Galvanic Skin resistance Meter

This is a circuit for measuring the skin resistance of a person. The resistance is converted into voltage and fed to a microcontroller.

The two dots represent the electrodes which the subject will touch with his fingers. The voltage follower on the right is used as an impedance transformer. Now my question is can anyone explain what will be the equivalent circuit of the voltage divider on the left? The capacitor is used as a low pass filter but i just cant understand the circuit operation. Can anyone please help.

It is simply a potential divider and unity gain buffer amplifier. The voltage at the output of the amplifier is the same as at pin 5 but drawing current from it will have less effect than if you measured pin 5 directly.

The two resistors normally set their center voltage to half the supply so you get 2.5V a the output. When skin resistance is in parallel with the lower resistor, it's value is reduced and the voltage drops. Use Ohms law to calculate the effective value of the lower resistor, (1M * Sr) / (1M + Sr) where Sr is the skin resistance.

Brian.

betwixt said:

It is simply a potential divider and unity gain buffer amplifier. The voltage at the output of the amplifier is the same as at pin 5 but drawing current from it will have less effect than if you measured pin 5 directly.

The two resistors normally set their center voltage to half the supply so you get 2.5V a the output. When skin resistance is in parallel with the lower resistor, it's value is reduced and the voltage drops. Use Ohms law to calculate the effective value of the lower resistor, (1M * Sr) / (1M + Sr) where Sr is the skin resistance.

Brian.

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Thanks Brian for the explanation i understood the resistive voltage divider part and the unity gain buffer but what is the capacitor doing here? How is it affecting the voltage input to the buffer? Please give detailed explanations and equations regarding the role of the capacitor. Thanks in advance.

PS: Please note that skin resistance is a varying quantity and will vary depending on the presence or absence of sweat on the body. And this variation has a low frequency. I want to know the filtering action of the capacitor here.

The capacitor is to try to remove mains hum, I would think. Otherwise there would be a lot of 50 or 60Hz on the output.

Keith

keith1200rs said:

The capacitor is to try to remove mains hum, I would think. Otherwise there would be a lot of 50 or 60Hz on the output.

Keith

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I understand that it is acting as a filter but i cant understand the mechanism. Can you please explain that to me. The upper half has a resistance and a capacitive reactance in parallel and the lower half has a fixed 1M resistance and a variable skin resistance in parallel. Up to this i understand. Now how do i derive the voltage equation and the cut off frequency?

The cutoff frequency will vary with the skin resistance so it is rather crude and probably not very effective. The resistance to calculate the cutoff will be the parallel combination of the two 1M resistors and the skin resistance.

Keith

keith1200rs said:

The cutoff frequency will vary with the skin resistance so it is rather crude and probably not very effective. The resistance to calculate the cutoff will be the parallel combination of the two 1M resistors and the skin resistance.

Keith

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Sorry Keith i don't understand. Actually im a newbie to circuit designing. I can't visualize the low pass filter from the circuit diagram. I mean where should i assume the noise source to visualize the filter? The noise source should be below the skin resistance right to form the RC low pass filter? But if that is true shouldn't the capacitor be connected to ground instead of +5V?

It can be connected to either +5 or ground. For sake of analysis and hopefully in real life, the ground and supply are shorted to AC by the supply decoupling capacitors.
If you assume +5 has zero impedance to ground, it makes no difference which way it is connected. As Keith pointed out, by far the greatest cause of inaccuracy is pick-up from AC mains power sources so the time constant should be very long to avoid 50/60Hz being picked up. As skin resistance change will be relatively slow, you can afford to use a capacitor which makes the voltage change slow as well and at the same time give adequate LF rejection. The actual time constant ranges from being set by 0.5M and 0.1uF to something shorter as the skin resistance drops the effective value of the resistors.

Brian.

betwixt said:

It can be connected to either +5 or ground. For sake of analysis and hopefully in real life, the ground and supply are shorted to AC by the supply decoupling capacitors.
If you assume +5 has zero impedance to ground, it makes no difference which way it is connected.

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Sorry i didn't understand this part. Can you please explain it again. Im a newbie.

To AC signals, the 5V and ground (0V) are connected. So a capacitor from the opamp input to 5V will have the same effect as a capacitor from the opamp input to ground.

Keith

Perhaps the concept of 'impedance' is the stumbling block.

The ground and supply are obviously not joined directly as that would short out the power source. But that only applies when you think in terms of DC. From a signal (AC) point of view, if there is no signal on +5 it looks exactly the same as ground. In good circuit design there is never signal on the supply line, it is rock steady so the capacitor will charge and discharge in exactly the same way if connected to ground or supply lines.

Brian.

Hey thanks guys. I understand now. +1 to both Keith and Brian.