3
\$\begingroup\$

I am trying to make a simple circuit to light an LED based on a 9 V supply. I only want this circuit to light based on a switch, which will close when a separate voltage supply is provided.

The trouble I'm facing is that I have no control over the switch's input source, and there's an LED (as part of an optocoupler) that will light if enough current allowed through the circuit I'm trying to build. I am trying to prevent that LED from lightning, but still use just enough of the current coming from that circuit to power a switch.

So, I am trying to put a pretty high resistance in front of my circuit that receives that current to limit it to microamps. That should prevent the source's LED from lighting, but I run into the problem of how to use this tiny current to drive the switch.

schematic

simulate this circuit – Schematic created using CircuitLab

I've included an example using a relay. I have no control over everything on the 12 V source before the SPST switch and cannot change it.

I'm looking for some help on what to use a a switching mechanism that can switch on this low of current. Do they make relays that will work for this application? Any other suggestions? I'm a complete layman and appreciate any help!

\$\endgroup\$
6
  • 2
    \$\begingroup\$ Use a MOSFET transistor. You won't need the high resistance. \$\endgroup\$
    – DKNguyen
    Sep 7, 2022 at 17:48
  • \$\begingroup\$ I've edited your question to use the built-in circuit editor (CircuitLab), as well as replaced the term "phototransistor" with optoisolator. An optoisolator is a pair of an LED with a phototransistor or with a photodiode, sometimes with additional signal-forming circuitry. \$\endgroup\$ Sep 7, 2022 at 18:49
  • \$\begingroup\$ Your problem it's not clear. You propose a half-baked solution, but we don't know your problem so we can't decisively provide a clear-cut answer. Are you perhaps trying to detect when another circuit is activated and display its state using an LED? If this is the case, it would be very helpful to describe what kind of circuit you are trying to monitor. \$\endgroup\$ Sep 7, 2022 at 18:56
  • \$\begingroup\$ @LorenzoDonatisupportUkraine The relay is there just for illustration purposes. OP is looking for something that would work like a relay but be practical, i.e. draw a minuscule current. \$\endgroup\$ Sep 7, 2022 at 19:01
  • \$\begingroup\$ @Kubahasn'tforgottenMonica Yes, I did understand that the schematic was a very crude attempt at explaining the situation. My comment still stands: why would the OP want to do that? He didn't explain the real problem he is trying to solve. \$\endgroup\$ Sep 7, 2022 at 19:10

2 Answers 2

11
\$\begingroup\$

An N-channel MOSFET will do the job as an "on-when-on" switch. The entire current draw comes from the gate "anti-float" resistor R2. Most optoisolators have current transfer ratio poor enough that the 1 MΩ load with a 12V supply is immaterial and won't trigger the optoisolated output.

The gate series resistor R3 is used to isolate the gate capacitance, so that when SW1 is depressed, the optocoupler's LED doesn't flash as the gate is charged.

schematic

simulate this circuit – Schematic created using CircuitLab

Alternatively, using a P-channel MOSFET gets you an "off-when-on" switch. Again, the entire current draw is from the anti-float resistor.

schematic

simulate this circuit

The gate pull-up/pull-down resistor R2 is absolutely necessary. A MOSFET gate is an open circuit and will pick up stray charge and change the transistor state randomly - never mind that after turning ON, the circuit would stay ON for a long time, as the gate capacitance is usually large enough to take minutes or even hours to discharge sufficiently.

If you'd want to further reduce the input current, R2 could be a larger value (say 10 MΩ) at the expense of increased sensitivity to electromagnetic noise and stray electrostatic fields.

To reduce such sensitivity, a current-driven BJT buffer would do.

Below, D2 keeps Q1 from saturating, keeping the base current under control. C2 shunts AC currents around the B-E junction, so that the transistor is not turned on by stray induced voltages, RF interference, etc. In the previous circuits, such shunting was provided by the gate-source capacitance of the mosfet.

schematic

simulate this circuit

Now the input current is 0.5 μA. AM2 and AM3 are just to illustrate the LED currents. In an actual circuit they'd be replaced by short circuits.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Thanks Kuba, this is what I was exactly hoping for and it actually helps me eliminate another physical switch. \$\endgroup\$
    – BudJB
    Sep 7, 2022 at 22:08
3
\$\begingroup\$

So, I am trying to put a pretty high resistance in front of my circuit that receives that current to limit it to microamps.

It sounds like you need to buffer that circuit using a MOSFET: -

schematic

simulate this circuit – Schematic created using CircuitLab

For 9 volt operation (V2) you would make R1 about 1 kΩ. This would then produce about 7 mA in your LED.

The input to the MOSFET (gate to source) is very high impedance and, it's always sensible to use a 1 MΩ resistor (R2) across gate and source. It's not all that clear what the gate signal connects to so, if you reveal that information in the form of a schematic I might be able to make further recommendations.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.