is between 0 and 5 voltage. I do not want to use a voltage divider for 2 reasons. A) I am only interested in values that the sensor gives me which are less that 3V. (Perhaps I should have mentioned this). B) I do not want to divide the signal because I am loosing sensitivity of the sensor for no reason (including A)).

All I need is to clamp the voltage at 3.3V, I understand that I am loosing all the data above 3.3V which I am fine with.

sreedevk:
You need to look into the datasheet and put a resistor which ensures that the current through the zener diode does not exceed its maximum current limits.

Thank you, this is what I needed.

Grumpy_Mike:
The problem with a Zener on an analogue input is that the knee is not very sharp, so it distorts your signal when close to the rail.

Use either an op amp or clamping diodes to the supply.

Can you provide me further information on this?

Can you provide me further information on this?

As a zener approaches its knee ( cut off voltage ) it starts to conduct. This is way before it clamps. This distorts the signal that you are trying to measure.

Grumpy_Mike:

Can you provide me further information on this?

As a zener approaches its knee ( cut off voltage ) it starts to conduct. This is way before it clamps. This distorts the signal that you are trying to measure.

Oh okay thank you. I have never heard the term "knee" makes sense as the cut off voltage.

I have never heard the term "knee" makes sense as the cut off voltage.

See picture.

Yes, the analog signal is between 0 and 5 voltage. I do not want to use a voltage divider for 2 reasons. A) I am only interested in values that the sensor gives me which are less that 3V. (Perhaps I should have mentioned this). B) I do not want to divide the signal because I am loosing sensitivity of the sensor for no reason (including A)).

It's possible to use a single op-amp, with 5V tolerant inputs, rail to rail, single 3.3V supply type. This op-amp could be connected as a voltage follower, so that 0-5V input will output 0-3.3V. The output will be clipped to 3.3V for any value of input from 3.3V-5V.

Another advantage is that this will buffer the signal and maintain the highest accuracy. Note that from the Due's datasheet, Table 46-35, the ADC's source impedance at 12-bit varies from 10K to 274K, depending on the ADC clock frequency.

i'm not sure what you are doing, what is the impedance and why can't you simply let the arduino decide which voltages you like or don't like after the voltage divider? Otherwise use an opamp or a comparator. The zener diode is something you normally use to get a stable reference voltage, but it needs a fixed current to become more stable. If you want a sharp knee then take a schottky diode

I built a zener power supply circuit using 3 zeners to produce a very stable voltage, pointless though may as well use lm317

If I knew how to attach photos I would attach the circuit.

dgelman:
All I need is to clamp the voltage at 3.3V, I understand that I am loosing all the data above 3.3V which I am fine with.

Ah well, it's pretty easy. Go back to Mike's first suggestion - a resistor in series with your source, and a Schottky diode to the 3.3V supply.

Solved!

(Note the Schottky needs to be one with a low reverse leakage.)