LM317 as a High Current Voltage Regulator.

Please let me know that I made this attached circuit but

10 ohms resistor gone very hot and I connect Halogen

bulb of 100W 12VDC but it can not glow.

If the 10 ohm resistor becomes hot, the transistor is either defective or wrongly connected.

100/12=8.3A
70mA should be for LM317 then PNP transistor will take the current over.
Could you implement the circuit correctly ??

Hi,

What are your actual component values? Have you copied this circuit identically? According to this calculator 100W/12V = 8.3A

The LM317 formula is:
Vout = Vref * (1 + (R2/R1)) + (Iadj * R2)
Vout = 1.25V * (1 + (5k/270)) + (100uA * 5k)
Vout = 1.25V * (1 + 18.51) + 0.5V
Vout = 24.887V

Maybe this is why the resistor is getting hot and maybe, but just a thought rather than a fact, the 317 is perhaps even shutting down due to over temperature or overcurrent...

That bypass circuit is proportional, the bypass PNP carries let's say 90% of the current and let's say the 317 will carry the other 10%.

Can you measure the voltage across the 10R resistor terminals or just show your actual circuit if it isn't identical to the bristolwatch one?

I measure voltage across 10R resistor without connect load is zero volt but when load was connected voltage again show zero across 10R.

And I use 330ohms instead of 270ohms.
And rest of the circuit remains same.

I read datasheet of MJ2955 TO-92 it is BCE configuration

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Please attached herewith pictures of my board transformer 220vac to 19v output 5A

But when I connect bulb directly on transformer it has glow.

Hi,

What voltage do you get at the output with maybe a ~2k resistor as load? By calculations with 5k/330R it should be about ~20.7V. (Doesn't the bulb only need 12V and ~8.3A?...)

Unless I'm calculating something very incorrectly, I'd change the 10R, hopefully 830mV is more than enough to turn the BJT on, but the datasheet says 0.6V min > 1.8V max Vbe turn on. 2R2 would give 1.8V at 0.83A into the regulator. 2 would give 1.66V. 22R would give 0.081A through the regulator to make the BJT turn fully on and the rest pass through it.

If you have two voltmeters, you can measure the voltage across the base/input resistor and across some resistive load and see if there's any current moving.

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...What is/are 19V 5A? The transformer secondary AC voltage and the maximum current output from the transformer?

Are you rectifying the o/p of the transformer and smoothing it with a large capacitor?

Yes it is rectified and 100uF 50v capacitor used.

Hi,

100uF is way too small...for an 8.3A load

You should rather use 100,000 uF.

Klaus

How can the 10 ohms resistor get hot when it has no voltage across it??
An MJ2955 is in a large TO-3 metal case but your photo shows an old Fairchild (now owned by ON Semi) TIP2955 in a small TO-220 case, but its pins are also BCE.

The transformer voltage is much too high and its current is much too low. The little heatsink on the transistor is much too small.

the 10 ohm resistor gets hot when the 2955 fails short ...