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I have a circuit where I sense an input which is a voltage divider with resistor.

Since the voltage divider needs to have high impedance, the current flowing through is in the range of 10's of uA.

I need to protect the the analog input so that the voltage doesn't exceed 5V. Maximum range is 200mV (so clamping anywhere between 250mV to 5V is acceptable).

The first choice would be a Zener diode, but the problem is the reverse leakage current of those starts at in the 1uA range, which would induce an error to the measurement.

Maximum allowable current leakage is 25nA.

I thought of using a regular diode in forward mode but they also leak below the conductive threshold.

Is there any other clever way to clamp over voltage while having minimum leakage current ?


EDIT

Is this solution a viable idea ? Transistor have a CE leakage in the nA range, cascading 2 allows to increase the voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

The simulation give a leakage current of 14pA at 200mV at the resistor node.

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  • \$\begingroup\$ What kind of failure do you want to protect against? Someone shorting some >5V power supply to your pin, or someone applying more than xV to the divider so the the divided result is >5V? In the latter case maybe two zeners in series, or multiple normal low leakage ones in series will be ok too \$\endgroup\$
    – PlasmaHH
    Sep 20, 2018 at 7:23
  • \$\begingroup\$ I want to protect if the voltage goes above 5V. 2 zener in series won't reduce the leakage current. \$\endgroup\$
    – Damien
    Sep 20, 2018 at 7:26
  • \$\begingroup\$ You might want to add a target leakage current then; Often you can find better leakage current zeners at a different voltage range, thus combining two can lead to just half the leakage current. \$\endgroup\$
    – PlasmaHH
    Sep 20, 2018 at 7:39
  • \$\begingroup\$ Target leakage is in the 25uA range. 2 zener with 500uA leakage each in serie will still have 500uA leakage. @PlasmaHH \$\endgroup\$
    – Damien
    Sep 20, 2018 at 7:48
  • \$\begingroup\$ quickly searching at digikey most 5ish Zeners I can find have about 20µA leakage, they should be fine then. I could also find 2.5V ones with 10µA. Putting them in series should not increase it to 20µA, should it? \$\endgroup\$
    – PlasmaHH
    Sep 20, 2018 at 7:54

4 Answers 4

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You can use a FET transistor like Siliconix 2N4117A or similar device. This kind of diode has a very low leakage current.

schematic

simulate this circuit – Schematic created using CircuitLab

Then you clamp the input signal

schematic

simulate this circuit

The next circuit is a low leakage clamp + input current limiter. NOTE: The circuit example by Analog Devices says that input transitors (current limit) are depletion mode P-ch JFET, but they depicted them as I did. You have to contact the author of this circuit to get the correct version of it.

schematic

simulate this circuit

Reference:

Analog devices

Burr-Brown

enter image description here

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  • \$\begingroup\$ Thanks that's a very good solution. It also has the advantage to protect in reverse polarity. \$\endgroup\$
    – Damien
    Sep 20, 2018 at 11:23
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The jelly-bean 1N4148 signal diode has only nanoamps of reverse leakage current, so the trick is to make sure it's reverse biased for all signals of interest. D1 is the doing the clamping here, and is suitably reverse biased:

schematic

simulate this circuit – Schematic created using CircuitLab

I use a couple of heavily forward biased diodes to create a reasonably steady +1.3V, which is able to sink any clamping current that might happen when I overdrive IN. You can use this current sink for multiple clamps.

Because I've raised the cathode potential of clamping diode D1, it will be reverse biased for all input signal potentials under +1.3V, and consequently diode leakage will be just a few nanoamps.

You get this kind of clamping behaviour:

enter image description here

For more hardcore clamping action (I mean a flatter response, sharper knee, what did you think I meant?), you can use diode-connected transistors to obtain all the 0.7V drops. Reverse collector-emitter current is still tiny compared to zener leakage. Below, I'll use a single transistor to obtain a 0.7V sink, which will lower the onset of clamping, but is still good for your 200mV signals:

schematic

simulate this circuit

enter image description here

Be careful not to reverse bias the base-emitter junction of Q1, by applying an input less than −4V or so. That could cause the junction to break down and begin conducting.

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  • \$\begingroup\$ Definitely an interesting original approach, thanks for the contribution! \$\endgroup\$
    – Damien
    Apr 3, 2023 at 8:47
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There are plenty of op-amps (that you could use as a unity gain buffer) that specify maximum input currents in the realm of 10 mA. Using one of these means that it will buffer the high impedance node to your measurement system and be able to handle any over current due to the attenuated input voltage rising above the 5 volt level - the op-amp input will act like a clamp.

Alternatively you could use a diode that clamps the attenuated voltage to the 5 volt rail so the maximum that can be achieved is 5.5 volts or thereabouts.

Or maybe if you look at the data sheet of your input device it will tell you how much current can be sunk into the input and this might be enough on its own with no further modification.

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  • \$\begingroup\$ The input is 800Mohm. op-amp induce noise into the circuit, and aren't cheap, I would like to find a simpler solution. \$\endgroup\$
    – Damien
    Sep 20, 2018 at 7:27
  • \$\begingroup\$ Sorry, I don't understand your comment. I also offered three solutions. If you wish you should provide more information into your question rather than scatter gun a comment here and there. \$\endgroup\$
    – Andy aka
    Sep 20, 2018 at 7:31
  • \$\begingroup\$ I need to have high impedance input, so a standard opamp by will introduce an offset error and induce noise. The input is actually an instrumentation op amp with high impedance. The diode solution is part of the question why it cannot be used and the third one is on my reply above. \$\endgroup\$
    – Damien
    Sep 20, 2018 at 7:47
  • \$\begingroup\$ What is the input current rating for the InAmp? Part number please. \$\endgroup\$
    – Andy aka
    Sep 20, 2018 at 9:00
  • \$\begingroup\$ AD8226. <5nA at 1V \$\endgroup\$
    – Damien
    Sep 20, 2018 at 9:08
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If the divider has high enough impedance, then it likely already limits the current, and the input protection diodes of the node that the divider output feeds into will do the job without coming even close to the maximum allowed input current.

The divider you show doesn't have particularly high impedance relative to the output voltage, though, so that won't be enough.

A very dependable clamp - dependable enough to use without sacrificing precious ppms in front of the slope comparators in the dark theme edition of Keysight 3458A - adapted to a smaller output range looks basically as follows, sans the input current limiter:

schematic

simulate this circuit – Schematic created using CircuitLab

The Zener diodes have very low reverse leakage current. Their voltage can be increased if needed, but even at 50C I'd expect the leakage to stay under 20nA. Of course test this on the particular diodes you happen to use.

Rs is the equivalent source impedance of the divider you show in the question.

The output voltage is limited to about +1.7V and -0.5V, with minimum leakage in a practical circuit expected around +0.6V.

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