Use current from SMD LED to switch larger current

Question

I've got a bluetooth receiver with blue SMD LED's on that flash to indicate connection status. I would like to take that status LED and replace it with the white LED that's in a switch I have.

The switch is designed for 12-24V circuits and has a series resistor (not sure the value), and I have a 25.2V 6S lipo powering the rest of my system, including a similar switch with LED.

I tried to attach the switch and LED to the receiver with flying leads, I piggybacked the DPST switch just fine and removed the blue LED and wired up the switch LED, it didn't light. So I removed the series resistor in the switch, the LED now lights up but is very dim. See below.

^{simulate this circuit – Schematic created using CircuitLab}

So I thought about using a transistor switch to activate a higher voltage circuit. The problem I have is that the blue LED appears to be switched on the cathode after a series resistor, I don't know where to attach the base of a transistor which would see the right voltage drop. Obviously there's a lot of stuff going on on that PCB I don't know about, or don't understand illustrated as the box around the switch.

I've updated the schematic to show how I thought it could possibly work, but I don't think it will.

I can't figure out what this means for what I want to do, does it sound possible?

Thanks.

^{simulate this circuit}

Answer 1

A less intrusive, less circuit dependent option is to use an ldr or photodiode in a light detecting circuit. Just glue/tape/locate the sensor near your led without any light contamination, and you can trigger your led. In essence, you are making your own opto-coupler, using the existing led as the light half.

Does not depend on the voltages or currents of your light source, and is electrically isolated, so no ground loop issues, which a audio system tends to hate.

A variable resistor to set the threshold, and a common transistor like the 2n3904 would be enough. Your existing switch led is likely under 20mA.

Answer 2

I am not really sure what you want to do, but have you thought about an optocoupler? You could replace this SMD LED with it and you will have an open Collector output

Answer 3

Firstly, good work on your diagram of the measurements. It's a big help and I wouldn't have understood or believed your numbers without your schematics.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A little tidy-up of the schematics and combination of the readings results in a more legible drawing.

There's something odd going on here. If switch 1 is open then we should see 4 V on the top of the switch as there is nothing to pull the voltage lower than the battery voltage. With 1.9 V across the switch there would be 2.1 V across the LED and red, yellow, orange and green LEDs would light up.

With SW2 closed there is 2.6 V across the LED. This suggests to me that it is a blue LED.

Figure 2. A 2.6 V LED forward voltage suggests a blue LED running at somewhere around 20 mA. It should be rather bright. Graph source: LED I-V curves.

^{simulate this circuit}

Figure 3. One small problem with your proposal is that you have no return path for the base current to get back to the left side of the circuit. You need to add a ground connection on the left side. The big problem is that the transistor only switches on the LED which might look nice but that's all.

^{simulate this circuit}

Figure 4. A possible working solution.

How it works:

• Opto-couplers use infrared LEDs. Looking at the graph of Figure 2 we can see that at 20 mA an IR LED will drop about 1.4 V or so.
• The infrared opto-LED will therefore hog the current and the SMD LED probably won't glow much, if at all.
• You haven't said how much current your device draws so I've assumed its more than the opto-transistor can handle so I've added a regular transistor to create a Darlington pair. Note that you will lose a volt or so over this arrangement.

There are several advantages to this arrangement.

• Optical isolation avoids connecting the grounds together and any associated power or ground-loop problems.
• You can solder flying leads onto the existing LED without disturbing it on the board. Simply remove the leads to restore original operation.
• The opto-transistor can handle high-side or low-side switching.

One thing to watch. There has to be (already) some current limiting in SW1. This is designed for the visible LED. The IR LED, with its lower V_F, may draw more current. Measure and compare with datasheet.

Answer 4

Since the cathode is only dropping to 1.4V, this implies that there is a current-limiting resistor between the LED and the switching device (transistor or microcontroller).

I'll make a better schematic when I'm at my computer, but it would be something like this:

Vcc -> LED -> Resistor -> Switch -> Gnd.

You could use the node between the resistor and switch as an input to your external circuit. When the node goes low, your circuit gets triggered.