Could you dedicate a SPDT set of relay contacts to the other indicator? A DPDT set of contacts would reduce this to a trivial exercise.

 

Yes, the LED lets you know whether the relay is switched on or off. It runs a "fat" switch on the plates of the tube. The ground reference needed to be elevated to reduce hum.

 

Could you dedicate a SPDT set of relay contacts to the other indicator? A DPDT set of contacts would reduce this to a trivial exercise.

I'm using a SPDT relay. The other set of contacts are used for a voltage control. Would I need to use two relays?

 

I messed up again! I'm using a DPDT relay. I'll attach the whole schematic. Maybe that will make it easier to understand. It's a musical instrument preamp.

 

All that's needed for that is a ground. The transistors serve no useful purpose.

A couple of transistors probably work out cheaper than upgrading to a DPDT switch.

The inverse parallel pair of LEDs need nothing more than a series resistor to limit the current.

In the standard Ohm's law formula - you just need Vcc minus LED Vf, and what current value to run the LEDs at.

 

Change the 82 ohm to 330 ohms.
Move the transistor grounds to the right side of the bridge rectifier.
Change the base resistors to 680 ohms or so.

 

Change the 82 ohm to 330 ohms.
Move the transistor grounds to the right side of the bridge rectifier.
Change the base resistors to 680 ohms or so.

That seems to work, but I have 3.5 VDC at the LED. Is that too much?

 

That seems to work, but I have 3.5 VDC at the LED. Is that too much?

Don't forget ground for your meter is now the right side of the bridge.

 

That seems to work, but I have 3.5 VDC at the LED. Is that too much?

That's near right for blue and white LEDs. Green LEDs; expect 2.0V or thereabouts, Red LEDs come in around 1 3/4V.

 

Heres LED wiring with DPDT switch

 

The LED circuit below is what I've decided that I need. The switch idea is fine, but I also have a foot switch option. I wonder what the ideal values for the resistors would be if I were to use this in the circuit. I'm Nbtone, by the way. I couldn't remember my password, and the password reset doesn't work. This way there are no transistors.

 

This way there are no transistors.

there was never a need for the transistors.

your approach is also inefficient: there is a constant flow of current from the ground / rail. it can be made a lot more efficient.

 

there was never a need for the transistors.

your approach is also inefficient: there is a constant flow of current from the ground / rail. it can be made a lot more efficient.

There is a constant flow of current when the relay is switched on. When it is switched off the only current flowing is for the LED. I'm not sure how it could be more efficient?

 

There is a constant flow of current when the relay is switched on. When it is switched off the only current flowing is for the LED. I'm not sure how it could be more efficient?

I'm not clear on the 2 ground symbols. Is there a reason for that?
There is no voltage shown in your circuit.

 

View attachment 113764
Here is a more complete schematic. The relay shows only one side, and there are two circuits that it switches. I was a little confused about where I was finding the voltage. I went and spent some more time studying, and these are correct voltages. The two 100 ohm resistors used to go straight to ground, but I've had to do some ground referencing. Now the LED won't work. Sorry about the earlier posts.

The SPDT switch is part of the relay?

 

View attachment 117812


The SPDT switch is part of the relay?

Yes.

 

I'm not clear on the 2 ground symbols. Is there a reason for that?
There is no voltage shown in your circuit.

The two ground symbols only mean that the current flows through the LED to ground from the opposite B+. I haven't built one with this design in this circuit yet. I know that it works, because a built it in a variation on the circuit that I'm going to use. The problem is that I'm getting a much higher reading for the forward voltage. I made a DC filament supply and tried it. The circuit I've shown uses an AC filament supply.

 

The two ground symbols only mean that the current flows through the LED to ground from the opposite B+. I haven't built one with this design in this circuit yet. I know that it works, because a built it in a variation on the circuit that I'm going to use. The problem is that I'm getting a much higher reading for the forward voltage. I made a DC filament supply and tried it. The circuit I've shown uses an AC filament supply.

I bet you are.
If you have 2 sets of contacts like @hp1729 shows you can make the circuit like what @eetech00 posted. Don't forget the resistor.

 

I bet you are.
If you have 2 sets of contacts like @hp1729 shows you can make the circuit like what @eetech00 posted. Don't forget the resistor.

There are two switches, but they aren't both connected to the LED circuit.

 

I just put everything back together with an ac filament voltage, and I have 7.2 VDC to feed the LED.