# Capcitor discharge problem

Discussion in 'Electronic Design' started by john, Oct 24, 2006.

1. ### johnGuest

Hi,
Please go the following link and then figure 16 page # 13.
I can not find a way to post the schematic on the web so i am
explaining that I what did I do with the circuit. The INA133 is getting

the input from a voltage ouput DAC ( span is +/- 2.5 volts) . DAC is
connected to pin# 3 of the INA133 and pin#2 in grounded. A capacitor of

47uf and is connected between the non-inverting pin of the OPA131 and a

relay1 ( which connects the output of the capacitor to two different
loads ). The capacitor provides AC coupling. The relay1 connects the
current source to one laod at one time. The power supply of the circuit

is +/- 16 volts. Inorder to discharge the capacitor, I connected a
relay2 in series with a resistor at the input of the capacitor or at
the node where the OPA131 and the capacitor connects together. So, when

the capacitor gets charged I just turn on the relay2 and connect the
node to zero volts. Resistor is grounded.
The problem is that when I switch from load A to load B then I get DC
offset acorss the load B . i thought that What if I just turn off and
on the +/- 16 volt supply of the INA2133
by using two relays simulataneously. The DAC will always be power up (
but generating
only zero at the output ). So, the current source always have input
voltage at its voltage inout pins. Can this voltage be harmful to the
chips if the power supply of the current source or INA 133 is OFF.? If
not then I think this will solve the problem. Turn OFF
the power supply, swith the relay to load B and then turn ON the power
supply again?
Regards
John

2. ### Tom BruhnsGuest

If you want to discharge the capacitor, why not connect the resistor
and RY2 contacts in series, directly across the capacitor? Make the
resistor small, just large enough to protect RY2 contacts and the
capacitor from excessive current. If at the time the capacitor is
being discharged, you also want to insure that the load current is
zero, use SPDT contacts for RY2, connect the moveable contact to the
output end of the capacitor (now goes to RY1 moveable contact), the
normally-closed side to the RY1 moveable contact, and the normally-open
side to the resistor, the other side of which goes back to the
capacitor's input. I suppose ten ohms would be OK, though considering
the activation time of any mechanical relay and the dielectric
absorption in the capacitor, a hundred ohms would probably be fine:
47milliseconds to discharge for ten time constants, but a lot longer to
get rid of the dielectric absorption.

Now just WHY you have a capacitor in there is a bit of a puzzle to me,
since ultimately it will just limit the maximum time that a current of
a particular polarity can be delivered to the load. But I'll leave
that puzzle to you.

Cheers,
Tom

3. ### johnGuest

Hi,

Thanks for the reply! I am using the capacitor for AC coupling. Can I
use another method to do it?
John

4. ### Tom BruhnsGuest

The point is that the capacitor at that point in the circuit won't
behave like an AC coupling capacitor driven from a voltage source,
driving a resistive load. Why? Because if the current source has a DC
component, that DC component will flow into the capacitor (and out the
other end of the capacitor, through the load), charging the capacitor
until the current source can no longer supply enough voltage. Then the
current source will no longer supply the requested current. If there
is a higher-frequency AC component in addition, it will become clipped
as the source approaches the limit of the voltage it can supply. In
contrast, a source with constant DC voltage just puts a charge of that
voltage on the capacitor, and after the initial transient, the load is
unaffected by the source's DC component. If you had a DC current
source with infinite voltage compliance and a coupling capacitor with
infinite voltage capability, the load would see a DC current through
the capacitor.

Another way to look at it (in the frequency domain instead of the time
domain) is that the capacitor's AC impedance is infinite at DC and is
inversely proportional to frequency. If the source driving it has
infinite impedance (a perfect current source), the capacitor will look
relatively like a short-circuit down to very low frequency. In the
limit as the frequency goes to zero, the current remains the same, so
the load continues to see the same current. That model assumes
linearity, which will fail as the current source output DC voltage
approaches the supply rails.

One way to insure that you don't need to reset the capacitor is to
monitor the DC voltage at the output of the current source and use
feedback to insure that it stays within bounds. Another way is to
insure that the DAC output does not have a DC component, or at least
that any DC component of its output is negligible. Then you should not
need a coupling capacitor at all, assuming low DC offset in the DAC and
op amp circuits. Since I don't know what it is you are trying
ultimately to accomplish, it's a bit difficult for me to give
recommendations.

Cheers,
Tom

5. ### johnGuest

Hi,

I am trying to provide constant current with no DC component to tissue
via electrodes. So, the circuit's load is tissue. I thought that the
capacitor will not let DC component pass through it and will let only
AC pass through it. you mentioned that DC component can be reomoved at
the DAC's output. Can you advice me how? Will a capacitor between the
output of the DAC and the input of the INA133 ( non-inverting input )
remove the DC component?

John

6. ### Ancient_HackerGuest

John, if this is going to HUMAN tissue, I suggest you stop right now.

You don't appear to know basic electronics well enough to come up with
a safe circuit.

Any circuit hooked up to a human needs to be very carefully designed
and tested.

Same goes for animals, unless you want the PETA types to get on your
case.

7. ### Jim ThompsonGuest

Unless you are _cooking_ them ;-)

...Jim Thompson

8. ### johnGuest

Hi,

Its not going to humans or animals. the tissues will be extracted from
rats and will be in the dish under the microscope. Now, yes i do not
know about electronics that much and i have not tested the design as
you can see that I am seeking advices and analyzing. So, if you have
advice for me then you are most welcome though you do not have to be
scared and tell me not to go for it.

Thanks

9. ### Tim AutonGuest

I like my PETA activists medium-rare, served on a bed of rice

Tim

10. ### Tom BruhnsGuest

John,

How much current do you want? Specifically, what is the instantaneous
peak value of current you want to achieve? What is the expected
maximum and minimum impedance represented between the electrodes in the
tissue? I suppose "constant current" means current with a constant RMS
value, or current essentially independent of the impedance of the
tissue. Clearly if there is no DC component, it's not constant in an
instantaneous sense.

If you just AC couple the signal from the DAC to the voltage-to-current
converter, then the output current (neglecting any small DC offsets in
the op amps) must also have no DC component. If it does, you could
easily null that out with a small current fed through a high-value
resistor from a potentiometer (variable resistor; "trim pot") connected
between the supply voltages. So the circuit would place a capacitor
between the DAC and the input to the current to voltage converter, and
omit the capacitor at the output. If the DAC is driven from a digital
source that can filter out any DC component in the signal digitally,
you could omit the capacitor. And you could (perhaps) even monitor the
current in the tissue with an integrator, fed back to some earlier
point in the circuit, to automatically insure that the DC component of
the current in the tissue is zero, or as nearly so as the integrator is
able to detect.

If the desired maximum current is small enough (a few microamps), it
should work fine to just use a voltage source, with a high-value
resistor between the voltage source and the tissue. If the resistor
can be 100 times as large as any variation in the tissue resistance,
you know the current will be stable over that variation to within a
percent.

Cheers,
Tom

11. ### Paul E. SchoenGuest

There is really no way to apply an AC current to any sort of load without
having some DC component, at least during the first several cycles.
However, it can be made arbitrarily small if you can delay the application
of the final value of AC current.

The DC component is really the average of the waveform over the period of
time from first application to the present. If you simply apply a sine wave
current, starting at zero crossing, the average current will first increase
until it reaches the peak, and then decrease as it goes through zero
crossing, and continue decreasing until the second zero crossing, at which
it will be zero. As continually more excursions occur, the true average
over time will decay toward zero.

This is why it is not good to use a zero crossing solid state relay to an
inductive load, as there will almost always be a net DC component that can
saturate or magnetize the inductor. Applying the waveform at the peak
causes the DC component to swing positive and negative as a damped
oscillation.

One way to minimize DC offset is to start the signal at a very low level,
and gradually increase its amplitude until the desired value is reached. If
this is acceptable, it can be done easily with the DAC by using a voltage
ramp as the reference, or by providing the correct digital signal to the
input.

If this is for a scientific study then it is imperative to have a full
understanding of the physiology of the electric current through the tissue,
and the effects of various DC levels over short periods of time, as well as
the magnitude, frequency, and length of time of application of AC signals.
It would be helpful to know the reason for the need to block DC offset, and
determine if it can be tolerated for a short time at the start of the test.

Paul

12. ### joseph2kGuest

Gee, and i always thought that transformers would not transport DC. Then
again transient conditions is not DC by definition. Then you wander and