Logic gates - open input ciricuit resistor needed

Thread Starter

BarryTron

Joined Nov 18, 2018
89
I have been trying to understand the logic gates of diodes. In trying to replicate the example in the video that starts in 03:14 the resistor i use gets extremity hot, and my LED never fully turn OFF. Any advice/help will be appreciated.

mmmmmmmmmmm.JPG
 

dl324

Joined Mar 30, 2015
12,927
What is the resistor value, supply voltage, and the low level that you're applying to the LED cathodes?

Which leads are for the DVM?

Didn't watch the video.
 

AlbertHall

Joined Jun 4, 2014
11,398
If you are using a 12V supply as in the video, the resistor should be between 470Ω and 10kΩ to keep the current within a sensible range. Such a resistor will not get too hot with a 12V supply.
 

Thread Starter

BarryTron

Joined Nov 18, 2018
89
If you are using a 12V supply as in the video, the resistor should be between 470Ω and 10kΩ to keep the current within a sensible range. Such a resistor will not get too hot with a 12V supply.
Hi Albert, I am using

Resistor = 10Ω
Supply Voltage = 3V

Per the posted picture the resistor starts to get got as soon as i connect the green wire. Without the green wire one LED is bright the other one is very dull.
 

AnalogKid

Joined Aug 1, 2013
9,290
A standard red LED has a forward voltage drop (Vf) of approx. 1.8 V. With a 3 V battery, that leaves 1.2 V across the resistor. For an LED current of 12 mA (a nice safe value), Ohm's Law says the resistor value should be 100 ohms.

Note - the resistor in the video is 470 ohms. With the 12 V source being used, this yields an LED current of around 22 mA.

ak
 

ebp

Joined Feb 8, 2018
2,332
When you have both inputs LOW, the LEDs are effectively connected in parallel. If you look at the voltage versus current curve for a LED, it is similar to other semiconductor diodes. There is very little conduction below a certain voltage (usually something around 1.8 to 2 V for RED LEDs - your LEDs are measuring about 2.1 with pretty high current; at 5 mA they'll probably be around 1.8-1.9 V). Once they start to conduct, the voltage only rises a little with rising current. They more or less act like voltage regulators. There is always some variation from one to another, so if you have two in parallel, the one that conducts at the lower voltage will take most of the current. As it heats up from the current, its "forward voltage" will drop further, so it will take an even greater portion of the current. This is why one lights up brightly and the other dimly.

[EDIT - I see AK was postin' while I was typin'
You can calculate the value of the resistor you should use by applying Kirchhoff's voltage law and Ohm's law. In the series circuit of the power supply, LED and resistor, the voltage across the resistor will be the supply voltage minus the LED forward voltage (from the datasheet, but use 2 V as a reasonable approximation for red LEDs). With a 3 V supply, this would leave 1 V across the resistor, so use Ohms law to calculate the resistance to give you the current you want - anything in the 1 to 20 mA range would be typical for a small LED.

A little experiment:
If you have some "freezing spray" ("canned air" used for blowing off dust will work if you turn the can upside down), with both LEDs connected try giving the brighter LED a little blast of spray and watch how the brightnesses shift.
 
It is not a good idea to put LEDs in parallel. They will share current unequally and one may be brighter than the other. Each LED should have their own resistor.

The amount of voltage that a LED drops is first dependent on the color of the LED. It also has a voltage range that it will drop. That parameter is Vf. Call it 2.1 V for a RED LED. There is a typical led current (call it 1 to 10 mA if you don't know.

The resistor is calculated by R=(V-n*2.1)/10e-3 where V is the supply voltage, 2.1 is the drop and 10 mA is the typical operating current and n is the number of LEDs in series.

You can place LED's in series, BUT the variation in Vf might cause intensity variations.

The required wattage of the resistor should be checked. i.e. The resistor should have a higher wattage than calculated and more likely close to a factor of 2. i.e if you get 0.24 W, use a 1W resistor.
 

Thread Starter

BarryTron

Joined Nov 18, 2018
89
A standard red LED has a forward voltage drop (Vf) of approx. 1.8 V. With a 3 V battery, that leaves 1.2 V across the resistor. For an LED current of 12 mA (a nice safe value), Ohm's Law says the resistor value should be 100 ohms.

Note - the resistor in the video is 470 ohms. With the 12 V source being used, this yields an LED current of around 22 mA.

ak
I have tried the 200Ω but cant get the LED to turn on as the example, as soon as i connect the green wire LEDs turn OFF.
 

Thread Starter

BarryTron

Joined Nov 18, 2018
89
It is not a good idea to put LEDs in parallel. They will share current unequally and one may be brighter than the other. Each LED should have their own resistor.

The amount of voltage that a LED drops is first dependent on the color of the LED. It also has a voltage range that it will drop. That parameter is Vf. Call it 2.1 V for a RED LED. There is a typical led current (call it 1 to 10 mA if you don't know.

The resistor is calculated by R=(V-n*2.1)/10e-3 where V is the supply voltage, 2.1 is the drop and 10 mA is the typical operating current and n is the number of LEDs in series.

You can place LED's in series, BUT the variation in Vf might cause intensity variations.

The required wattage of the resistor should be checked. i.e. The resistor should have a higher wattage than calculated and more likely close to a factor of 2. i.e if you get 0.24 W, use a 1W resistor.
I am not sure how the person in the video is able to get both LEDs to have same brightness. In my case I only get one LED bright and the other one is dull.
 

dl324

Joined Mar 30, 2015
12,927
I am not sure how the person in the video is able to get both LEDs to have same brightness. In my case I only get one LED bright and the other one is dull.
It's probably because they don't have the same forward voltage. The one with the lower forward voltage will be brighter.
 

Thread Starter

BarryTron

Joined Nov 18, 2018
89
It's probably because they don't have the same forward voltage. The one with the lower forward voltage will be brighter.
You are right, put in brand new two LEDs and they are both same brightness. Thank you!!!

Now all i need is to be able to get them to work with the green wire
 

ebp

Joined Feb 8, 2018
2,332
Schematic came up as I was typing - yes, the green wire short circuits the LEDs so they should not / cannot light.

If the green wire is connecting the bottom end of the resistor where it joins to the LEDs to "ground" (negative side of the power supply), it should prevent the LEDs from lighting - it short circuits them. They may have lit with the 10 ohm resistor simply because there is enough resistance in the green wire to allow enough voltage to light them.

Those clip leads are notorious for being high resistance from the start and getting worse as they are used and strands in the wires break, at least in my experience. I used to buy cheap clip leads. With the better quality ones I'd just replace the wire with 22 AWG wire properly soldered to the clips. With the really cheap ones, I'd keep the insulators for the clips and scrap both the clips and the wire.
 
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