Circuit protection

Thread Starter

Mathews M John

Joined Jul 14, 2015
8
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?
 

Hypatia's Protege

Joined Mar 1, 2015
3,200
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.
As I understand your question/concern incorporation of redundancy would seem your clearest solution...

Best regards
HP
 

Thread Starter

Mathews M John

Joined Jul 14, 2015
8
That was pretty stupid from my side. Here's the circuit
circuit1.png
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
 

OBW0549

Joined Mar 2, 2015
3,075
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?
 

AnalogKid

Joined Aug 1, 2013
8,212
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
 

OBW0549

Joined Mar 2, 2015
3,075
I need to ensure that it doesn't go above 5mA
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.
 

Thread Starter

Mathews M John

Joined Jul 14, 2015
8
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
 

OBW0549

Joined Mar 2, 2015
3,075
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.
Did you mean 5 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
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.
 
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