Precise Voltage Shifting

This circuit below is of a milliOhmeter. After the Resistance under test, a high pass filter is implemented which servers as a voltage shifter.

What are the solutions for shifting the square wave without applying high pass filtering?



http://cappels.org/dproj/dlmom/dlmom.html

It is possible to shift the square wave a specified amount by using an op-amp and some resistors. But that would not be of any use in this milliohm meter. That is because we don't know a-priori exactly what voltage shift is needed. The goal is to get the square wave to be symmetrical around 1.8 volts, going just as far above 1.8 as it goes below 1.8. A high-pass filter does that very well and very economically.

However there is something that bothers me about this circuit. It looks like it is an attempt to do a classic 4-point resistance measurement. That requires 2 contacts to carry the current to the resistor under test, and 2 other contacts to measure the voltage across the resistor. This removes the contact resistance as a factor because the voltage measuring contacts have essentially no current flowing through them. No current means no offset voltage due to those contacts. But this circuit violates that principle. It shows the bottom current-carrying contact from the resistor going to the ground symbol. It also shows the lower of the two voltage-measuring contacts going the same ground symbol. Therefore there is nothing to prevent the current from the resistor from going through the voltage-measuring contact, especially if that contact resistance was less than the resistance in the current-carrying contact - thus inducing an offset voltage. This circuit does not appear to be a faithful implementation of the 4-point resistance measurement.

U2A has a .33 MF + 1K as a high pass, and a .047 MF +4K7 as a low pass, its just got a restricted bandwidth.
Frank

thanks for that insight, I didn't notice that was making a bandpass filter because in the description of the circuit it only mentions a high pass filter. Do you think that is necessary? The advantage of that configuration is that the potentiometer can be adjusted to remove the offset.

I simulated this circuit and I found a significant nonlinearity from measuring range from 10mOhms to 100mOhms. I used AD8655 for the opamps in order to minimise offset errors etc..

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Actually no, both of them are high pass. The description specifies that it has a pole for 60Hz mains and another somewhere near 700Hz for the Harmonics.

jonnybgood said:

I simulated this circuit and I found a significant nonlinearity from measuring range from 10mOhms to 100mOhms. I used AD8655 for the opamps in order to minimise offset errors etc..

Click to expand...

This circuit is inherently non-linear, so that is not surprising. The firmware in the Atmel processor should compensate for the non-linearity. As I see it, this circuit operates by generating a square wave whose amplitude is supposed to be proportional to the resistance under test. That assumes a constant current through that resistor, which is pretty much guaranteed by the 220 Ohm resistor in series with the resistor under test. So the rest of the circuit is an attempt to measure the amplitude of the square wave.

The high-pass filter centers the square wave around 1.8 volts. Then U2B and U3B form a synchronous demodulator, so we have a DC voltage at pins 1 and 9 of U3. This voltage differs from 1.8 volts by an amount proportional to amplitude of the original square wave.

U2C has a nominal 1.8 volts at its positive input, so its output will do whatever its needs to do to make the negative input also be 1.8 volts. So think of pin 9 as a virtual 1.8 volts. If the "dump" switch, U3D, is closed, this is easy. All three pins will be at 1.8 volts, and the capacitor will be discharged. When the dump switch is opened, U3D will inject whatever current is necessary into the capacitor to keep the negative pin at 1.8 volts. This current is exactly the same as the current flowing out of the capacitor, which is related to the demodulated voltage. Since this current is also charging the capacitor, the voltage on the capacitor will be changing accordingly. The comparator in the Atmel senses when this voltage crosses the threshhold of 1.8 volts. I suppose that it measures the time from when the dump switch was released to when the threshhold was crossed, and uses that to calculate the resistance under test.