Led circuit help

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chris2001net

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Hi

I was hoping someone might be able to check my circuit, it splits 3v between 2 leds of 3v and 2 leds of 2v (with resistor), & uses an NPN transistor to switch the leds on and off from a 1.8v source (via resistor).

It seems to work but just hoped someone might point out any problems. thanks!

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Don't take the 200Hfe number for granted. It will inrease as temperature rise. Then short.

Also, two diodes in direct parallell is a bad idea since one always will take precedence and conduct most current.

Baseline: One of the two upper led's is prone to die from overload.
 
You need separate resistors for each LED and you're forgetting about the 0.7V junction drop through the transistor.
 
The transistor will probably drop only about 0.1V (or whatever the saturation voltage is for the transistor you're using), not 0.7V (that's the Vbe drop). If you've only got a 3V supply it probably won't light the 3V LEDs properly. As the others say, use a separate current-limiting resistor for each LED.
You have a potential load of 80mA. As a rule of thumb, and to allow for gain variations etc, I would use a base current of ~ 8mA. That implies a base resistor of ~120 ohms, so your 91 ohms should be fine, if a bit wasteful on current.
 
Put a 47 ohm resistor in series with the top two leds.
Put a 100 ohm resistor in series with the bottom tow leds
Most leds don't have to run at exact 20 mA.
 
The transistor will probably drop only about 0.1V (or whatever the saturation voltage is for the transistor you're using), not 0.7V (that's the Vbe drop).

Oops, yes, at Ic of 80mA, the Vce will only be ≈0.2V for a saturated BC547.
 
Thanks for the replies very useful, ok I'll put a resistor per led and replace R1 to 120ohm for NPN transistor. How do you calculate 47ohm for 3v leds, and 100ohm for 2v?

I've used this led resistor calculator: http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator
In parallel: 0ohm for 3v leds, and 25 ohm for 2v led (note circuit shows 1 resistor)

The formula for the transistor resistor was in this pdf.
http://techhouse.org/~dmorris/projects/tutorials/transistor.switches.pdf
R1 = switching voltage * HFE / (1.3 * load current)
it states 1.3 to allow for a little extra load current. top of page 4.
(I think it calculated about 92ohm)
 
Your LED calculator program is stupid and is completely wrong!
Where can you buy a 2V or a 3V LED? Nowhere.
A 2V LED has a range from 1.7V to about 2.3V and a 3V LED has a range from 2.7V to 3.3V due to manufacturing tolerances.

If one 2V LED is actually 1.7V and the other is 2.3V then only the 1.7V LED will light since they are in parallel.

If the 3V led is actually 2.7V then the 3V supply will immediately burn it out.
If the 3V LED is actually 3.3V then the 3V supply will cause it to be very dim or not light.
 
Basic ohms formula used
I used your led values. a red led is actually 1.7 Volts not 2 Volts as Audioguru already said

in your case.

U = I * R
U supply - U Led = U
3 - 2 = U
therefore U = 1
at 20 mA = 0.02 Amps.
from U - I * R
1 = 0.02 * R
R = 50 ohms
Take 47 or 56 as nearest value.
 
Thanks for the formula Rodalco.

Audioguru, i see what you're saying i think as the leds dont light correctly with revised circuit.

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As suggested I've put a resistor per led, 47ohm for 3v and 100ohm for 2v. 120ohm for transistor.

I should mention each led is a different colour, all ultra brights, so after reading Audioguru's post the two 3v leds are probably slightly different voltages, and same with the 2v leds.

What I see is the ultra bright red 2v is lit nice, 2v yellow not so good, 3v green dim, and 3v blue very very dim.

I thought having them in parallel would supply 3v to each.

Is there something else I can try? thanks.

I'm wondering if amplifying the 3v source and running them in serial might work any better.
 
With 4 different LEDs you need to work out the right resistor for each to get proper brightness. Blue LEDS need more than 3 volts so may not be a good choice here. The 3 volt supply could be boosted with some difficulty.
Do you have 5 volts avalible?
What is this for?
Maybe we can come up with some alternatives if you let us know what you are doing.
Andy
 
Obviously the dim green LED has a voltage close to 3V and the very very dim blue has a voltage of more than 3V.
Maybe the blue LED needs 3.5V then the 47 ohm current-limiting resistor needs an additional 1V so the supply must be 4.5V, not 3V.
 
Its to light the buttons for a wireless xbox 360 controller
**broken link removed**
The led's are glued inside the clear buttons (right)

I've only seen people light all the buttons with the same LED colour/spec and use a resistor. (Either clear buttons or coloured plastic)

Power points
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I use the 3v bottom right, (which is 2.4v when controller is off, so to avoid this led drain I use a transistor switch on the 1.8v point which is 0v when the controller is off, which is marked as 2v on this photo)
 
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Two 1.5V battery cells are 3V when new and drop to 2V when drained. Then only 1.7V red LEDs can be powered but they will look dim when the battery is not new.
 
Is there space to insert a voltage-boost circuit? If so, it would enable all flavours of LED to be powered (albeit with increased current drain on the batteries).
 
An ICL7660 voltage boost IC has an output current that is much too low to power the green and blue LEDs with 20mA each.
A Joule Thief circuit also has a low output current.

If a voltage boost circuit can be found with an output of 4.5V at 40mA then the drain on a new 3V battery is about 72mA and mightr be more as the battery voltage runs down.
 
I doubt the OP really needs to run the LEDs at anything like 20mA each. They will be aimed directly into the eye through the translucent buttons. At currents as low as 5mA modern LEDs, particularly blue or white, can be surprisingly (even painfully) bright. Lower current requirements would ease the construction of a suitable compact boost circuit.
 
thanks for the replies I've perseverared with trying diffierent resistors and reached a point where all are lit equally and reasonably bright enough. I'll get a video when I've put the case back together which is going to be difficult
 
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