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Topic: Amplifying current from microamps to milliamps (Read 342 times) previous topic - next topic

JohnRob

I kind of understand what your goal is, however I'm not sure why an additional resistance in the circuit would diminish your results. 

If you know the voltage (i.e. the power supply)  and you know the value of the fixed resistor, can you not back calculate the change in resistance in the piezo element?

I'm assuming you are making measurements when different forces (actually impacts) are applied to the piezo element.

John
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wvmarle

Only if you mean, a sense resistor in series with the piezoresistive material.
A fixed series resistor, measure the voltage in the mid point. Doesn't matter which side you measure, the result remains the same. As said, standard voltage divider, nothing special about it.
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ReverseEMF

#17
May 11, 2019, 11:35 pm Last Edit: May 11, 2019, 11:40 pm by ReverseEMF
A fixed series resistor, measure the voltage in the mid point. Doesn't matter which side you measure, the result remains the same. As said, standard voltage divider, nothing special about it.
Well, actually, there is something "special" about it.  A sense resistor, I suppose, could be considered a portion of a "Voltage Divider".  It does, indeed, divide voltage.  But, it also subtracts voltage--taking it from an existing voltage. Basically, it parasitically consumes some of the existing voltage, for the sake of measuring the current flow in whatever that voltage is across. 
The typical concept of a voltage divider, is a series pair of resistors, placed across a point in a circuit, to divide the existing voltage down.  It doesn't need to consume any of the original voltage to do this, and it's not in series with the original circuit--like in the case of the sense resistor--but is, instead, in parallel with the original circuit.  It will, however, parasitically consume current -- and even, the consumption of that current, may cause a rise or drop in the voltage, the divider is applied across, but that's secondary to the parasitic current.
So, it's not really the same.

Now, if you're advising the use of a voltage divider to measure the PRM's current, I see no way for that to work.

Also, it does matter which side you measure from, in both cases.  The result is not  the same.  You will, possibly, be able to use math to convert from one result to the other, but in the case of the series sense resistor, measuring any other way, but across the sense resistor is a ridiculous proposition.  Also, often, a voltage divider is used to reduce a larger voltage down to a voltage in the common mode input range of some device, such as the input of an ADC, or an OpAmp, or MCU input.  Connecting these things to the other side, would cause component failure.  So, not sure what you mean.

Are you thinking of a Wheatsone Bridge?
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wvmarle

I was thinking of a standard voltage divider - the common way to measure an unknown resistance such as also used for NTCs and FSRs.

A Wheatstone bridge may also work, depending on the actual (change in) resistance. It's a common way resistive load cells are measured.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

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