How to protect Sensitive Op-Amp circuit from huge over voltages while turned off?

Hi All,

I have a problem that has me stumped, was wondering if anyone out there had any suggestions.

I have an application where I take the output of a zero flux current transformer to measure current flowing in a conductor. The circuit is battery operated(3.6V), it is implemented using single supply Op-Amp circuitry and the input is referenced to VCC/2. The current I measure is typically

Reply to
PFITZ
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You have to limit the current into the input with a resistor and clamp the input to Vcc and gnd with schottky diodes. The diodes will conduct before anything in the op amp gets forward biased.

Reply to
Raveninghorde

Direct lightning strikes are more than any reasonable circuit can handle. Nearby strikes are something that can be dealt with with care.

Schottky diodes and resistors will work in many cases. JFETs can only pass about 100mA so they can't do it all on their own. They can work as a second layer of protection if you cant stand to have the actual diodes to rails. The idea is like this:

---- ! ---/\/\---+---/\/\---+----/\/\--+--!-\ ! ! ! >-- [TVS] !- !+/ ! -V-->! GND !---GND

The JFET will take about 30mA away from the op-amp when on but is just a small capacitance when off.

Reply to
MooseFET

Your best fix would probably be a relay. A small DPDT relay could disconnect/short both the signal input and Vcc, both of which may be problems here.

John

Reply to
John Larkin

What's the largest signal of interest? Can you short it with back-to-back schottky diodes (fairly large ones, if the capacitance isn't a problem)?

Tim

-- Deep Friar: a very philosophical monk. Website:

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Reply to
Tim Williams

What voltage do you think schottky diodes will start conducting at?

Reply to
Raveninghorde

Raveninghorde a écrit :

Hint: a zero flux current transformer has zero output voltage (since it's working at zero flux) and a pair of schottkies is perfect in that case. The pb will be more to protect the amplifier's output which is driving the compensating winding.

If the magnetizing current isn't compensated, then there's still the possibility to set the opamp as a transimpedance amplifier driven by the current transformer burden resistor, and still have it's minus input clamped to ground by a pair of schottkies (mind the diodes capacitance).

--
Thanks,
Fred.
Reply to
Fred Bartoli

o-back

)?

se.

also consider the speed of the desired signal vs the spikes. If the desired signal is 60Hz or lower, and the spikes are ligtning induced, you can add R C's to all the inputs and slow down aand reduce the spikes that you have to deal with...

Mark

Reply to
Mark

ack

Hi, Thanks for the suggestions, the signals of interest will never exceed a couple of millivolts so this may indeed work, I hooked up some BAT54s back to back to see if it would work, but it seems to attenuate my input. What parameter should I be looking for in a potential diode, I know I'm really looking for a low leakage current in the region before it starts to conduct and most diodes would be trying to go in the opposite direction in terms of desirable attributes, I've never seen a mention of this in datasheets etc. any ideas?

P
Reply to
PFITZ

-to-back

em)?

case.

e
.

This is true, I have implemented this in the design but perhaps I need to set the cut off a bit lower.

Reply to
PFITZ

What is the smallest DPDT type of relay thats out there? I'd be concerned about the effects of the magnetic field around the conductor on the magnetics in a relay though, I have seen them do funny things before like spurious opening & closing etc.

Reply to
PFITZ

I like the FUJITSU FTR-B3GA4.5Z series, 2nd sourced as OMRON G6K-2G-Y-DC5.

John

Reply to
John Larkin

What frequency or impedance? If it's HF and high ohm, maybe you should put a sacrificial FET follower in front, or something like that (with suitable adjustments for controlling offset, noise, etc.).

If you're working with DC, you screwed something up bigtime. The current at

0V obviously is zero, and the current a few mV either side of zero follows the diode formula perfectly: I = I_th * (exp(V/k_B*T) - 1). I_th can be estimated from other data points, like Vf at a useful current, e.g. 1mA at 0.3V, room temperature, gives 9.3nA. The leakage at less than 20mV is going to be less than 9nA (roughly 2Mohm?).

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Reply to
Tim Williams

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