J
john
- Jan 1, 1970
- 0
Hi,
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.
John
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.
John