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# noob question: resistors needed to hook up LED to 36 volt source...

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#### therealciviczc

##### New Member
Complete rookie here. I know absolutely nothing about electronics, but I'm trying to learn as I go.

I'm trying to build an LED that replaces the buzzer on my Golf Cart. The problem is that everything on a golf cart is 36 volts, including the buzzer. I removed the buzzer and just left the leads disconnected, and everything works fine on the cart, but I can never tell when I'm in reverse, so I want an LED.

Here was my brilliant idea, feel free to frame this artwork...

The volt meter is only there for testing. So before I hooked up the LED, I tested as the voltmeter shows, from the resistor to the negative, to see what voltage the LED would be seeing. To my surprise, it was still the full 36 volts. It didn't matter if I had the resistors in there or not, the volt meter still sees 36volts.

The resistors are 1kohm, 1 watt, 5%, and wired in a series.

Any sort of direction is a step forward because I'm clueless.

Thanks, and sorry to clutter up the forum with elementry questions.

The meter will see 36V because it's only drawing a tiny current. Look up Ohm's law, V = IR, the current through the 2k resistor (two 1ks in series) is negligible so the voltage across them will be nearly 0V leaving 36V for the meter.

Look at it as a potential divider consi sting of 2k in series with 10M (the typical impedance of a DVM).

To calculate the resistor use the following formula:
$R = \frac{Vin-Vf}{If}$

In your case the current will be 16.25mA, assuming you're using a blue LED with a forward voltage, which is fine.

$I = \frac{36-3.5}{2000} = 0.01625A = 16.25mA$

The power dissipated by each resistor is equal to the following formula:

$P = I^2R = 0.01625^2 \times 1000 = 264.06mW$ which is fine, indeed you could use a single 2k 1W resistor.

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I think I see your problem. At full charge a 36V lead acid battery will be slightly greater than 36 V probably around 39V. The LED has about a 3V drop. Where you are probing it will be the voltage drop of the resistors which in this case is 36V

With a light bulb, you present it with the rated voltage, and then it takes a current related to the wattage.

With an LED, it's the other way around. You present it with an appropriate current, and then the LED 'drops' a voltage. (The drop voltage is dependent on the colour of the LED).

If you give an LED much too much current, it nearly always turns into a 'wire link' (no voltage drop). Current still passes through it OK, but it doesn't do anything. You can give an LED too much current in many ways, one of them is to accidently connect a meter set to 'current' accross the resistors.

There is not much that can go wrong with your circuit, just connect it to 36 V and the LED lights.

Finally, an LED only works when conventional current passes in the direction of the arrow in the LED symbol. If it doesn't light, try connecting the LED the other way around.

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He's measuring the voltage across the LED.

The LED must be disconnected or broken or he's measuring the wrong voltage.

Complete rookie here. I know absolutely nothing about electronics, but I'm trying to learn as I go.

I'm trying to build an LED that replaces the buzzer on my Golf Cart. The problem is that everything on a golf cart is 36 volts, including the buzzer. I removed the buzzer and just left the leads disconnected, and everything works fine on the cart, but I can never tell when I'm in reverse, so I want an LED.

Here was my brilliant idea, feel free to frame this artwork...

The volt meter is only there for testing. So before I hooked up the LED, I tested as the voltmeter shows, from the resistor to the negative, to see what voltage the LED would be seeing. To my surprise, it was still the full 36 volts. It didn't matter if I had the resistors in there or not, the volt meter still sees 36volts.

The resistors are 1kohm, 1 watt, 5%, and wired in a series.

Any sort of direction is a step forward because I'm clueless.

Thanks, and sorry to clutter up the forum with elementry questions.

DID you see that the LED is straight across the charger output?
had you connected like that, it would have died immediately !!!!!

This is a huge help and very informative. Thanks!!

*** Sorry for the bad picture. The LED wasn't hooked up when I tested because I didn't want to burn it up.

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DID you see that the LED is straight across the charger output?

No? What charger output?

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This is a huge help and very informative. Thanks!!

*** Sorry for the bad picture. The LED wasn't hooked up when I tested because I didn't want to burn it up.

You mean the LED that was very clearly shown in the diagram.... wasn't actually there?

I'd never of guessed that one!

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Oh!!!
The other item shown was Volt meter?
it should be indicated as V symbol to feel it as voltmeter. i too it for a charger.
even though, how come you get 36 there?
whether the LED is connected in reverse?
in such case there would be no current through the circuit, and the meter would read the battery potential thro the two resistors.

Sorry. I was trying to show the idea and how it was tested both in the same picture. So this is what it looked like last night.

I do feel like an idiot by the way.

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So this is what it looked like last night.

So?

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Yeah, meater reads 36 V! So?

I'm beyond clueless on this, but learning slowly. I think last night I expected the resistors would drop the voltage down, which I guess really doesn't make sense, but I think I expected they'd use up the voltage and the meter would show like 5volts for the LED.

I'm beyond clueless on this, but learning slowly. I think last night I expected the resistors would drop the voltage down, which I guess really doesn't make sense, but I think I expected they'd use up the voltage and the meter would show like 5volts for the LED.
if it is glowing, it should drop around 3.4 to 3.6 V if it is blue.
if NOT glowing, 2 reasons
1. it is connected in reverse.
2. the led is already damaged and is open circuit.

Resistors drop voltage you are right. BUT only when a current passes through them? Read fundamentals

P S: the meter will have its resistance in MEG ohms. when you connect 2 nos of 1K in series to the meter, it doesn't matter for the meter and it will still read same voltage , through theoretically less by few micro volts. so in you new diagram without LED also the 36 is read a almost 36.

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I'm beyond clueless on this, but learning slowly. I think last night I expected the resistors would drop the voltage down, which I guess really doesn't make sense, but I think I expected they'd use up the voltage and the meter would show like 5volts for the LED.

I see clearly what you mean now! Good plan of yours that is, checking the circuit first before connecting the LED. Yes, that's a reasonable assumption to make, that the resistors 'will drop voltage', just like a pressure regulator will drop PSI. Wrong of course, but reasonable. ***The resistors only drop voltage when there's a significant current flowing through them*** A volt meter takes hardly any current at all (that's the whole idea of a voltmeter by the way)

You'd do better switching the meter to ohms and measureing the resistance instead.

Also another test, if you're feeling brave, connect the meter, set to milliamps where the LED is meant to go. Check the resisters, like resist the battery current according to ohms law. You can also test ohm's law too. I did before, it works, but we all do at some time.

Beware though if you're a newbie with a meter set to amps, that it's easy to blow things up. I remember when I was a newbie and I was often blowing fuses inside the meter.

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It won't be, as it's still in its bag, unopened. The OP is fearful that it would explode if he connected it to his first ever electronic circuit.

LED

short lead is negative ( also indicated as the flat side of led )

When all is connected correctly you should be able to read the LED zenervoltage which is between 1.7 and 3.3 volts depending upon colour, and LED glowing.

The schematic is here . I hope the therealciviczc is accustome to read the electrical schematics.

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Thanks sama with your schematic which is correct.

one error (34) should be 3.4 Volts.

Cheers

Raymond

Sorry. I was trying to show the idea and how it was tested both in the same picture. So this is what it looked like last night.

I do feel like an idiot by the way.
It will read 36V because there's no current flowing through the resistors. I tried to explain this in my previous post.

Your meter probably has a resistance of 10M, look up potential divider on Wikipedia and you'll discover that this is what you should expect if you connect a 10M digital multimeter to a voltage in series with two 1k resistors.

https://en.wikipedia.org/wiki/Voltage_divider

$V_\mathrm{out} = \frac{R_2}{R_1+R_2} \cdot V_\mathrm{in}$