DAQ input protection (overvoltage & overcurrent) question

[ricardosilva]

I apreciate the reply. As I said I can only test the circuit on Monday, I will tell then if it fits my needs.

The DAQ board will be a simultaneous acquisition card, with 4 channels each one wwith it´s own AdC, much like the NI USB9215. So I think that there won´t be any problem with the discharge of the ADC input capacitance

[ricardosilva]

Hello again

At the moment I don´t have access to the components, so I simulated the circuit in Multisim.

I have a small problem, I simulated the circuit that Lynn suggested, but there is a voltage drop with the diodes connected to GND when the input voltage goes above 10V, for smaller voltages the circuit works well. If I put +-20V in the input, there should be +-5V at the diodes terminals,after the voltage divider but there is 1V less, could it be the effect of leakage current of diodes like Mark suggested. If I disconnect the diodes from ground, the circuit works as I pretend, and the effect of those leakage currents are not noticed.

I´ve attached the multisim file and a picture of want happens with the different circuits.

[ricardosilva]

More updates, I have tried with a DC voltage and it worked, with the diodes connected to GND

[johnsold]

I do not have multisim.

Certainly leakage currents can be a problem.  Try with a higher zener voltage, say 15-16 V. This should still protect your amplifier.

I could not find a good data sheet for your diode.  I looked at the 1N5235, 6.8 V, 0.5 W.  The curve only goes down to 10 uA and that occurs at about 6.2 V, so it is possible that it could draw something near 1 uA at 5 V.

After you get a circuit which works at 20 V, try connecting an additional 230 V AC source at the positive end of your signal source with the other end of the 230 V source at ground.

Lynn

[ricardosilva]

The diode was only an example, I´ve been trying with several diodes, from the database to see which works best.

On the simulations, it does not appear that the leakage current has any effect, if I put a +-20V input, at the diodes there will be +-5V on the output, only if I connect the diodes to GND, it appears not working.

I have tried adding a DC component to the AC wave, and the circuit with the diodes connected to GND works. If I connect the negative end of the power supply to the circuit, it only works when removing the GND connection from the diodes.

I have used a 10V DC + 10V AC, to test if the output was what I pretend. With the 230V AC source like you said, the diode limits the voltage.

[johnsold]

Can your simulator report the current in the diodes?  You might need to put a 1 ohm resistor in series with the diode and monitor the voltage drop across it.  The voltage will only be microvolts, but simulators do not mind.  I am interested in the difference of the currents between the grounded and ungrounded cases.

Lynn

[johnsold]

Here is why the diode conducts.  I re-drew the circuit to emphasize the loop current.  The calculated currents are based on the resistor values you used.  All calculated voltages and currents are those which would occur if the diodes were disconnected.  For the floating voltage source the ground has no effect on the current but it does affect the voltage magnitude and polarity on the diodes.

Lynn

[ricardosilva]

Using the first circuit with a DC voltage source=20V

Isource=8.31uA

Idiodes=6.44uA

Voltage drop = V+-V- = 2,76-(-0.620)=3.38V

I was obtaining this value with an AC voltage source and the diodes connected to GND, I suspect that those 620 mV corresponds to the voltage drop of a direct polarized diode, that will consume some current.

In the second circuit without the diodes referenced to GND

Isource=7.5uA

Idiodes=9pA

The diode model that I´m using is the 1N5235.

As you said in the second case the diodes are not conducting (they have a small leakage current 9 pA, provoking a small voltage drop across the 1Mohm resistor, not important). With this circuit there is the corret voltage drop after the diodes.

 Voltage drop = V+-V- = 2,5-(-2,5)=5V

I think that I will test the second circuit, I will try to search for some zenner to test the circuit, even if it isn't the 6.8V.

Another thing that I noted is that the diode has an built in capacitance that will affect the frequency at the input, and it will form a pole with the input resistance. The DAQ will have a sampling rate of 100kHz, but must allow also some higher frequency signal, because of the undersampling techniques (500kHz tops because of the ADC used). So I have to balance the RC values so that the cuttoff frequency doesn´t affect the signal amplitude and most importantly the signals phase. In this case I have a C=1pF and a R=1Mohm, giving a pole at ~160kHz. 

Shouldn´t I lower the input resistance, allowing more current into the circuit, or just limit the maximum input frequency.

[ricardosilva]

Sorry, the resistive divider give us a equivalent resistance of 250kohm, so the frequency is ~640kHz.

Should I still lower the resistances values, and if I can, by how much? until the current limit?

[ricardosilva]

Sorry for the late update

I´ve tested the proposed circuit, with the diodes. The circuit works with the 1/4 resistive divider and the 5.1 zener diodes, conected as in the atachment. DO NOT CONNECT the GND between the 2 diodes, or else the negative part of the input signal will be cutted, the diode will always conduct.