LEDs running on AC

[F0427ea1]

and here's what I did with the function lights . used a bright orange for the SQ decoder light and a bright red for stereo beacon light , both of these lens ground flat and needed black sharpie treatment also (too bright as you can see in post 20) these were running at 23 volt dc so only needed a dropping resistor

 

[F0427ea1]

I used ultra bright white for dial and meter lamps and ultra bright orange and red for the function lamps , all in all I think it turned out very nice , been going on 3 years now and this receiver gets pretty regular use , all leds holding out fine . currently set up with a Fisher 332 doing front channel duty and the 304 running 2 channel strapped doing rear channel duty , using a Lafayette SQ-M decoder for quad duty , and all driving 4 Fisher XP 7b's . sounds real nice IMHO

 

[F0427ea1]

for comparison the Fisher 332 in post #22 has stock lighting and here's a pic of a 404 receiver with stock lighting also (one meter lamp out , its getting a led conversion someday soon too)

 

[NnionzngEMF]

Hi Mike. That's the way I replaced the burned out vu meter lamp in my Kenwood 5500. Both LED's fit fine in the original bulb mount. Used a dropping resistor, small cap and it's working fine.

I have a KR-720 I'm working on troubleshooting and restoring (most likely improving also to study/practice on).
http://audiokarma.org/forums/index.php?threads/semiconductor-sources.370721/page-2

I just realized the Lamp (I see online termed as a "grain of wheat lamp") that looks like a 6 or 8V 70mA is burnt out and that feeds into the Power Indicator transistors (I think as a switch and maybe an amplifier for the LED's if not just power intensity from the S-meter... I need to read into more)... so wondering if that is causing the Power Indicator to not function other than the one middle LED.

I'm thinking since I have some old LED TV's that had broken LCD panels and I salvaged the LED strips... to use one of those or maybe two since they're rated at 6V and ~65mA each.

Looks like the line in to the Lamp is coming from the -19.7VDC (measuring ~-24VDC) single diode rectified line. Thinking I just need a resistor... though if there is flashing a capacitor would smooth out the DC ripple.

Any thoughts and/or suggestions?

Thanks!

 

[elnaldo]

If you see a bulb is part of an electronic circuit, replace it with the same kind of bulb. The circuit is designed with that current draw in mind, so fitting a led can cause problems. Even a light bulb with different current will cause some problem. I'd stay with the original design and avoid a new troubleshooting.

I recently fixed a Nakamichi 730 receiver, it was not working properly. The problem was the light bulbs somebody replaced, causing a lot of problems (transistors not biased due low current).

 

[NnionzngEMF]

If you see a bulb is part of an electronic circuit, replace it with the same kind of bulb. The circuit is designed with that current draw in mind, so fitting a led can cause problems. Even a light bulb with different current will cause some problem. I'd stay with the original design and avoid a new troubleshooting.

I recently fixed a Nakamichi 730 receiver, it was not working properly. The problem was the light bulbs somebody replaced, causing a lot of problems (transistors not biased due low current).

Thanks for the feedback. I was thinking this also since the Lamp acts like a resistor and an LED acts like a resistor and diode. I'm sure there is a math balance equation to make sure the circuit is balanced properly to make the equivalent electromagnetic component.

Logically thinking... I have to determine what components are the simplest to make the equivalent Lamp component and same function. My thoughts are using Ohms and the Power Law's to calculate out with the known values. Might need some Kirchoff's logic also... though figured I'd ask to save time and simplify.

Right, though... "like for like" component replacement is the way to go. I just have a bunch of free salvaged components and not that one. Since, if not like for like... reminds me of validated methods and alternate validated methods I used to develop... and to get Regulatory Approval we had to make sure they were "equivalent" or "equivalent and better than" the standard method.

 

[elnaldo]

Thanks for the feedback. I was thinking this also since the Lamp acts like a resistor and an LED acts like a resistor and diode. I'm sure there is a math balance equation to make sure the circuit is balanced properly to make the equivalent electromagnetic component.

Logically thinking... I have to determine what components are the simplest to make the equivalent Lamp component and same function. My thoughts are using Ohms and the Power Law's to calculate out with the known values. Might need some Kirchoff's logic also... though figured I'd ask to save time and simplify.

Right, though... "like for like" component replacement is the way to go. I just have a bunch of free salvaged components and not that one. Since, if not like for like... reminds me of validated methods and alternate validated methods I used to develop... and to get Regulatory Approval we had to make sure they were "equivalent" or "equivalent and better than" the standard method.

The problem when you replace a bulb with a LED, is the current. A 70mA bulb lets 70mA to "flow", but a led can't do that. Max current is around 20mA for a Led, and I prefer to run them at less, like 10-15mA. You should add a dropping resistor to run the led at the desired current, and add a resistor in parallel to the led-resistor network to draw 70mA in total. But electricity takes always the easiest path, so perhaps the electrons prefer your parallel resistor and not the LED, perhaps you need to add a trimpot in parallel, or try different resistors until you have 70mA total, and the desired LED brightness.

To me, is too much trouble just to replace a bulb. I'd fit another bulb and use my time in something else.

It could be also that the 70mA don't play a role in that circuit, and fitting a led-resistor is enough.

 

[NnionzngEMF]

The problem when you replace a bulb with a LED, is the current. A 70mA bulb lets 70mA to "flow", but a led can't do that. Max current is around 20mA for a Led, and I prefer to run them at less, like 10-15mA. You should add a dropping resistor to run the led at the desired current, and add a resistor in parallel to the led-resistor network to draw 70mA in total. But electricity takes always the easiest path, so perhaps the electrons prefer your parallel resistor and not the LED, perhaps you need to add a trimpot in parallel, or try different resistors until you have 70mA total, and the desired LED brightness.

To me, is too much trouble just to replace a bulb. I'd fit another bulb and use my time in something else.

It could be also that the 70mA don't play a role in that circuit, and fitting a led-resistor is enough.

Thanks for the more detailed feedback. I found the LED's I have from the broken LCD panel TV look like a higher voltage and current. I wasn't sure so I did some calculations and experimented (otherwise I was thinking digging out the TV boards and measuring to determine specifications of the LED's).

Quick math... assuming the stock Lamp is 8V and 70mA and has the following basic electromagnetic characteristics (please note if I am interpreting anything wrong as I am still not confident about voltage or current use, and whether fixed value, devices based on operation other than transistors/MOSFETs):

P(W) = V(V) × I(A)

8V x 0.070A = 0.56 W(VA)

V(V) = I(A) × R(Ω) ... solving for R(Ω) = V(V) / I(A)

8V / 0.070A = 114.285714286Ω


The LED specs appear to be somewhere in the range of (guessing mine are Sharp spec):
LG INOTEK LATWT391RZLZK 3535 3.5 x 3.5 x .65 mm ~150 250mA @ 6.0 – 6.8V

Sharp GM5F20BH20A 3537 3.5 x 3.7 x 0.8 mm 73.7 – 97.7 lumens 160mA @ 6.1 – 6.9V

6V x 0.160A = 0.96 W(VA)

V(V) = I(A) × R(Ω) ... solving for R(Ω) = V(V) / I(A)

6V / 0.160A = 37.5Ω

x2 LED's in Series:

~12V, ~0.166A, ~2W and 75Ω

or 6V x 0.250A = 1.5 W(VA)

6V / 0.250A = 24Ω

x2 LED's in Series:

~12V, ~0.250A, ~3W and 48Ω

Therefore... in this higher current LED scenario, I just need a:

114Ω - 48Ω = 66Ω resistor

The negative voltage kind of confuses me a little more. So I wonder what the calculation looks like this way:

-24 - 12 = -36V

R(Ω) = V(V) / I(A)

36V / 0.250A = 144Ω

or if 0.160A = 36V / 0.160A = 225Ω

This is where the negative voltage is confusing me. I figured I'd wire the ~ -24VDC input to the lamp to the negative terminal of the LED and have the resistors on that side.

I did a quick test last night and found the Voltage measure ~ -15VDC just past the two resistors and ~ -10VDC after the two LED's with two 100Ω resistors in series.

I added another 100Ω before the two 100Ω resistors already soldered to the negative terminal for a total of 300Ω and the voltage dropped to -13.6VDC and -8.4VDC after the two LED's.



Anything else I am missing to be safe before I start trimming this down to package?

I don't think I need a capacitor since the LED's were stable and earlier in the circuit there are after the rectifier diode.

Do you think a resistor is still needed in parallel since these LED's are higher current?

I'm guessing I need to perform KCL so I can mathematically visualize better the circuit effect at the different nodes/section. I am wondering if the specs I have for the LED's noted above are not correct since there is only about a 5V drop in voltage across the two LED's.

 

[elnaldo]

measure the voltage across the resistors, V / ohm = A. That will give you the actual current draw of your Leds. If leds are not too bright, they are in a safe operating range. If you can clip or solder some extension wires and test the circuit. You won't damage the amplifier if current is around 70mA or less , that would be around 21V across 300 ohm (21 / 300 = 0.07 A) or less.

 

[NnionzngEMF]

measure the voltage across the resistors, V / ohm = A. That will give you the actual current draw of your Leds. If leds are not too bright, they are in a safe operating range. If you can clip or solder some extension wires and test the circuit. You won't damage the amplifier if current is around 70mA or less , that would be around 21V across 300 ohm (21 / 300 = 0.07 A) or less.

Across the 200Ω resistors the Vin reading was -23.3V and Vout -15.3V. Across the LED strip Vin -15.3V and Vout -9.8V.

15.3V / 200Ω = 0.08A



Adding the additional 100Ω resistor for a total across of 300Ω the Vin reading was -23.3V and Vout -13.6V. Across the LED strip Vin -13.6V and Vout -8.4V.

13.6V / 300Ω = 0.05A


All voltage measurements were made with the ground clipped to the chassis and in V DC.

Any more details or methods I'm better consider?

 

[elnaldo]

You have 8V across 200 ohm, that's 8/200= 0.04A (40mA) and 9.7V across 300 ohm, 32mA in this case. I'd give it a try. It won't hurt to have less current. Worst case is some circuit without enough power.

Btw, leds should be backwards in your drawing if voltage is negative.

 

[NnionzngEMF]

You have 8V across 200 ohm, that's 8/200= 0.04A (40mA) and 9.7V across 300 ohm, 32mA in this case. I'd give it a try. It won't hurt to have less current. Worst case is some circuit without enough power.

Btw, leds should be backwards in your drawing if voltage is negative.

I see now... I need to calculate based on the difference in voltage for "across". I'm still struggling with visualizing easy in my thoughts.

I'm visualizing the resistors like a section of water pipe (diameter) to restrict water (charge) pressure (voltage) and flow (current). Eventually, the math will be more natural as isn't yet.

Thanks... I should have known the LED cathodes as I am remembering long term finally the component schematic (going by "C" the line is what the arrow points to and A is positive, C is negative in grade review)... I just didn't review the drawing.

I wasn't sure about measurements though since I'm self taught in electronics and RF engineering for the most part with some circuit analysis and study during my first undergrad degree Physics electricity and magnetism (which I didn't do that great... was working full time plus and student full time) and Instrument Analysis courses prior to working on the Spectrometer and Chromatographs. Then some more starting a few years back working on the FCC Amateur Technicians License. I was about to test on the General and Extra and am enjoying learning more for now higher level and recently digging into the intricate details more. I probably need to read through a book in theory again versus learning as I come across what isn't working and what I need to apply. Is more a hobby... though I do want to learn more if I can stay busy and applying more to re-enforce the memories long term. This hands on method is re-enforcing better than book only. I see why Elmer's are great to have and clubs for the radio work.

Thanks again for your support and clarification!

 

[elnaldo]

Every time you see an instruction to check a voltage "across" a resistor, it means one probe of your multimeter at each end of the resistor, so you measure the voltage drop on that resistance and you can calculate the current.

 

[NnionzngEMF]

Every time you see an instruction to check a voltage "across" a resistor, it means one probe of your multimeter at each end of the resistor, so you measure the voltage drop on that resistance and you can calculate the current.

OK, thanks for the clarification. I get nervous on live systems and typically to make sure I don't damage anything (including myself) have built the confidence to use an alligator probe clip on the ground (chassis), or if AC then I get more risky with what is testing continuity powered off, and then probe around at the different node/via/location. I figured I wouldn't be at any risk to fry anything or have to use two hands on a live system.

Directly measuring current is different and I have to read into before I do anything since I haven't in a year or so.

I like the calculating method. That is making more sense. I am eager to practice more.

Do you think there is a better more equivalent voltage to have at the end of the LED strip since the lamp out was 8V 70mA going to three transistors emitters and one of the transistors in the middle collector of the Power Level Indicator?

Would I be better to consider a stable 8V drop from the lamp or is that surge to start it and then the difference of the ~ -24VDC isn't ~-18VDC leaving the lamp?

 

[elnaldo]

To measure on a live system is not risky if you clip your probes firmly with the unit OFF, and the turn ON the amp to read the voltages, and turn it OFF again without touching anything.

I think at this point you should start a new thread with your particular problem so we don't write more pages here in a generic LED conversion thread. You can post a link here to the new thread if you like.