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Need help to decipher indicator lamp specifications

ionbogie said:
Wow, I work on my A-X9 in same time like you , last year, burned unlimited sets of LEDs with 510ohms resistors on them , also new 10uf caps just installed
wend bad in a week!
I would go for the R481 and R482 but my question is ; do you have 14V LED in it at this time, or you have just regular LEDs?
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Ionbogie, the LEDs I am using have a 25ma steady current rating. As Ylli said, these are current driven devices, so the voltage drop they see individually doesn’t matter.

For most applications a resistor in series with the LED will cause a current drop (by loading the circuit), but in this case a resistor does not do that. The whole circuit needs to be modified to lower the constant current.

I have about 40 hours on my new LEDs so far with the replacement of R481 and R482. This is much longer than the LEDs lasted previously when I tried to install them.

Because LEDs are very efficient, there are probably a lot of resistor values that will work. I just used 470 ohm because that is what I had. When I measured the current after installing them it was good.

Ylli said:
Theoretically, you do not need a resistor. In practice, I think I would add series resistance so that the collector voltages of X437 and X438 were as specified on the schematic (about 14.6 volts). With 5 mA, perhaps start at 1.8K.
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Measuring the voltage won’t work because the LEDs have a different resistance than the incandescent lamps. There are several good places on the circuit boards to measure the current, though (it has some jumpers which are easy to remove). So the strategy would be to try a resistor value and measure the current in the lamp circuit. When it is in the range that the LEDs selected can handle, then you’re all set since it is a constant current circuit. Having more lamps lit doesn’t change it.
 
Corktowner said:
Measuring the voltage won’t work because the LEDs have a different resistance than the incandescent lamps. There are several good places on the circuit boards to measure the current, though (it has some jumpers which are easy to remove). So the strategy would be to try a resistor value and measure the current in the lamp circuit. When it is in the range that the LEDs selected can handle, then you’re all set since it is a constant current circuit. Having more lamps lit doesn’t change it.
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The Constant current sources will set the current. With no resistors in series the voltage at the transistor collectors will be the forward drop of the LEDs. (Without looking too close I think there are always 2 lamps in series) So assuming an LED drop of 2.0 volts, that would be 4.0 volts. If you want the transistor collectors to be at 14 volts, then you need to drop an additional 10 volts. If you have the current source set to about 5 mA, that means you would need to add a series resistor of 2.0K.
 
Ylli said:
The Constant current sources will set the current. With no resistors in series the voltage at the transistor collectors will be the forward drop of the LEDs. (Without looking too close I think there are always 2 lamps in series) So assuming an LED drop of 2.0 volts, that would be 4.0 volts. If you want the transistor collectors to be at 14 volts, then you need to drop an additional 10 volts. If you have the current source set to about 5 mA, that means you would need to add a series resistor of 2.0K.
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That’s not correct. First, all of the lamps are in series, and there can be anywhere from 1 to 5 lit on the first circuit and 1 to 4 lit on the second circuit. One could have a 10V or a 2V drop, but you can’t keep changing resistor values. The transistors change the voltage in the circuit based on the load, but keep the current constant.

If you look at the circuit diagram closely you can see that the voltage drop across R481 and R482 is 5.1V. With the resistance equal to 150 Ohms, this give a current of 34mA. That’s I e. I b is going to be really small because there is a 22k Ohm resistor connected to the base, so I e is approximately equal to I c, and that is what I measured.

Now, the part I don’t know is how to calculate is what the load of the entire lamp circuit, transistors and all. But I correctly guessed that the current could be reduced by increasing the value of R481 and R482. I also know from experimentation that adding the same resistor on the collector side of X437 and X438 did not change the current flowing through them. Luckily in my case the 470 Ohm resistor reduced the current to about 5mA
 
The wiring of all those bulbs is convoluted and I was taking the "two lit at a time" from earlier in the thread. Story changes a bit if 1 - 5 can be lit.

The constant current sources will have an output current of 5.1/Re, and a compliance of 0 - 45 volts.

[expanded description: Emitters will be 0.7 volts above the base voltage. On the schematic it is listed as 49.0 volts on the bases, so 49.7 volts on the emitters. The high end of the emitter resistors is connected to a 54.8 volts line, which gives a voltage across those emitter resistors of 5.1 volts, as you said. I = E/R = 5.1/150 = 34 mA. If you increase the emitter resistors to 470 ohms, the voltage across them will still be about 5.1 volts, and the current will be I = E/R = 5.1/470 = 10.9 mA. That is a reasonable current for a typical LED. (1K would get you down to 5 mA)]

The constant current source transistors are fed from a reasonably high voltage, so they have a compliance (output voltage capability) of 0 - 45 volts. Series resistors are not really needed.
 
Ylli said:
The wiring of all those bulbs is convoluted and I was taking the "two lit at a time" from earlier in the thread. Story changes a bit if 1 - 5 can be lit.

The constant current sources will have an output current of 5.1/Re, and a compliance of 0 - 45 volts.

[expanded description: Emitters will be 0.7 volts above the base voltage. On the schematic it is listed as 49.0 volts on the bases, so 49.7 volts on the emitters. The high end of the emitter resistors is connected to a 54.8 volts line, which gives a voltage across those emitter resistors of 5.1 volts, as you said. I = E/R = 5.1/150 = 34 mA. If you increase the emitter resistors to 470 ohms, the voltage across them will still be about 5.1 volts, and the current will be I = E/R = 5.1/470 = 10.9 mA. That is a reasonable current for a typical LED. (1K would get you down to 5 mA)]

The constant current source transistors are fed from a reasonably high voltage, so they have a compliance (output voltage capability) of 0 - 45 volts. Series resistors are not really needed.
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Thank you very much guys, I order the LEDs from dgwojo. will replace all of them I hope for the last time
Instal the two mentioned resistors and put it back to daily play duties.
By the way my amp has all the output transistors replace as an upgrade I understand from previous owner, I can look to see what they are and post back if somebody is interested in restoring an A-X9.
 
ionbogie said:
Thank you very much guys, I order the LEDs from dgwojo. will replace all of them I hope for the last time
Instal the two mentioned resistors and put it back to daily play duties.
By the way my amp has all the output transistors replace as an upgrade I understand from previous owner, I can look to see what they are and post back if somebody is interested in restoring an A-X9.
Click to expand...

Good luck with the replacement, and let us know how it works! And of course if you’ve got more upgrades to share, please do! Just start a new thread
 
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