LED Power & Standby Indicators

[TheTinMan]

I'd like to have two LED indicators for my next build: RED to show power on and GREEN to show standby-off (IOW "play").

The red LED can be run off of the 6.3 volt heater winding with a current dropping resistor. No problem (I think).

How do I run a green LED off of B+? It's a DC voltage and way higher than anything needed for an LED. I thought of a voltage divider to get down to something reasonable (5-9 volts for example), but it seems like a zener to regulate the voltage might be needed.

If this is dead simple, please forgive my ignorance of solid state electronics.

Thanks for any advice,

Chip

[Enzo]

You don;t need the LED to run off the B+, you need it to indicate that the B+ is present where needed.

Go look at the schematic for a PV 5150. Steal their circuit, or something like it.

Power indicator is simple. put your LED and a series rectifier and a limiting resistor across the 6VAC heater winding. Same as a regular bulb really.

Then the standby light, instead of trying to run the LED directly off the B+, use some large value resistors to step it down to turn on a transistor that controls the LED.

[tyru007]

Use a DPDT or DPST switch for the standby. Use one pole to connect the B+, and use the second pole to connect the led and resistor to the 6VAC (same as the power). The poles in the switch are not connected.

[TheTinMan]

Use a DPDT or DPST switch for the standby. Use one pole to connect the B+, and use the second pole to connect the led and resistor to the 6VAC (same as the power). The poles in the switch are not connected.

Bewdy!

I knew there had to be a dead-simple way to get there. Thanks very much.

Enzo - I really appreciate the Peavey 5150 reference. That's a pretty clean way to skin the cat too.

Chip

[Enzo]

I guess the question is what do you want it to indicate, really.

If you just want to know which way your switch is set, then adding the second section to the switch is fine. The PV arrangement actually detects the B+ at the screen node to turn on the standby lite - they call it a status lite.

And nothing prevents you running the LED from the B+. The difference between running an LED off 5 volts and 50 volts is the value of the current limiting resistor in series with it. I prefer the transistor method because it keeps high voltage away from the control panel.

[Steve Conner]

I've used the two-pole approach before. The second switch pole also ran a 12V cooling fan for the power tubes in this case.

Modern high-brightness LEDs will give a decent light output with only 5mA. You can run that straight off B+ with something like a 100k, 2 watt resistor.

The advantage of this is that the LED actually tells you that the B+ is present, which is one more bit of troubleshooting information. With the two-pole switch, the LED will still come on even if the B+ is faulty.

[R.G.]

And nothing prevents you running the LED from the B+. The difference between running an LED off 5 volts and 50 volts is the value of the current limiting resistor in series with it. I prefer the transistor method because it keeps high voltage away from the control panel.

On the Workhorse series of amps, I put an LED in series with the ground side of the bleeder resistors on each filter cap. The currents were compatible, and this would let me tell at a glance whether there was voltage on the main filters, screen supply, PI, preamps.

Then we never had a power failure in an amp to use it. All dressed up and nowhere to go.

The bottom end of a bleeder resistor is near ground, and can be kept near ground by a zener diode to ground back at the bleeder resistor. That lets you run the ground side of the bleeder out to the front panel in relative safety. Zeners typically fail shorted unless there are huge currents available to vaporize bonding wires, which there is not through a big bleeder resistor, so this is pretty safe. (Nothing is absolutely safe!)

[tyru007]

Slip one of these into the front of your amp chassis to monitor B+ volts. I've been think about doing this along side a VVR control.

[TheTinMan]

You all are GREAT!

The question "What do you want the light to indicate?" isn't one I'd thought of. Frankly, I'd like one light to show that 120v AC is going to the PT. Heater supply will work well enough for that.

R.G. wins the elegance award though! A bleeder resistor will show me that there is B+.

I'm guessing that a 5 volt zener rated at 1 watt would be ok for an amp with max draw current draw of 100 ma. Is that right? Also, I still need a current limiting resistor between the zener/bleeder junction and the LED right?

An interesting side effect may be having the LED dim as the caps are drained...

Thanks again for your help,

Chip

[TheTinMan]

Does this look workable?

I tried to keep the parallel resistance between the power rail and ground at about 220K (circuit for heater center tap voltage reference plus dropping resistor for zener/green LED). Don't know if that was necessary. 470K probably should be 1 watt.

Chip

[Enzo]

If your B+ is about the 388v or so I calculate, then with 470 ohms, down to that 5v zener, you are only getting less that a milliamp of current. I don't think that is enough for the zener to zene, is it? And that is a 1W zener, even a half watt zener would be starving, I think. Then when you parallel that LED and resistor, you will be sharing that current. I don't think you will be getting 2ma through the LED, let alone 20ma.

Remember, your zener is just there to get you in the ball park, it really doesn't matter how close it regulates the voltage there.

And your drawing has the LEDs reversed. Cathodes should be towards ground with positive power supply.

With 388VDC B+, for 5ma through an LED, I get a dropping resistor of about 78k. And that is without any other parts like zeners.

[TheTinMan]

I freely admit that I don't know what I'm doing regarding LEDs and zener diodes.

The power indicator should be between the two heater supplies just like any other pilot light.

The B+ indicator LED is upside down as Enzo points out.

The value of the 470K bleeder resistor seemed reasonable to get a combined resistance between B+ and ground of about 220K. I do know that a 78K resistor between B+ and ground in parallel with the 460K formed by the voltage reference for the heater ground would be about 67K, and that would be a problem for the amp's operation.

I thought that the whole point of the zener was to maintain the voltage at 5 volts where the zener and the bleeder resistor come together.

However, I do see what Enzo is looking at. I'm guestimating B+ of about 355 volts. If I need 20ma of current with a 350 volt change, does that mean I need a 17.5K resistor? (V/I=R) I don't see how that would work.

Please help me out here.

Thanks,

Chip

[Enzo]

Ohm's Law explains the relationship between voltage, current, and resistance. If you have 355v and want 20ma, you can;t get it through a 470k resistor.

If you want to string an LED and series limiting resistor across the 6VAC heater supply like a common bulb circuit, do yourself a favor and add a regular diode in series with them. That will block reverse voltage - the LED is a dioed, but is not designed to handle much reverse voltage.

If you ONLY wanted to get a resistance to ground - a bleeeder - from the B+, then you can use the resistance you want. But when you ALSO want that resistor to power something, then you have to consider the current.

Yes, the whole point of the zener was to establish a 5v reference point, but you do have to consider that a zener diode needs a certain minimal amount of current flowing through it before it will "zene" at its voltage reliably. Now it is one thing to establish a voltage reference, but we are qadding the need to supply 20ma, or even just 5ma, for the LED. Your LED current has to come through that huge resistance, and it can;t.

SO you can look at the individual parts and get all these various answers, but a circuit is not just a list of parts, a circuit is all the parts in a relationship, and that relationship cannot be ignored. (or bunnies boil)

First, do an experiment with a battery and a pot and an LED. Find out how small a current is needed to light that LED. You don't really need 20ma.

[Steve Conner]

I think RG meant for the zener just to be a safety backup. It wouldn't conduct at all in normal use. It's just there to make sure that the bleeder resistor still functions, and that B+ doesn't appear on the LED wiring, if the LED or its wiring get disconnected somewhere.

So, you don't need to calculate anything about it at all. Just choose its zener voltage to be comfortably higher than the forward voltage of the LED you're using. Say 6.8V, because even a blue or white LED has a Vf less than 4V.

The bleeder resistor is the current limiting resistor for the LED, and also for the zener when the LED is disconnected. Either 47k or 100k should do fine.

PS: A LED used as an indicator on AC will flicker at 60Hz, which I personally find a tad annoying.

[TheTinMan]

First off, I really appreciate you all continuing to try to get me on the right track here and apologize for my lack of knowledge.

Enzo - Your battery test with the LED is just the kind of practical approach that works for me - thanks! I "get it" that everything in a circuit must be considered as a whole. I just don't know squat about solid state electronics and forgot Ohm's law.

Steve - assuming 5ma will light the LED, then a 68K bleeder/current limiting resistor between a B+ of 355 and zener voltage of 6.8v seems to meet the criteria:

(355v-6.8v)/.005ma = 69,640 ohms

However, that 68k resistor needs to be able to handle almost 2 watts right?

348.2v * .005ma = 1.74 watts

Last, I don't know what the zener's specs need to be. 6.8v * 20ma = slightly less than 1 watt, right? I guess it needs to "zene" at something less than 1/4 of that...

Also, would Enzo's diode in the power indicator circuit limit the flickering? It's not much rectification but might help. IOW:

Heater(a) > 1N4007 > 230 ohms > LED > Heater(b)

I really appreciate you mentioning flickering with AC through an LED because that would drive me nuts.

Last but not least, no one seems to be concerned about having less than 68K of resistance between the power rail and ground. It just seems like a bad idea to me. Plus, whenever I forget to disconnect my 100K cap-draining resistor it seems to affect both tone and voltages throughout the amp.

Please bear with me, and thanks again.

Chip