Powering AC LED - old stereo

[frusciante89]

Hi everyone, I have an old Marantz amplifier with a broken lamp. The lamp is rated 8V AC 200mA. However, I used my multimeter to measure voltage and current: Voltage is 7.33 V AC and current is 5.68 mA (first silly question: do I have to use the AC amp or DC amp setting? cause the AC amp setting is reading a zero)
I want to replace the buld with an led, whose specifications are these:
https://www.sparkfun.com/datasheets/Components/YSL-R547W2C-A13.pdf
I suppose I have to use a diode, a resistor and a capacitor, but how can I choose the values?

Thanks for you help

 

[betwixt]

The voltage is OK, it is/was quite normal to under-run incandescent lamps when full brightness isn't needed. It greatly extends their life which is of course important when the lamp isn't easily replaceable. I'm not sure how you took the current measurment though. You would normally use an AC current range and on a lamp with those ratings I would expect the current to be around 150mA but if it's burned out the current would be zero. You can't just put a current meter across the lamp connections because the lamp itself has to limit the current, without it in series you run the risk of damaging your test meter.

Anyway, yes, that LED or just about any other "hi brightness" one will work, I suggest you add a diode in series with it, even a small signal diode like a 1N4148 will work and also add a resistor in series with them. So you have the three components in a chain across the original lamp connections. The resistor should be 180 Ohms.

The data sheet doesn't say which pin pf the LED is which but usually the cathode side has the small flat section of the body beside it. The LED and the normal diode should be facing the same way so the banded end of the diode goes to the side of the LED WITHOUT the flat on it. If in doubt, wire it up and connect it across a 9V DC battery, if you have it right, it will light up when the battery is one way around but not light when it's reversed. Because the lamp runs on AC, it doesn't matter which way round the connections to the lamp holder are made.

Brian.

 

[frusciante89]

The voltage is OK, it is/was quite normal to under-run incandescent lamps when full brightness isn't needed. It greatly extends their life which is of course important when the lamp isn't easily replaceable. I'm not sure how you took the current measurment though. You would normally use an AC current range and on a lamp with those ratings I would expect the current to be around 150mA but if it's burned out the current would be zero. You can't just put a current meter across the lamp connections because the lamp itself has to limit the current, without it in series you run the risk of damaging your test meter.

Anyway, yes, that LED or just about any other "hi brightness" one will work, I suggest you add a diode in series with it, even a small signal diode like a 1N4148 will work and also add a resistor in series with them. So you have the three components in a chain across the original lamp connections. The resistor should be 180 Ohms.

The data sheet doesn't say which pin pf the LED is which but usually the cathode side has the small flat section of the body beside it. The LED and the normal diode should be facing the same way so the banded end of the diode goes to the side of the LED WITHOUT the flat on it. If in doubt, wire it up and connect it across a 9V DC battery, if you have it right, it will light up when the battery is one way around but not light when it's reversed. Because the lamp runs on AC, it doesn't matter which way round the connections to the lamp holder are made.

Brian.

I used a IN4937 and a 180 ohms resistor but the light is very dim... do you know why?

- - - Updated - - -

should I use a transistor to amplify the signal?

 

[betwixt]

Based on the values you gave it should be OK. A transistor is not the solution because you actually have to drop the available power to a level the LED can withstand rather than increase it. The original lamp was 8V 200ma = 1.6W whereas the LED is 3.3V 0.02A = 0.06W meaning it uses 24 times less power. The efficiency is better and the brighness of the LED is much higher because it doesn't waste as much power in the form of heat.

Tell me what the model of the amplifier is and I'll try to track down a service manual for it. From that I can check the optimum values for you.

Brian.

 

[frusciante89]

Based on the values you gave it should be OK. A transistor is not the solution because you actually have to drop the available power to a level the LED can withstand rather than increase it. The original lamp was 8V 200ma = 1.6W whereas the LED is 3.3V 0.02A = 0.06W meaning it uses 24 times less power. The efficiency is better and the brighness of the LED is much higher because it doesn't waste as much power in the form of heat.

Tell me what the model of the amplifier is and I'll try to track down a service manual for it. From that I can check the optimum values for you.

Brian.

Thanks for your help...
It's model 1030. I have the manual in pdf, if you want I can send it to you...

 

[betwixt]

I've found the schematic. It doesn't specify the voltage supply to the lamp and as you only measured very low current I suspect the transformer secondary feeding it is faulty.

I suggest doing this instead:

1. forget the wires going to the original lamp holder, if they are in the way, cut them and insulate them so they can't short to anything.
2. Connect new wires to the LED, one from the cathode (flat on body side) to the ground rail which you should be able to reach at J703, J723, J724 or J725 on the P700 board.
3. Connect the other side of the LED through a resistor of 1.5K Ohms to J709 or J710 on the P700 board.

That should give you about 18mA directly from the DC supply.

Brian.