Electronics coursework help

Thread Starter

dazzer7

Joined May 7, 2016
26
My coursework was to work out a parallel circuit on paper with values then to take it on to a breadboard to test , I can have 12v power and three LEDs and one resistor only . I have guessed the red LEDs VF and If as i don't have the information , VF 3v and If 20mA or 0.02a) The resistor value it unknow yet , so to work out R each LED has 20mA going though it, the current out thought the resistor is 3 x if so 3x20 = 60mA or 0.06a , next we have 12Vs the voltage across the Led is Vf 3v , R= Vs - Vf = / 3iF 12-3= 9 then 9 divided by 0.06 = 150 ohms . I am new to Electronics and the problem is the resistor gets very hot i am not sure what i am doing wrong .20211211_105637.jpg
 

bertus

Joined Apr 5, 2008
21,861
Hello,

When you have a 12 Volts powersupply and 3 leds, put the leds in series with a smaller resistor.
The voltage accross the resistor will be much less, so less power.

Bertus
 

Thread Starter

dazzer7

Joined May 7, 2016
26
Hello,

When you have a 12 Volts powersupply and 3 leds, put the leds in series with a smaller resistor.
The voltage accross the resistor will be much less, so less power.

Bertus
the problem is i have to do this in parallel as more of a test not that i would do LEDs like that , If the calculations are correct and ohm's law then why is the resistor so hot ,
 

crutschow

Joined Mar 14, 2008
29,520
the problem is the resistor gets very hot i am not sure what i am doing wrong .
You are doing nothing wrong.
Resistors get hot depending upon the power dissipated.
Do you know how to calculate the power dissipated by that resistor?
If so, you will see why it is very hot (note the resistor power rating)
 

dl324

Joined Mar 30, 2015
14,325
I am new to Electronics
You should start posting well drawn schematics, as they're easier to read than tracing a breadboard. The only time we need to see a breadboard is if incorrect wiring is suspected.

When posting pictures, well focused are better. I can't read the color code for the resistor. Use a high (larger number) f-stop for a wider depth of field.
 

MrChips

Joined Oct 2, 2009
25,930
You are not doing anything wrong, except that your resistor is underrated for the power it has to dissipate.

Calculate the power that the resistor has to dissipate. This is emitted as thermal power. Besides thermal capacity, what you are sensing with the tip of your finger is temperature rise. The temperature will be lower if you can remove the thermal energy fast enough.

Take the calculated power dissipation and then double that for your resistor power spec required.

Your resistor is rated for only 0.25W.
 

ericgibbs

Joined Jan 29, 2010
15,360
hi d7,
As you choose 3Vfwd for the Red LED, that means the current when using a 150R series load would give 60mA, as you stated.
But if you had selected the usual 2Vfwd for a Red LED, means when using that 150R,
the current is now 66.6mA.

So calculated power dissipation in the 150R for @ 60mA.
ie: 9Vdrop across the 150R = 540mWatts.

But actual is 10v/150R -= 66.6mA, so resistor dissipation is 10V* 66.7mA = 670mW

A percentage increase of 100 * (670mW/540mW) = 1.24 an increase of ~25%

So you can see that difference of 1volt the LED Vfwd can increase the resistor dissipation a lot.

Use a 1Watt resistor…

E
 

Thread Starter

dazzer7

Joined May 7, 2016
26
just reporting back to say thanks for all your help lots of good information here, I completely forgot about the watt's for the resistor , Because i don't have any resistor with the correct wattage for this I decided to change a few things so i can say i did something that worked or didn't - I changed power supply to 6v - 6 LEDS and i did use IF 15 mA instead of 20mA, i did test a single red LED on my power supply and it was ok at 2v 15mA but i could have used 20mA . Sorry if the circuit schematic is not great i am still learning i use letters A. B. C . for the voltage difference, for the 66ohm's i use 100 ohm's as i didn't have a 66 ohm's resister, although the watts calculated to 0.36w would be to hot again i reduced the amps to 0.04a which gives 0. 16w with still enough brightness on the LEDs and the resister was cool . pc.jpg
 
Last edited:

bertus

Joined Apr 5, 2008
21,861
Hello,

It is not wise to put leds direct in parallel.
They will never have the same properties.
The one with the highest current will fail first.
Than all others will get more current and again the one with the highest current will fail, etc.
it is like a chain reaction, until all leds are dead.

Have a look at the attached PDF.

Bertus
 

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dl324

Joined Mar 30, 2015
14,325
Because i don't have any resistor with the correct wattage for this I decided to change a few things
You can put lower wattage resistors in parallel to increase power handling capability. For example, 5 330 ohm 1/4W resistors in parallel gives you 66 ohms and would let you dissipate around 1W. We have to derate power dissipation ratings at higher temperatures.
I changed power supply to 6v - 6 LEDS and i did use IF 15 mA instead of 20mA,
When you operate LEDs in parallel without ballast resistors, the one with the lowest forward voltage is going to hog current. That can set up a cascading failure; particularly if you operate the LEDs at their maximum continuous current rating. If one LED fails, the rest will now be operating at 120% of their maximum current. Then another will fail and the remaining will be operating at 150%. And so on...

Unless you have LEDs that have matched forward voltages, this is a typical variation (from Kingbright):
1639326169620.png
1639326121191.png
1639326149929.png
Sorry if the circuit schematic is not great i am still learning i use letters A. B. C .
It's much easier to read than a breadboard. We don't normally label nodes. When you have wires that cross and connect, we use a connection dot. If you don't use dots, you have to use "humps" for crossings that don't connect. That style has fallen out of favor.

1639325896299.png
The connection dots on LEDs 3-5 are optional. My style is to use them.

When you have multiple components of the same type, component designators are required.
 
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