hi t,
As you say, a series resistor will 'modify' the charge/discharge current.
What value of series resistor do have in mind.?

The other factor is the source impedance of the 1kHz Sine source.

E

 

Is this for my attention.???

Edited reply to re-include TS’S question.I don’t know where it went, but strange things can happen when working on a iPad.

 

hi t,
As you say, a series resistor will 'modify' the charge/discharge current.
What value of series resistor do have in mind.?

The other factor is the source impedance of the 1kHz Sine source.

E

hi eric,
I thought to use 1k series resistor and source impedance is 50R (I have added this one into simulation as well)
From the simulation with 1k resistor the voltage difference across resistor would be around 175mV amplitude or 124mV RMS

 

If I will use larger value resistor, then the voltage waveform across the capacitor will reduce its amplitude and will be shifted, probably that is fine, but ideally I want to keep certain amplitude of the voltage for a device

If you have 10V peak drive signal, and use 100mV across the sense resistor that is only 1% loss.

 

If you have 10V peak drive signal, and use 100mV across the sense resistor that is only 1% loss.

Yes, that is fine
The other issue in here is that for the capacitor I am using a device (which is basically 2 conductive plates isolated by dielectric material) and its capacitance value can be changed from 1nF upto 50nF depending on how dielectric material is changing, so I will get a loss of 1% at 1nF, then I will get more at higher capacitance values, but I want to keep waveform fairly constant across the device (hope that makes sens)

 

hi t,
This is a Draft option, uncalibrated
E

 

hi t,
This is a Draft option, uncalibrated
E

View attachment 252136

Hello eric,
Thank you for your reply. I will have a look at this solution and play around
But my idea was to place a resistor before the capacitor and use differential amplifier across resistor to measure voltage difference (current draw). Would my idea be difficult to implement or there are some drawback using it?

 

Would my idea be difficult to implement or there are some drawback using it?

Hi t,
I could explore options.

It always helps if we can have a discussional circuit, albeit not the best solution, so that we can consider options.

Are you interested in measuring the instantaneous or average current through the cap.?

E

 

Hi t,
I could explore options.

It always helps if we can have a discussional circuit, albeit not the best solution, so that we can consider options.

Are you interested in measuring the instantaneous or average current through the cap.?

E

Hello eric,

At the moment I haven't considered any specific circuitry, trying to find different solutions and analyse them.
Average current, as I need to monitor if the device is in place (there is no loss connection)

 

hi, t.
Is the 1nF a cap, a component that is already installed in a piece of equipment, with one end to connected to 0V.? and the other end of the cap free, so that a 1k can be added.?

More info on the actual application would help.
E

 

hi, t.
Is the 1nF a cap, a component that is already installed in a piece of equipment, with one end to connected to 0V.? and the other end of the cap free, so that a 1k can be added.?

More info on the actual application would help.
E

Hi eric,
So the application is I have this capacitor (it is a device which acts as a capacitor, but its capacitance value will vary depending on the dielectric material used (physical tests), so we can call it cap). You do install this capacitor into the cartridge in the equipment and want to make sure that the cap is electrically connected. The idea was to use a resistor in series with cap, supply AC waveform and measure current, so you would know that cap is electrically connected. In terms of connection the the AC supply it can be either with a resistor to ground or cap to ground (there is another interface board, which allows to do this connection)

 

Hi t,
This an IA built using LM324, if you can find a IA that will tolerate a CMV close to the power rail levels, that would be better.
E

 

The average current is easy. It is zero.

Bob

 

Hi t,
This an IA built using LM324, if you can find a IA that will tolerate a CMV close to the power rail levels, that would be better.
E

Hi eric,

Thank you very much
This is what I've been looking for, I need to have a look and try to implement it

 

The average current is easy. It is zero.

Bob

I meant RMS, not average)

 

Thank you very much
This is what I've been looking for, I need to have a look and try to implement it

hi t,
Post your circuit design when ready, we can give a run through with LTS.

If you decide to make your own IA, ensure that the resistors are better than 0.1% tolerance, for matching.
R3, that 5k sets the Gain, use a say a 5k trim pot.
You could add a comparator in the circuit that lights a GO/NO Go LED when testing.

E

 

hi t,
Post your circuit design when ready, we can give a run through with LTS.

If you decide to make your own IA, ensure that the resistors are better than 0.1% tolerance, for matching.
R3, that 5k sets the Gain, use a say a 5k trim pot.
You could add a comparator in the circuit that lights a GO/NO Go LED when testing.

E

Hello eric,

I will port it for definite
Comparator you mean between U1 and U2 outputs?

 

Hello eric,

I will port it for definite
Comparator you mean between U1 and U2 outputs?

Sorry U1 and U4?

 

The problem is how to measure current? Using differential amplifier across resistor?

The simplest way is to measure the current directly with an op amp connected as a transconductance amp (below), assuming that connection is not a problem in your application.

A 10kΩ feedback resistor gives an output of 10mV/µA.
The feedback resistor can be selected to give whatever output gain you want (within the op amp limits of course).

Since the capacitor is going into the virtual ground of the op amp, this circuit also has the advantage that there is not the small capacitor voltage error which a series shunt resistor would cause.

 

The simplest way is to measure the current directly with an op amp connected as a transconductance amp (below), assuming that connection is not a problem in your application.

A 10kΩ feedback resistor gives an output of 10mV/µA.
The feedback resistor can be selected to give whatever output gain you want (within the op amp limits of course).

Since the capacitor is going into the virtual ground of the op amp, this circuit also has the advantage that there is not the small capacitor voltage error which a series shunt resistor would cause.

View attachment 252175

Hi,
What do you mean by connection is not a problem?
From the circuitry Where is the return bath for the capacitor, is it going to be a virtual ground on the inverting input?