Reg:Darlington Pair Transistor

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PARTHIBAN31288

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Hello Friends,
we are using Below darling-ton pair of transistor in our circuit. By using this we will get 200mA Peak Current. for switching purpose (TR1) we use BC847B transistor and BCX5616 power transistor for amplification. Our problem is , the power transistor get heated.
it's around 50-60 C heat. i dont know why it's heated up..
i'm, using BCX5616 power transistor of DC Current 1A and Peak current of 1.5A.
So i cant understand why this transistor gets heat.
what is the wrong in my design....?

Note: i drive the BC847B transistor from Micro controller....

 

Most of the 30V is dropped across the BCX56 which is why it gets hot. When the current through R1 increases it reduces the base-emitter voltage of the transistor so the transistor limits the current. Presumably that is what you want it to do. But you need to look where the voltage is being dropped. Why use 30V for driving the LED? I don't know what the forward drop of the LED but normal LEDs are a few volts.

Keith.
 

[ When the current through R1 increases it reduces the base-emitter voltage of the transistor so the transistor limits the current. ]

We are not able to reduce/ increase this R1 value.. Because if i increase the value means emitter optical power (i.e) lumex get lower. intensity reduced.
if i lower the value means intensity get higher and also heat get higher.

[But you need to look where the voltage is being dropped.]

I'm not having any idea about this..can you tell me little bit clearly..?

[Why use 30V for driving the LED?]

Because my application need to operate in 10-30 V

[ I don't know what the forward drop of the LED ]

Forward voltage is 2.5Volt

Keith.[/QUOTE]
 

If you want to run from 30V and the LED only needs 2.5V then the rest has to go somewhere. So 27.5V at 200mA = 5.5W peak which is either going to be dissipated in either the resistors, the transistor, or a combination of the two. Unless you use a switching regulator style circuit using an inductor then you will have to live with the heat.

Keith
 
You can control LED brightness (if it is what you want to) by quickly switching the transistor on and off (PWM modulation). This way the power dissipated in the transistor will be minimum. Anyway, why have you a resistor in the emitter?
 


Emitter resistor is used to control the current flow through the transistor.It controls the intensity of my emitter led...
 

Dear PARTHIBAN
Hi
If you want decrease the emitter current it isn't good idea to use a resistor in emitter . because you'll have power dissipation across it . you can limit base current to control emitter current as well .
Best Wishes
Goldsmith
 

 

You can reduce the transistor heat by allowing voltage to drop in a resistor instead. It won't reduce the overall power dissipation - you will just be moving the heat around.

If you want a low loss drive where your supply voltage (30V) is considerably more than the voltage required by the device (2.5V) you need an inductor based switching regulator type of solution.

Keith
 

keith1200rs said:
[You can reduce the transistor heat by allowing voltage to drop in a resistor instead. ]

Can you tell me how to do this.../? if i do like this. whether i need to go for high wattage value of resistor......
Click to expand...
 

Hi again
If you have to drive it via 30 volt , try to drive your transistor with a simple PWM and add an LC filter in collector and then give it to your LED without any resistor , if you change the D.C ( duty cycle ) you can control the voltage across the LED as well . and for PWM you have many ways . such as : a simple 555 . or .... etc .
Best Luck
Goldsmith
 

 

No that capacitor isn't enough . that filter called butter worth and as you probably know , the average of each square function will given from : 1/T integral over Vs dt from zero up to ton and after solving that equation you'll have D.C*VS . so to get a perfect average you have to use that filter . BTW : don't forget to add a diode to prevent from high voltages .
Best Wishes
Goldsmith
 

i just have a feeling that the capacitor 22uF used will charge to 30V when the LED is switched off and once the Darlington is turned on (almost fully) the current flowing through the Transistor would be (without the current control through the base) (30V-2.5(led forward voltage)-0.5(approx drop across Darlington))/10E=2.7A which makes your transistor hot.
This condition happens only for a short time(say 100us) till the 22uF capacitor is discharged......
i mean to say even though the design might be for 1A due to the high value capacitor which acts as a temporary storage might play a role....you shall reduce the value to 1uF which can bring the heat down...

Regards
EA
 

Thanks for your reply.....

i will check this and reply..

Thank you....
 

Increasing your 56 ohms to 1k or more will reduce the transistor power dissipation. However, the best you can do with such a circuit is re-distribute the power dissipation - you will not reduce it. To reduce the overall power consumption (and therefore improve efficiency) you need a different approach.

Keith.
 

[Increasing your 56 ohms to 1k or more will reduce the transistor power dissipation. ]

Thanks Keith...
i change this 56 ohms to 1k... Now the dissipation comes little bit lower than the past...But heat will be there....

[To reduce the overall power consumption (and therefore improve efficiency) you need a different approach.]

Can you suggest me few methods to do this.....
 

To be honest, the simplest would probably be a switching regulator to reduce the voltage down to something like 7V. The 7V will then be constant across a wide range of input voltage but your power dissipation would then only be based on 7V not 30V. Something like the LM2841X-ADJ may do

Keith.
 

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