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lm 317 tip 3055 constant current

rahulrahullll

Nov 27, 2017
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i am using this circuit------working ---------

https://hassanulmakers.files.wordpress.com/2014/12/lm317-5a-variable-adjustable-supply-circuit.gif



-------circuit exact same as above---few----modifications--done----

i am using 0-40 v ac transformer---5a---using tip 3055 plastic package----10 ohm/10 w resistor----10 k potentiometer


------sharing current concept lm 317 and tip 3055----------

As current starts to flow through the 10 ohm resistor, a voltage drop will appear across it. When this voltage drop is higher than 0.6 volts, the 2N3055 starts to conduct and supply current to the output of the power supply. This starts happening when the IC LM317 is outputting 60 mA, as ohm's law show us: 0.6/10= 0.06 a . so lm 317 outputs only 60 ma while rest current is handled by 2n3055 or mje 3055 or tip 3055 or npn transistor.power dissipated by 10 ohm resistor = 0.6*0.6/10=0.036 w.



-----------------i have few queries------------------



1---------is current sharing between lm 317 and tip 3055 okay ---in future i will put 8 pcs more tip 3055 with small output equal balancing resistors as 0.22/5 w for equal sharing of power between transistors ?



2--------------i want add constant current --in circuit---potentiometer value/watt( current adjustment should be sensitive i mean i can adjust from 0.5 to 0.78 a)----what extra modification circuit needed---a diagram would be very helpful if possible--------?


3-----------weird problem----whenever i connect 10 k pot----it burns out--why-----using smallest wattage pot--i guess 1/2 w pot-----how exactly do i connect 3 legs of this kind of pot--a figure would be helpful-- ?


--https://www.build-electronic-circuits.com/wp-content/uploads/2015/11/potentiometer.jpg

----------------------thank you------------------
 

AnalogKid

Jun 10, 2015
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Not the best circuit, because the 3055 is outside the regulator control loop. So you have a regulated 317 output followed by an unregulated 3055 current booster. A couple of things ...

The 3055 may start conducting at Vbe = 0.6 V, but as the collector current increases so does Vbe. Don't be surprised if the output voltage changes by several tenths of a volt as the output current goes from 0.1 A to 5 A.

A pot's power rating is based on the internal heat being dissipated across the entire resistive element. For any output voltage setting you can calculate the pots % rotation and hence the % of the element dissipating the heat.

ak
 

Audioguru

Sep 24, 2016
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The 40VAC transformer might produce 42V with not much load current. Then its peak is 59.4V and is reduced to 58V by the bridge rectifier. Then the LM317 and the TIP3055 might be destroyed by too much voltage if anything shorts the output.

I agree that the circuit will provide poor voltage regulation. The datasheet of the LM317 has much better circuits.
 

kellys_eye

Jun 25, 2010
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Maximum input for the LM317 is 37V if I recall correctly????
 

rahulrahullll

Nov 27, 2017
4
Joined
Nov 27, 2017
Messages
4
----------------i have few queries-----------------


2--------------i want add constant current --in circuit---potentiometer value/watt( current adjustment should be sensitive i mean i can adjust from 0.5 to 0.78 a)----what extra modification circuit needed---a diagram would be very helpful if possible--------?


3-----------weird problem----whenever i connect 5k or 10 k pot----it burns out--why-----using smallest wattage pot--i guess 1/2 w pot-----how exactly do i connect 3 legs of this kind of pot--a figure would be helpful-- ?


--https://www.build-electronic-circuits.com/wp-content/uploads/2015/11/potentiometer.jpg

----------------------thank you------------------
 

rahulrahullll

Nov 27, 2017
4
Joined
Nov 27, 2017
Messages
4
Not the best circuit, because the 3055 is outside the regulator control loop. So you have a regulated 317 output followed by an unregulated 3055 current booster. A couple of things ...

The 3055 may start conducting at Vbe = 0.6 V, but as the collector current increases so does Vbe. Don't be surprised if the output voltage changes by several tenths of a volt as the output current goes from 0.1 A to 5 A.

A pot's power rating is based on the internal heat being dissipated across the entire resistive element. For any output voltage setting you can calculate the pots % rotation and hence the % of the element dissipating the heat.

ak





------------so enlighten me by your wise circuit --------analoog
 

AnalogKid

Jun 10, 2015
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Maximum input for the LM317 is 37V if I recall correctly????
Not quite. The max. input-output differential voltage is 37 V. That can be sitting on 1000 V if the shunt gain set resistor is around 1.7 M.

ak
 

kellys_eye

Jun 25, 2010
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....the OP doesn't indicate the inclusion or use of a limiting resistor in the pot circuit....
 
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