Ohms, help me understand where I'm going wrong?

Thread Starter

MissOrange

Joined Nov 7, 2014
18
Ohms, can you help me understand why I'm getting incorrect numbers?

So... I've taken apart a simple 3led touch button light. It has a 10ohm resistor (colours Brown, black, black, gold) 3 triple A batteries are used to power it.

What I don't understand is. If I have 3 white LEDs, gonna assume they have a voltage drop of 3.3 each. So power supply 4.5v - (3.3v x 3) / 20ma (assumed current of led) gives me -270ohms. What am I doing wrong?

What I want to do...is take these three LEDs and use a USB power bank output is 5.3v 1000ma and capacity is 2700ma.

I'm trying to work out what resistance I need but obviously doing it wrong.
 

Audioguru again

Joined Oct 21, 2019
5,162
The LEDs are in parallel which is bad unless they are tested and selected to have exactly the same forward voltage.

You need to measure the forward voltage of the LEDs or find it on their datasheet. If they are 3.6V then the current was (4.5V - 3.6V)/10 ohms= 90mA. There were three LEDs in parallel so each one got 30mA. You wrongly assumed the LEDs were in series which needs 9.9V that you did not have.

With a power bank output of 5.3V and a total current of 90mA then the resistor must be (5.3V - 3.6V)/90mA= 18.9 ohms.
 

Thread Starter

MissOrange

Joined Nov 7, 2014
18
3.3 x 3 = 9.9V ??? Are the LEDs in series or parallel?
Assuming white LEDs in parallel, the voltage is 3.3.
4.5-3.3=1.2 volts across the resistor.
3.3 x 3 = 9.9V ??? Are the LEDs in series or parallel?
Assuming white LEDs in parallel, the voltage is 3.3.
4.5-3.3=1.2 volts across the resistor.
Thanks for your reply. So I was calculating it for a series.

Is it correct to add the voltages together for a series circuit?

I assumed the cheap torch was in series because it contained 3 leds, a push button, and a 10 ohm resistor. I thought for paralell each led needs its own resistor?
 

Thread Starter

MissOrange

Joined Nov 7, 2014
18
The LEDs are in parallel which is bad unless they are tested and selected to have exactly the same forward voltage.

You need to measure the forward voltage of the LEDs or find it on their datasheet. If they are 3.6V then the current was (4.5V - 3.6V)/10 ohms= 90mA. There were three LEDs in parallel so each one got 30mA. You wrongly assumed the LEDs were in series which needs 9.9V that you did not have.

With a power bank output of 5.3V and a total current of 90mA then the resistor must be (5.3V - 3.6V)/90mA= 18.9 ohms.
Thanks for your reply. Would you mind explaining why they are in parallel? WhenI took apart the cheap torch there were three LEDs, a push button and a 10 0hm resistor. I assumed this was in series. I thought in parallel each led needs it own resistor, is that correct?

With the power bank setup, is it 90mA total because I add up the current if each led? Is this also a parallel circuit? Would I need a resistor on each led?
 

Ian0

Joined Aug 7, 2020
6,286
I assumed the cheap torch was in series because it contained 3 leds, a push button, and a 10 ohm resistor. I thought for paralell each led needs its own resistor?
Absolutely correct - each should have 30Ω.
But you did say it was a "simple" light, by which I guessed you meant "cheap"!
A manufacturer can ask the LED supplier to supply LEDs which are closely matched in voltage, and that's probably cheaper than buying (and fitting) the extra two resistors.
Also, even if the LEDs didn't share the current evenly, they would be unlikely to be so far out that any one of them exceeded the maximum current.

If you look at the V vs. I curve for a typical LED, firstly you can work out the LED current by plotting the voltage vs. current for the resistor (just like biassing a triode valve!). The point where the two lines intersect is the LED current.

Secondly, you can measure the slope around the operating current and work out the LED's incremental resistance from ΔV/ΔI, which I get to about 15Ω (one square up for about 0.4 squares across). So, there is effectively a 15Ω resistor in series with each LED, so mismatched forward voltages won't be too much of a disaster. LED life might be reduced, but how long is it going to operate for? At 90mA it will only operate for 11 hours on a set of batteries. It will take 4500 sets of batteries before it gets to the normal rated life of the LED.

Screenshot at 2021-08-22 07-29-06.jpg
 

sparky 1

Joined Nov 3, 2018
731
Ohms law is based on proportional relationship between Voltage current and resistance.
It is expressed that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.

After the standard unit resistor was established it got easier to understand. If the resistor measurements was always true then and the other two units were established there was enough information to complete Ohm's study. If he only had an led there could have been problems and precision was primitive outside the better labs.

Georg Ohm wanted to know what would happen when he changed the number units of voltage while holding the resistor constant. The question was is there a proportional change in current ? There was also a need for precision standards and academic agreement that would follow we can see that solving for voltage knowing current and resistance were derived in what is now called Ohm's law 1827

https://en.wikipedia.org/wiki/Ohm's_law

After the units became more established the replications increased the study became more familiar with accepted and repeatable results.
We can see a standard resistor 34 years later in 1861 with capability of higher precision, Ohm's law had become accepted.

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


But the experiment still has to be done very carefully. The mulimeter will have % error and the diode unlike a conductor will introduce nonlinearity.
The (light emitting) diode's I /V curve equivalent model and thermal drift can be used otherwise % error multimeter and using an led is just something to live with.

https://en.wikipedia.org/wiki/Diode_modelling
 
Last edited:

DickCappels

Joined Aug 21, 2008
9,299
I have had some cheap LED torches in which all of the white or UV leds were in parallel and powered by 3 x 1.5 volt batteries. Leds with that high of a voltage drop are much better at sharing current than (for example) red LEDs. It is possible that they also bought LEDs binned by voltage, at least the better would...maybe.
 

Audioguru again

Joined Oct 21, 2019
5,162
I also have a cheap "electric torch" that has 24 white LEDs in parallel. The factory must have someone measuring the forward voltage of all the LEDs and piling them in 3.0V, 3.1V, 3.2V, 3.3V, 3.4V etc groups so that each LED does not need its own resistor.
See MissOrange? You assumed too many things like the forward voltage and the battery voltage.

I think your torch was designed to use cheap "super heavy duty" poor quality and weak batteries like mine then the 4.5V is actually only 4.0V. When I replaced the batteries with good powerful ones the current was so high that the LEDs almost quickly burned out. I needed to increase the resistance of the resistor. Most ordinary 5mm diameter LEDs have a maximum allowed current of 30mA. The math is (4.5V - 3.3V)/10 ohms= 120mA total or 40mA for each LED. Then maybe the torch factory selected a different resistor value to match the forward voltage of the group of LED plus the reduced voltage of the weak batteries.
 

ronsimpson

Joined Oct 7, 2019
2,386
When I designed LED torches, make many millions/year, parallel LEDs were chosen by their voltage. ( binned ) I can get the LEDs sorted from the factory. PART_NUMBER-3.1 or -3.2
Using 3 white in series LEDs you need a voltage source of about 12 Volts. This requires a PWM circuit that makes a constant current not really a voltage.
Because you want simple and 3 batteries, parallel LEDs with 3 resistors is where you are going to go. Maybe you can use one resistor and not care that one LED is pulling more current than another. I find that your LEDs might have close to the same voltage because they were made together. When getting 10s of millions in a year you will find the voltages are not the same. When you get the lowest cost LEDs you will find that I got all the 3.1V and AudioGuru got all the 3.2V and Dick ordered all the 3.3V which leaves the 3.0V and 3.4V which gets pushed off on ebay and aliexpress. (yes mixed together)
With a resistor (or 3) the light will be bright with fresh batteries and dim as the batteries age. Years ago I made LEDs for shoes. The Boss wanted to save the cost of a resistor. (fraction of a copper coin) So we chose a low end battery with high internal resistance. The good news is that children's feet grow fast and no one cares if the LEDs don't last.
 

ronsimpson

Joined Oct 7, 2019
2,386
one bowl of rice per day
To save one bowl of rice, a machine sorts the LEDs by voltage and color and lumens in the testing process.
Point taken, some people do dangerous work, in very hot weather, no AC, for one bowl of rice because 1 bowl is better then 0 bowls. In some places the government collects people off the street, calls them criminals, and makes them work years for free.
 

Thread Starter

MissOrange

Joined Nov 7, 2014
18
Thanks to everyone for their help. For anyone reading who is asking the same question with tips from here I then learnt about how to find the forward voltage which was a process of… Measuring my test battery (10.3v) place resistors value 1150 ohms in series at the positive end, then place the led and wire to negative.Then measure the led - 2.7v. Then 10.3v - 2.7v = 7..6v /1150 ohms = 6.6 ohms. So 2.7v is the forward voltage and (I think!?) 6.6 ohms is the forward current. Which then allowed me to put these figures into a series resistor calculator to find that in series I needed a 680 ohm resistor when using 12v. (I’ll be trying the 5.3v with a parallel circuit and different resistors.

in case anyone is interested, my purpose here is a couple of ideas. One is creating a light powered by USB. The other is creating coloured shadows. I’ve made a lens for each led one in red, one blue and one green. I’ll be attaching these to an arduino which is set up with potentiometers to change both pwm and duty cycle (if I haven’t got the terms right there, one is changing the length of on off - in time, one is changing the amount of current (width of pulse). This produces a strobe effect. I’ve got this strobe effect hooked up to a water system where a stream of water becomes sample aliased (stroboscopic efffect) to produce levitating water.
 

eetech00

Joined Jun 8, 2013
3,418
Thanks to everyone for their help. For anyone reading who is asking the same question with tips from here I then learnt about how to find the forward voltage which was a process of… Measuring my test battery (10.3v) place resistors value 1150 ohms in series at the positive end, then place the led and wire to negative.Then measure the led - 2.7v. Then 10.3v - 2.7v = 7..6v /1150 ohms = 6.6 ohms. So 2.7v is the forward voltage and (I think!?) 6.6 ohms is the forward current. Which then allowed me to put these figures into a series resistor calculator to find that in series I needed a 680 ohm resistor when using 12v. (I’ll be trying the 5.3v with a parallel circuit and different resistors.
The best approach, IMO, would be the one described by lan0 in post #6. Use LED's that have known data as in the LED forward I/V chart.
 

eetech00

Joined Jun 8, 2013
3,418
How would you advise going about it if you didn’t have the data sheet?
If I were going to use a small quantity of unknown LEDs for a "one time" project, I would make measurements of each, as needed, using different resistor values and a regulated power supply, until I felt confortable with a consistent value. Easiest would be to use a regulated DC power supply with adjustable current regulation.
But If this were for a large quantity of product, I wouldn't use or buy LED's that don't have a manufacturers datasheet.
 

Audioguru again

Joined Oct 21, 2019
5,162
Voltage divided by resistance equals current which is measured in Amps, not ohms. Then (10.3V - 2.7V)/1150 ohms= 0.0066A which is 6.6mA.
Most LEDs are rated with a current of 20mA so with your 10.3V power supply the resistor will be (10.3V - 2.7V)/20mA= 380 0hms. 390 ohms is the closest common value.

For a 12V supply the resistor will be (12V - 2.7V)/20mA= 465 ohms. 470 ohms is the closest common value.
 
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