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Blinking LED’s

Baldhead

Nov 29, 2018
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AF011084-3879-4D79-AE92-1F6CAB361CFB.gif This is my very first (probably stupid) question so please go easy on me!
I’ve copied this circuit onto my breadboard, but the LED’s don’t light up when connected in series as the circuit shows, however if I connect them in parallel the LED’s work perfectly.
Is the circuit wrong or am I doing something wrong?
The circuit is designed so the LED’s flicker when the jack plug is connected to a music source (in this case it was my iPod), I intend to use it with a prop for Halloween next year.
Thanks
Stew
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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There are several problems:
  1. The forward voltage of the LEDs when added together is probably higher than the battery voltage.
  2. You have nothing to limit the current to the LEDs (and also through the transistor)
  3. Placing LEDs in parallel without individual current limiting resistors is a bad thing.
  4. You should have a resistor in series with the base connection to the transistor. This will both protect the transistor and cause the sound to be less distorted.
 

Baldhead

Nov 29, 2018
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Thanks Steve.
I expected the LED’s to ‘pop’ when the circuit was completed because there were no resistors to limit the current flowing through them.
How can I work out what size of resistor I should connect in series with the base to protect the transistor?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Firstly, determine the maximum current through the LEDs.

Let's say you find they have a Vf of around 4V and you want 20mA through them. Ignoring the voltage drop across the transistor, you need a (12 - 4)/0.02 Ω resistor in series with each LED (that's 400Ω -- use the closest standard value, 390Ω 1/4W).

So, with 4 of these in parallel, the transistor has 80 mA flowing through it.

Look up the datasheet for your transistor for HFE (preferably at approx 80mA). Let's say it claims 110 to 800 at 100mA. Use the lower end, and let's just call it 100.

So, to switch 80mA, you need 0.8mA of base current. (Ic/HFE). Let's assume your signal amplitude is 1V, and that Vbe is 0.7V. The series resistor required is (1 - 0.7)/0.0008 Ω, or 375Ω. I would use a 390Ω resistor again.

In this case I might be using a 2N2222 or a BC548.

This should provide you with a starting point.

Practically speaking, because you are only driving the LEDs on positive going half cycles, you can reduce the series resistors to maybe 220Ω, and the base resistor could be anything from 100Ω to 1kΩ depending on the volume level from your signal source and the actual gain of your transistor. You could start with 220Ω and vary it if necessary.
 
Last edited:

73's de Edd

Aug 21, 2015
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Sir Baldhead . . . . .

Here is referencing to familiarize you with the suspected voltage thresholds needed to be met ,for individual different types / colors of LEDs.
Utilize the FAR right column.

SUBMISSION . . . . . (Translated . . . . . from Olde Tyme English to Amerikanski . . .Colour . . .Color )

upload_2018-11-30_19-19-14.png


73's de Edd
. . . . . . . . . .


Doughnuts versus a salad . . .
You know when you buy a bag of salad and it starts getting brown and has that stinky gross water in it and slime on the lettuce ?
Well . . . . . Doughnuts never do that.

 

Baldhead

Nov 29, 2018
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Thank you 73’s de Edd, that’s one to print off for future reference.
Baldhead
 

Morgann

Dec 4, 2018
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You have to calculate the working voltage of the LED. If the voltage of the four LEDs exceeds 12V, the LED is impossible to light up.
 

davenn

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Sep 5, 2009
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You have to calculate the working voltage of the LED. If the voltage of the four LEDs exceeds 12V, the LED is impossible to light up.


yup, that was stated way back in post #2 :)
 

Audioguru

Sep 24, 2016
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That wrong circuit with the TIP31 power transistor was an Instructable made by a kid who was only 10 years old and who knew nothing about electronics.
 
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