# Inductive Circuits

Discussion in 'Electronic Design' started by john, Jun 7, 2007.

1. ### johnGuest

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

2. ### John O'FlahertyGuest

If it's a constant DC 240 nA, maybe just a DC constant current source
of that magnitude in parallel with the load.

3. ### John DevereuxGuest

Why don't you put a capacitor in series with the load, instead? A few
microfarads should do it, use an unpolarised type (i.e. not an
electrolytic).
50H is very large you know! You would be better off designing a
current source without the leakage.

4. ### Guest

Erm, that's 240V .... Isn't the load impedance for a current source
usually 0 ohms???

5. ### John DevereuxGuest

Although 600uA at 20K is 12V. Perhaps the limits are not independent

No, I don't think so. It is supposed to stay constant over some range

6. ### John PopelishGuest

That is like saying that the normal load impedance for a
voltage source is infinity ohms.

7. ### Guest

Hm, I was thinking transducers...

8. ### Tony WilliamsGuest

[snip the inductor solution. Probably not viable.
It sounds like you need a low frequency, (<10Hz),
dc voltage offset servo loop. An inverting opamp
integrator with 10meg resistor looking at the o/p,
balancing the cc source towards zero voltage offset.
Use a low offset fet input opamp. Reducing to <1mV
dc offset voltage would be equivalent to <50nA into

9. ### johnGuest

Hi,

What I understood is that I have to add another opamp at the output of
my current source or build a new one...

John

10. ### johnGuest

Hi,

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!

John

11. ### Tom BruhnsGuest

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
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
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.

Cheers,
Tom