I designed a constant current source that can produce current in the range of 0 to 600uA with frequency range of 10 to 32kHz. The current source has a leakage current of about 240nA which I do not want to go to the load. My load impedance is between 20Kohm to 400kohm. Inorder, to protect the load from the leakage current I placed an inductor in parallel with the load. I did not know the exact inductor value, found some old inductor in the cabnet. The inductor did get rid of the leakage and let the ac waveforms appear across the load. I started calculating the inductor value that I needed using formula X = 2pi FL.
I choose X = 10 Mohm for which L = 50 H at 32 Khz but X drops to 159 Kohm when frequecy drops to 10 Hz. which is less than the load impedance so the current will choose the least resistive path. can anyone adivce me that how to compensate this drop of impedance ( X )?
I can not trimmed the offset of the constant current source using potentiometer due to mechanical problems.
I am also thinking to add a switch in series with the inductor. When DC current flows into the inductor, the switch should get closed and when AC apperas across the inductor then switch should offer very high impedance ( like in mega ohms) so the reactance of the inductor will not be a issue anymore. Can anyone suggest what kind of circuitry I need to do it!
600uA into 400kohms is 240 volts. RMS or peak or what? In any event, it's quite a lot. Does your current source really have that much compliance?
Keeping DC out of the load it almost trivial: just add a blocking capacitor. Its reactance at the lowest frequency of interest should be small compared with the load, so that you don't have too much drop across it, though since it's a constant current source, it doesn't really matter that it be especially low reactance... except that you also need a place for the leakage current to go. It's seriously impractical to use an inductor that will be megohms reactance at low frequency and not show some potentially bad resonances and the effects of shunt parasitic capacitance at higher frequency. Instead, just use a resistor. The DC drop in 22 megohms for 240nA current is only a little over 5 volts, and any current source that can deliver 240 volts should have no trouble with that. You can always return the resistor to a voltage that keeps the output of the current source centered near
0, if you want. I'm assuming that the error caused by 22 megohms in parallel with your load is not significant in your application, but perhaps that's not correct; in that case, you can use a more complicated boot-strapped circuit to accomplish the task. If you use such a resistor (say 22Mohms) to absorb the DC current, be aware that stray capacitance from the output node of the current source to ground will represent a significant load: even just 10pF is only about
500kohms at 32kHz. People who try to keep high impedances at high frequencies tend to use guards that are driven to the same voltage as the node they guard, so that the effective capacitance from the guarded node to ground or common is very much less than the capacitance from the guarded node to the guard. Another way of saying that is that the driven guard keeps capacitive currents from the guarded node to very low levels, compared with what they would be if the node were not guarded.
Sounds to me like you're trying to learn to swim in water that's pretty deep. I wish you luck. For example, 50 H at 32kHz is (theoretically, but difficult to achieve in practice) about 10Mohms as you say, but at 10Hz, 50H isn't anything close to 150kohms.
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