LM3914 based LED VU Meter Using Transistor for multiple Leds

Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
Hi i am working on a LM3914 VU Meter with a student and he wants to have about 4or5 leds on each output pin of the Lm3914 could a Transistor be used for this and if so what one. or is there some other way of doing this thanks

 

Dodgydave

Joined Jun 22, 2012
8,663
You dont need the series 1K resistors from the led outputs, the current limit is set by the resistor on pin 7, as for extra leds, use a PNP transistor with a 1K ohms base resistor from the ouptuts and feed the leds with 100 ohms series resistors.
 
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Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
You dont need the series 1K resistors from the led outputs, the current limit is set by the resistor on pin 7, as for extra leds, use a PNP transistor with a 100 ohms base resistor from the ouptuts and feed the leds with 1K series resistors.
Hi Dogydave

Would a BC557B Diotec Bipolar PNP Transistor -45V or BC556B Diotec Bipolar PNP Transistor -80V work with this.

so pin 1 to 10 on the IC use a 100 ohm on each pin out to the base of the PNP TRANSISTOR and then a 1k from the emitter or the collector
 

Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
What is the typical Vf of the LEDs?
Is a supply voltage greater than 9V available to power the LEDs?
Hi Alec_t
Sorry I missed your post thanks for the replay. The leds are Piranha
Red- Forward Voltage: 3.0-3.4V (at 20mA)
Yellow Forward Voltage: 1.9-2.1V (at 20mA)
Green Forward Voltage: 2.8-3.6V (at 20mA)
Blue Forward Voltage: 2.8-3.6V (at 20mA)

and I will use 12v dc power supply if that helps
 

Dodgydave

Joined Jun 22, 2012
8,663
Paddy, if your leds have a volt drop of 3V,then four in series is 12V volt drop, may be better to put 3 in series with 150 to 330 ohms resistor, that would give an led current of 10 to 20mA.
 
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Alec_t

Joined Sep 17, 2013
10,685
As Dodgydave says, 4 LEDs drop too much voltage (unless the supply is ~15V). The resistors in series with each LED string are rather high in value, unless you want only a couple of mA through each string?
 

Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
Paddy, if your leds have a volt drop of 3V,then four in series is 12V volt drop, may be better to put 3 in series with 150 to 330 ohms resistor, that would give an led current of 10 to 20mA.
Sorry to keep asking but could I just up the power supply to 12v. And when you say 3v drop do you mean that's 3v max down to the lowest working voltage
 

Dodgydave

Joined Jun 22, 2012
8,663
Yes you can use a 12v supply , looking at the voltage drops of each colour, the yellow have the lowest and the blue the highest, so four yellow leds will be 8 V, whereas four blue leds will be 14V,

So its better to put 3 leds in series,with resistors of 330 ohms for the yellow, and 100 ohms for the other colours,otherwise the brightness will be different on each colours.
 
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Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
Yes you can use a 12v supply , looking at the voltage drops of each colour, the yellow have the lowest and the blue the highest, so four yellow leds will be 8 V, whereas four blue leds will be 14V,

So its better to put 3 leds in series,with resistors of 330 ohms for the yellow, and 100 ohms for the other colours,otherwise the brightness will be different on each colours.
Hi Dodgydave
Just have a question on working out the resister values for each led string.
Red- Forward Voltage: 3.0-3.4V (at 20mA)
Yellow Forward Voltage: 1.9-2.1V (at 20mA)
Green Forward Voltage: 2.8-3.6V (at 20mA)
Blue Forward Voltage: 2.8-3.6V (at 20mA)

if the voltage is 12v how are you working out the resister value and the mA

thanks
 

Dodgydave

Joined Jun 22, 2012
8,663
Hi Paddy, each led has a different voltage drop, so 3 reds will give a voltage drop of approx 9v, whereas 3 yellows are 6v, so 20mA for the reds needs a 300 ohms resistor,and 20mA for the yellow needs a 150 ohms.

Red resistor is 6v/20mA = 300

Yellow resistor is 3v/20mA =150
 

Thread Starter

paddyhughes086

Joined Jan 30, 2011
48
Hi Paddy, each led has a different voltage drop, so 3 reds will give a voltage drop of approx 9v, whereas 3 yellows are 6v, so 20mA for the reds needs a 300 ohms resistor,and 20mA for the yellow needs a 150 ohms.

Red resistor is 6v/20mA = 300

Yellow resistor is 3v/20mA =150
thanks dave

so 3 red at 3v is 9v at 20mA... 9v/20mA =450

if the mA of led is between 20mA-30mA would i have the value to 25mA
 

Dodgydave

Joined Jun 22, 2012
8,663
three reds is 9v, subtract from the supply so 12-9= 3v, the resistor drops the remaining voltage and sets the current,

thats 3v/20mA = 150 ohms
 
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Reloadron

Joined Jan 15, 2015
4,850
Just a simple question here. You say "LM3914 based LED VU Meter" VU is Volume Units which are Logarithmic and the LM3914 is a linear chip. If you do in fact want a VU meter representation wouldn't the LM3915 be the better choice? That would afford 3 db log steps.

Ron
 

Dodgydave

Joined Jun 22, 2012
8,663
what 300 resistor? if you mean the resistor to select for the red current, then as explained previously

3 yellow leds in series gives a volt drop of 2Vx 3= 6V, its run from a 12V supply so we need to drop 12-6 = 6V at 20mA thats 6V/20mA = 300 ohms.
 
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