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I am designing a linear step-down regulator using a PNP pass transistor. The output of the regulator is good at no load.

When I connect my regulator to a battery supply, leave it for 2 seconds and then I connect my load (say a 12 V DC motor or a microcontroller), it works good.

But when I attach my load to the circuit and then supply power, my output drops to almost 0 V.

Here is my regulator circuit...

enter image description here

There is a problem in my design. I am trying to analyse it. It would be appreciated if someone could help me.

enter image description here

Above one is a reference image from Texas Instruments.

Input supply (battery) = 40-50 v

Output volt (for controller) = 12 v (500 ma approx)

Edited:

enter image description here I connected the input for 7805 in source side by using a transistor to get a constant supply as shown in the image. Now my circuit works better.But pnp transistor heats quite a lot by connecting load. Since it's a low dropout regulator, why it heats much.

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    \$\begingroup\$ I guess, the 7805 or similar device has to be powered from source side, not from load side. You'd better to replace the 7805 (higher voltage) and place it on the source side. \$\endgroup\$ – Marko Buršič Sep 3 '19 at 8:49
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    \$\begingroup\$ By feeding the 7805 from the output voltage (instead of the input voltage) you created a potential chickin-egg situation: If the 12 V output is loaded such that the 7805 cannot get enough input voltage, it cannot make 5 V. That 5 V is needed to make a reference voltage and open the PNP pass transistor. If the 5 V is not there the PNP pass transistor cannot open meaning there is no 12 V, meaning there is no 5 V. \$\endgroup\$ – Bimpelrekkie Sep 3 '19 at 8:52
  • \$\begingroup\$ Apart from the other comments, the pass element needs DC bias for the base - emitter junction from the source (which would also need protection to avoid over-stressing that junction). \$\endgroup\$ – Peter Smith Sep 3 '19 at 10:37
  • \$\begingroup\$ Thanks for your response guys.. yes i understand that the problem is on 7805 side. But what is the solution for it. If i keep that on input i want some other one which is compact able for 50v, which is completely waste for this step down regulator. \$\endgroup\$ – Bud Sep 3 '19 at 11:46
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    \$\begingroup\$ Also, consider using a different pass transistor: The BD140 you're using would be dropping 38V at your 500 mA, that's 19W it has to dissipate. It is not rated for that. \$\endgroup\$ – Richard the Spacecat Sep 3 '19 at 12:19
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I Assume there's no other loads for the 7805.

You can take the input to the 7805 from +50V through about 40V zener diode. I guess the 7805 sinks few milliamperes only, so the dissipation in the zener diode is well below one watt. It would be one watt if the current were 25 mA.

To be sure how much the zener dissipate, measure the current consumption of the 7805+what's now connected to it.

I guess you want to use the 7805 because you have them freely available. If a zener diode is an impossible option, you can surely replace it with a voltage divider.

ADD: Just saw the edit which says the PNP transistor gets hot. If you take continuously 500 mA through it, the dissipated power is about UI = (50V - 12V)500mA = 19W. I guess it will burn soon because in a datasheet the absolute maximum allowed dissipation is 12,5W. That means you should have infinitely large and thick heatsink to avoid burning when the dissipation is 12,5W.

My final guidance: Get a switching regulator. It dissipates 90% less. If that's impossible, do as the commentators already have said - get a bigger transistor and a big enough heatsink.

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  • \$\begingroup\$ Sure i will try with zener diode \$\endgroup\$ – Bud Sep 3 '19 at 17:33
  • \$\begingroup\$ when my load current is 73ma and 38 volt drop across pnp, and therefore 2.7 watts. For 2.7 watts itself ,pnp transistor get more heat >70 degree. I think there is also a problem in my circuit too. \$\endgroup\$ – Bud Sep 4 '19 at 6:17
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Here is why it doesn't work. The reason it works if you apply the load after the supply has stabilized is because of the 3.3k resistor across the PNP. Without the 3.3k it wouldn't work because your 2.5V reference would never come up. With a load applied before the power, it draws all of the power and reference stays at zero. Remove the 3.3k and get the power for your reference at the PNP emitter. This will ensure the reference always has power. You can also replace the 7805 with a 2.5V reference.

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  • \$\begingroup\$ Yes rob, i did chances to my circuit and i edited my post. I use a transistor to drop voltage for 7805 input supply. \$\endgroup\$ – Bud Sep 3 '19 at 17:33
  • \$\begingroup\$ You ask why does the PNP heat up so much. Power = voltage X current. The voltage is the drop across the PNP, Measure from emitter to collector. If you can measure the current you can calculate the power. You may need to heat sink the PNP. \$\endgroup\$ – Rob B. Sep 3 '19 at 19:10

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