2
\$\begingroup\$

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.

schematic

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.

schematic

simulate this circuit

\$\endgroup\$
  • 1
    \$\begingroup\$ What colour is the SMD LED? I ask because the forward voltage drop varies with colour. \$\endgroup\$ – Transistor Sep 19 '18 at 16:03
0
\$\begingroup\$

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.

enter image description here

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.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ I like the idea of not messing around with the original circuitry! seems very simple and flexible. does anything change if that circuit has a 24V source? \$\endgroup\$ – Robyn Williams Sep 20 '18 at 9:37
2
\$\begingroup\$

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

\$\endgroup\$
  • \$\begingroup\$ Sounds like an elegant solution, as with most of this, I don't know how I would specify what opto to buy, how would I know it would work in the place of the SMD LED? \$\endgroup\$ – Robyn Williams Sep 19 '18 at 15:05
  • \$\begingroup\$ Hmm, sounds like there is a resistor in series with that LED, desolder it and replace it with the coupler input diode or put a small potentiometer between to set a good operation point. Can you post a picture of the real circuit? \$\endgroup\$ – Jan Rosum Sep 21 '18 at 19:11
1
\$\begingroup\$

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.

schematic

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.

enter image description here

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.

schematic

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.

schematic

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 VF, may draw more current. Measure and compare with datasheet.

\$\endgroup\$
  • \$\begingroup\$ Assuming it is set for 20mA. And my reading of OPs switch description is that it's independent, a typical automotive accessory switch. He's not trying to use the led to turn on the load, just the accessory light in the switch. Nice addition tho. \$\endgroup\$ – Passerby Sep 19 '18 at 16:53
  • \$\begingroup\$ Wait, you have npn transistors as a high side switch in figure 4. \$\endgroup\$ – Passerby Sep 19 '18 at 17:32
  • \$\begingroup\$ I sure do. Since the opto is photo-triggered there is no base voltage for the emitter to follow. It just conducts from C to E. You don't get the usual problems with emitter-follower circuits and you can level shift from 5 to 24 V. \$\endgroup\$ – Transistor Sep 19 '18 at 17:42
  • \$\begingroup\$ Thanks for your response! It is a blue LED, I'm going to edit my post for more clarity on the problem. it looks like an opto is actually a great solution though! @passerby It is an automotive switch yes. \$\endgroup\$ – Robyn Williams Sep 20 '18 at 8:30
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Finding the nodes is the problem I have, I don't know where to go digging to find that node! \$\endgroup\$ – Robyn Williams Sep 20 '18 at 9:27
  • \$\begingroup\$ @RobynWilliams Can you post a photo of the LED and its surrounding components? (If so, please edit it into your question and then leave a comment here to notify me!) \$\endgroup\$ – bitsmack Sep 20 '18 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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