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| Author: | Pelliott321 [ April 9th, 2022, 3:49 pm ] |
| Post subject: | Re: need help |
ok so this is working now but I still do not understand why. The 100k parallel resistor is needed, without it the LED doesn't last a millisecond. The voltage drop across the LED is 1.96 Vdc on my Fluke 77. Should I add a bit of resistance in series with the LED to increase longevity |
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| Author: | tomp [ April 10th, 2022, 12:21 pm ] |
| Post subject: | Re: need help |
Attachment: I must say that I don't understand that either. With an LED, there is very little flow through the diode until about 1.5V. Then the current increases but not in a linear fashion until it is high where, like a regular diode, it becomes more resistive. You should only have to worry about the correct value of the series resistor to limit the current to acceptable levels, not a parallel resistor.Tom |
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| Author: | Pelliott321 [ April 10th, 2022, 1:05 pm ] |
| Post subject: | Re: need help |
yes, I have both monoblocs repaired. I settled on 150 ohms in series with the B+ LED. It is still plenty bright. the only thing is that since the voltage source is coming from the B+ filter caps, when you turn off the B+ it takes a few mins for the lamp to be completely off. Thanks you for your help. If I was keeping these, I would probably go for a delay circuit, but since I am selling them I wanted it to be stock, or as stock as possible. |
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| Author: | GaryB [ April 10th, 2022, 3:02 pm ] |
| Post subject: | Re: need help |
tomp wrote: I must say that I don't understand that either. With an LED, there is very little flow through the diode until about 1.5V . . . It must be a transient turn on effect where the voltage across the LED rises above the breakdown voltage before it has a chance to turn on. I'm just speculating, but my guess is that before the LED turns on, its impedance is very high - many megohms. So the voltage divider of the two 200K resistors (100K effective) and the LED (megohms) puts most of the B+ across the diode until it turns on. The 100K resistor in parallel with the LED drops the voltage across the LED by a factor of 2 (100K / 200K). As the B+ ramps up, that must be enough margin to keep the LED from blowing up before it can turn on. If you wanted even more margin, you could lower the value of the 100K resistor in parallel with the LED. |
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| Author: | Roscoe Primrose [ April 10th, 2022, 3:14 pm ] |
| Post subject: | Re: need help |
GaryB wrote: tomp wrote: I must say that I don't understand that either. With an LED, there is very little flow through the diode until about 1.5V . . . It must be a transient turn on effect where the voltage across the LED rises above the breakdown voltage before it has a chance to turn on. I'm just speculating, but my guess is that before the LED turns on, its impedance is very high - many megohms. So the voltage divider of the two 200K resistors (100K effective) and the LED (megohms) puts most of the B+ across the diode until it turns on. The 100K resistor in parallel with the LED drops the voltage across the LED by a factor of 2 (100K / 200K). As the B+ ramps up, that must be enough margin to keep the LED from blowing up before it can turn on. If you wanted even more margin, you could lower the value of the 100K resistor in parallel with the LED. Or put a cap across the LED.... Roscoe |
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| Author: | Pelliott321 [ April 10th, 2022, 3:15 pm ] |
| Post subject: | Re: need help |
what size cap? |
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| Author: | GaryB [ April 10th, 2022, 4:17 pm ] |
| Post subject: | Re: need help |
Pelliott321 wrote: what size cap? Something small - 0.01uf to 0.1uf would do the trick. |
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| Author: | tomp [ April 10th, 2022, 4:19 pm ] |
| Post subject: | Re: need help |
GaryB wrote: tomp wrote: I must say that I don't understand that either. With an LED, there is very little flow through the diode until about 1.5V . . . It must be a transient turn on effect where the voltage across the LED rises above the breakdown voltage before it has a chance to turn on. I'm just speculating, but my guess is that before the LED turns on, its impedance is very high - many megohms. So the voltage divider of the two 200K resistors (100K effective) and the LED (megohms) puts most of the B+ across the diode until it turns on. The 100K resistor in parallel with the LED drops the voltage across the LED by a factor of 2 (100K / 200K). As the B+ ramps up, that must be enough margin to keep the LED from blowing up before it can turn on. If you wanted even more margin, you could lower the value of the 100K resistor in parallel with the LED. Not true. When in the forward condition which is what happens in the normal operation of the LED, there is no high impedance related to turn on speed. There is also no breakdown voltage in the forward condition. It behaves like any other diode in that sense. We used to use protection diodes in our commercial electronic flash units that were subject to very high rates of rise of both voltage and current as the flash tube was triggered with no ill effects. The major problem of any PN junction in the forward condition is current high enough and long enough to damage the chip. There is a different condition for switched devices like SCRs that have a di/dt rating where after triggering, applying a current at too high a rate can cause the area of the die around the gate to overheat because the die has not had a chance to fully turn on in the short term. That does not happen with diodes since they are not switched. |
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| Author: | GaryB [ April 11th, 2022, 11:00 am ] |
| Post subject: | Re: need help |
tomp wrote: Not true. When in the forward condition which is what happens in the normal operation of the LED, there is no high impedance related to turn on speed . . . I probably wasn't explaining myself well. There is clearly some sort of destructive effect going on since P Elliott reported the LEDs were blowing up on turn on without the parallel 100K resistor. The question is what is the cause of this. I was merely pointing out that the impedance of any diode is a function of the current passing through it. If you look at I-V curve of the LED that you attached a few posts earlier, the impedance of the diode is just the inverse of the I-V slope and that varies with current. When the LED diode is turned on and light is emitted (current around 20mA or higher), it's a relatively low impedance. When the voltage across the diode is lower, say <1.5v, then the current through the diode will be <<1mA and the impedance of the diode will much higher. But as I think about it further, the more important factor is the change in current through the LED with time. When the LED turns on, there is a sudden increase in current through the diode. That in turn could give a voltage spike due to the inductance of the dropping resistors and the associated wiring. It's the standard effect where V = L dI/dt. Experimentally, a parallel resistor solves the problem. And I suspect that Roscoe's parallel capacitor would also be highly effective. One can do both if one believes in belt and suspenders. |
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| Author: | SoundMods [ April 11th, 2022, 12:32 pm ] |
| Post subject: | Re: need help |
Or put it another way -- there is no such thing as a simple LED pilot light. 
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