Any as simple current source for optocoupler?

Hi,

Draw a sketch.
What optocoupler? Especially V and I rating
What current and what precision?
What switching frequency?
What are the control voltage levels?
Does efficiency matter?

And please use numbers and units. Don´t use phrases like "low", "best", "as fast as can be"

The simplest could be a bjt and an emitter resistor ... but I guess it won´t work for 3V.

For low speed you may switch the current source, for higher speed you may want to short circuit the optocoupler..
For high efficiency you may step down to - let´s say 2.7V - and use a simple resistor and a MOSFET as switch.

There are hundreds solutions .. each will have it´s pros and cons. I don´t know what´s important for your application.



Klaus

I'm sorry .
The optocoupler will be some common 4N35, PC817.
Switching frequency max 1000Hz.
The current solution is for a fixed voltage of 24V. Take it schematically, instead of the LED is an optocoupler and instead of V2 the output of the processor or some logic.
In terms of accuracy, the current should not fall below 6mA and exceed 12mA

There should be some kind of current feedback. To start with something simple

You presumably also want constant current vs temperature and other
influences.

A JFET with G=S will give you a pretty flat current above its VPO, but
a strong tempco can be expected.

The desired outcome, beyond fixed current in the LED, is unclear.

There are optos which contain two, matched channels such that
you can use one for forward and one for feedback in closed loop.
If you want to get linear current control (or, control to some reference,
like a LT1009 - maybe the resources exist "off-page") these could be
the ticket. But I will leave the specific-parts-search to you.

Hi,

I refer to your schematic from post#3.
Let´s calculate with 9mA LED current.

Change R1 value to 100 Ohms and move it to the emitter path.
On 9mA there is a voltage drop of 9mA x 100 Ohms 900mV across R1.

So the LED current path is:
3.3V ..24V Supply -> LED -> BJT -> R1 -> GND (All is 9mA by ignoring the base current)

If supplied by 3.3V, we have the follwing voltages: 900mV across R1, 1.2V across LED, remaining 1.2V across CE. This should be enough for regulation.

Base voltage (wrt GND) is 900mV + 550mV V_BE = 1.45V

Now we need a voltage divider from V2= 3.3V -> 1.45V -> GND.
The voltage across R3 should be 1.45V and the voltage across R2 should be 3.3V - 1.45V = 1.85V
If we allow a current of 1mA, then resistance in kOhm is the same as the voltage. Thus R3 = 1.45k Ohm, R2 = 1.85 kOhm

Done.

For sure the current will drift with temperature, with V1 voltage and with V2 voltage. But I expect it to be within your 6mA ..12mA range.

Klaus

There are lots of ways to make a constant current sink.
Two transistors are better than one in some cases, but now here at low currents.

With negative feedback, R1,R2,R3,R4 ratios all affect current limit which I chose as 7.5mA nominal but with hFE expected between 50 and 150 which affects the tolerance.

The feedback Q1 is more hFE sensitive so the two diode method is preferred here, but then the choice of diode is critical due to the Vf / If controls Vbe which controls the current.

Re basically determines the Icc but at these low base and diode currents, the voltage drops are not the same. If you want to assume 0.65V , that will be close but the diode current is controlled by R1 and Ib is controlled by R1 & R4.

1N4148's 523 mV @ 175 uA x2= 1.04V
PN2222A 645 mV @ Ib = 140 uA Thus (1.04-0.65V)/ 56 ohms (Re) = 7 mA


https://tinyurl.com/ymwpcoex Simulation.

Rev 3 final
Eliminated NFB fancy hFE dependancy

LM334 current regulator, TO92, SMD, 3 pins + 1 resistor.



Regards, Dana.