I just have resistors for Led's they are higher in values i think.. well, its a visit to the electronics shop tomorrow then!
Then comes the big challenge taking this from the breadboard and making a real circuit board..
But that should'nt be so hard, the connections and circuits are easy tp place...
I just have resistors for Led's they are higher in values i think.. well, its a visit to the electronics shop tomorrow then!
Then comes the big challenge taking this from the breadboard and making a real circuit board..
But that should'nt be so hard, the connections and circuits are easy tp place...
Yes!, thanks for the reminder, i will put two pins there which can be connected by a black "cap" sort of, well you know... so its easy to change if necessary..
The LM3914 will be extremely hot if it is set for BAR MODE, the supply for the LEDs is 12V, the LEDs are 2V red ones and all 10 LEDs are lighted:
1) LED current= 12.5/1.21k= 10.3mA.
2) Dissipation in each output transistor= (12V - 2V) x 10.3mA= 103mW.
3) Total dissipation for the LM3914= 103mW x 10= 1.03W.
If a 47 ohm/1W resistor is added in series with the LEDs then it will dissipate 499mW and the LM3914 will dissipate only 531mW. Then the 2.2uF capacitor is used as shown in the datasheet.
Ah ok, that wont change the leds light in any way i understand, so they still light up with full strenght?
I just add the resistor directly after the battery going to the leds then..
The capacitor as i understands it will take care of the "powerloss" and make them go as bright as they would with out the resistor and the capacitor....?
The resistor values of the circuit give an LED current of only 10.3mA so they will not be "full strength". Most LEDs have a max allowed continuous current of 30mA, some 40mA.
The 47 ohm resistor reduces the heat in the LM3914, not the brightness of the LEDs.
I have a Sound Level Indicator project that operates its LM3915 with 32mA in its LEDs. My supply voltage is 9V, my LEDs are 3.6V each and a 10 ohm/1W resistor is in series with all the LEDs to share the heat. It is very bright.
You need to re-calculate the values for the two resistors from pin 7 to 0V. Their ratio should remain the same.
Then you need to reduce the supply voltage or limit the voltage to the LEDs with a pretty big resistor.
Instead of making a lot of heat, why not make a lot of light?
Connect LEDs in series at each output of the LM3914. If the LEDs are 2V red ones then up to 5 can be connected in series at each output since the supply is so high at 12V. Then the LM3914 will not get hot.
The LED current is determined by the 1.21K resistor to about 10mA, this is only approximate.
The point thats been raised is the heat generated within the LM3914 when all the LEDs are lit when using the 'bar' mode.
A solution to reduce the heat within the LM3914 is to fit a series resistor in series with each led. The led current will be the same.
What colour of led to you plan to use.?
Another project I am helping with has two blue leds in series, a blue led can have a voltage drop of 3 to 4 volts, so when the two are in series the voltage drop is say 7v, so the other 5v at say 10mA is dissapated as heat in the LM3914. Thats approx 5 * 0.01 50mW/led output.
A single resistor for the led's or do i have to put a resistor at every led? on the + or - ? they are joined togheter with the kathode so there or from every "numbered" pin from the lm3914
A single resistor for the led's or do i have to put a resistor at every led? on the + or - ? they are joined togheter with the kathode so there or from every "numbered" pin from the lm3914