Current sensing Optocoupler

[JonPyro]

Hi All,

I need a little help with my maths if you may.

I have a circuit that either supplies 18V 25mA or 18V 1A+, I want to trigger a circuit when the previous circuit supplies the 1A+ pulse not the 25mA pulse. I believe the circuit attached should be able to do that with a low current sense resistor and then a current limit resistor for the LED connected to the opamp. How do I work out the values I need?

I have tried experimenting in LTSpice but its all guess work as I don't know the formulas to work this out and that's not science!

Many thanks,

Jon

[Ian.M]

You need a voltage drop across the current shunt resistor (R2) of at least the opto LED's max Vf at its desired If, +20% (to allow for drop across R4).  That means the circuit will drop a couple of volts and you wont get 18V out.  Also, if the max current isn't well defined and reasonable close to the desired detection threshold, this simple circuit is likely to be unsuitable.

[JonPyro]

Sorry I might have been unclear.

The circuit to the right of the transistor will actually be powered by a 12v supply.

The circuit on the left is supplied by the 18V from another system. I basically want to know when the current is greater than 1A supplied from the left side. Or infact anything above 500mA would be fine.

Vf is 1.3V and max 60mA

[Paul Moir]

As Ian.M said, Vf of the LED drives the design, and to a place that might not be acceptable to you.  To light the LED you must exceed it's Vf by the voltage dropped across R2.  For example, say you want the LED on at 750mA:
V=IR
1.3V=0.750A x R2
R2 = 1.7\$\Omega\$
Now you can't get a 1.7\$\Omega\$ resistor and like Ian.M said you'll need some extra voltage headroom for R4, so you select a R2 to be 2 .  And while you want the LED to be on for sure below 1A, 1A is what you typically draw.  So the actual drop across R2 is:
V=1A x 2
V=2V. 
Since those 2V of your 18V supply are going away in R2, your load is now only going to see 16V.  Is this OK?   Also R2 is going to be burning:
P=VI
P=2V * 1A
P=2W
Is it acceptable in your application to be burning away 2 watts of power?
If the answer to either question is "No", then this circuit won't work for you.   

[Ian.M]

Its worse than that - depending on how much the "18V 1A+ pulse" goes over 1A.  e.g. if it reaches 1.5A, the voltage drop with a 2Ω current sense resistor will reach 3V which only leaves 15V for the load, and results in 4.5W dissipation in the resistor.

If this circuit is unsuitable, the first question to ask is: Does the output need to be isolated, or can it share a common ground with the 18V load?

[JS]

Use a smaller sense resistor, an opamp feed from the 18V. I guess your 18V are more stable than the precession of the threshold you are after, so you can get the reference from there with a resistive divider. You could add some hysteresis so if the current is around the threshold you get a stable reading. Then feed the optocoupler from the opamp and you have a nice circuit. The opamp's inputs should be able to go to the negative rail and be fed from 18V, LM358 would do but you have an extra opamp inside the IC.

JS

[JonPyro]

Thank you all for your input, I understand a bit more now and carry on working on the design