# Circuit protection

Discussion in 'General Electronics Chat' started by Mathews M John, Jul 22, 2015.

1. ### Mathews M John Thread Starter New Member

Jul 14, 2015
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I am working with a modified Howland circuit and I need to make sure that the constant output current does not increase over a certain limit. I wanted to know what the best way to do this was.
Since the Howland circuit is technically a voltage dependent current source I could have a overvoltage protection circuit before the signal enters the circuit. However, I am concerned that something could go wrong in the Howland circuit, that would lead to an unwanted increase in current. I am not so sure about this and wanted to discuss about this. Can someone help?

2. ### ScottWang Moderator

Aug 23, 2012
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Where is the circuit?

3. ### Lestraveled Well-Known Member

May 19, 2014
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As Scott said, Please post your beginning circuit and then the discussion will begin.

4. ### Hypatia's Protege Distinguished Member

Mar 1, 2015
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As I understand your question/concern incorporation of redundancy would seem your clearest solution...

Best regards
HP

5. ### OBW0549 Well-Known Member

Mar 2, 2015
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My approach would be to current-limit the (+) and (-) DC power going into the Howland pump's opamp (and/or the opamp's power booster stage, if you're using one).

Hypatia's Protege likes this.
6. ### DickCappels Moderator

Aug 21, 2008
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Or monitor the voltage across the resistor in series with the output of the opamp.

7. ### Mathews M John Thread Starter New Member

Jul 14, 2015
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That was pretty stupid from my side. Here's the circuit

The first part of the circuit removes the DC offset from a voltage signal and feeds it into the Howland circuit. I believe this will produce around 3mA of AC current. The blue resistor of 200Ohm is the load (which will be varying in my actual circuit). Hope now there will be a proper discussion

8. ### OBW0549 Well-Known Member

Mar 2, 2015
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What current level do you need to limit the Howland pump's output to? Keep in mind that even if you do absolutely nothing at all, the maximum amount of current this circuit can possibly deliver to your load under worst-case conditions, even if it is shorted, is only 15 mA (+ or - 15 volts divided by R5). Do you really need the current limited to less than that?

9. ### Mathews M John Thread Starter New Member

Jul 14, 2015
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I need to ensure that it doesn't go above 5mA

10. ### AnalogKid Distinguished Member

Aug 1, 2013
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An external peak current limiter can be done, but that's just more parts that can fail. And since none of your components are anywhere near their operating limits, a direct-to the-rail chip internal failure will be not just rare, but a notable feat. But, continuing on with the question...

If the opamp is a rail-to-rail type, you might increase R5 to 3 K. I say "might" because we're still short some data. Is 5 mA the worst case fail current or the largest desired regulated current? If the opamp has a volt or three between its max output voltages and the rails (as is common) then the headroom gap is a problem for the increase-R5 solution.

ak

11. ### OBW0549 Well-Known Member

Mar 2, 2015
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Then I'd say increase R5 to 3.0K. The worst-case current will then be 5 mA (15V/3K ohms). The ratios R3/R8 and R4/R2 will have to be adjusted to compensate, if you want to maintain the same scale factor of output current to input voltage.

Under normal operation, with +/- 3 mA peak output, the opamp will have to output +/- 9.6 volts, which an AD713 should easily be able to do with +/- 15 volt supplies.

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12. ### Mathews M John Thread Starter New Member

Jul 14, 2015
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Thanks @AnalogKid and @OBW0549 , I understand what you guys are talking about and how to ensure that the maximum current I can get even in a failure in 1 mA. I think I'll go with an overvoltage protection on my power supply to ensure that the maximum voltage is around 15V. Should be able to do that with a couple of zeners

13. ### OBW0549 Well-Known Member

Mar 2, 2015
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Did you mean 5 mA?

My choice would be to put a pair of back-to-back 15V zeners from the opamp output to circuit common; depending on the output current capacity of the supplies putting the zeners across the supplies risks serious overheating if the voltage increases.