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I’m trying to build a 5V output @ 200mA. I have a full wave bridge rectifier attached to a transformer with a capacitor reservoir of 100uF, using a LM7805 voltage regulator and with a 0.1uF cap as well. To get the 200mA that I needed I used ohms law to find I needed 25ohms (5V/0.2A=25ohms.) But that power will burn out my resistor so I thought that I should use 4x 1/4W 100ohm resistors in parallel to achieve the same result. When I turned on my circuit, my voltage regulator became super hot and the voltage I recorded was lower then 5V (3.8V).

I've attached the schematic I drew to show how I wired it. Can you tell me what I have done wrong ? Sorry for the crappy drawing in advance.

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  • \$\begingroup\$ You'll find a vast amount on the subject in this SE answer My linear voltage regulator is overheating very fast \$\endgroup\$ – Russell McMahon Apr 12 at 13:57
  • \$\begingroup\$ Apart from the regulator heating, what are you trying to achieve? What are you wanting to do with the 5V @ 200 mA? I ask as it is not obvious that what you propose has any function apart from heating resistors. eg if you want to power something other than the resistors with a 5V supply then you do not need the load resistors AT ALL. \$\endgroup\$ – Russell McMahon Apr 12 at 14:00
  • \$\begingroup\$ How are you heatsinking the regulator? \$\endgroup\$ – Brian Drummond Apr 12 at 14:30
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    \$\begingroup\$ Duplicate of My linear voltage regulator is overheating very fast \$\endgroup\$ – brhans Apr 12 at 15:02
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The transformer secondary is 12 - 14 volts AC so, assuming 14 volts, it will have a peak voltage of 19.8 volts. After bridge rectification this becomes about 18.4 volts DC and the 7805 regulator is dropping 18.4 volts DC to 5 volts DC whilst taking 200 mA.

The 7805 power dissipation is (18.4 volts - 5 volts) * 0.2 amps = 2.68 watts. So, without a heatsink, the 7805 will boil.

You might want to read this article that describes the problem in more detail and offers a switched mode buck regulator as a solution: -

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The device above is from TI and is called a TPSM84205. It will efficiently deliver 5 volts without the excessive over-heating: -

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  • \$\begingroup\$ Thanks. Would using a half wave rectifier and increasing the reservoir cap, fix the problem ? \$\endgroup\$ – Tony Apr 12 at 8:20
  • \$\begingroup\$ No, because the DC voltage is dictated by the peak of the AC waveform from the secondary and the peak will be roughly the same whether full or half wave rectifying. \$\endgroup\$ – Andy aka Apr 12 at 8:29
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    \$\begingroup\$ If you can't change the transformer or anything else in the circuit, add resistor or multiple resistors in series between the 100uF cap and regulator input. The system will still dissipate the same amount of watts as heat, but not all of it is dissipated in the regulator. I did not calculate suitable resistance value, but just make sure that at 200mA current you still have 7 to 8 Volts at the regulator input when the bulk capacitor has minimum voltage. Adjust bulk cap value if necessary. \$\endgroup\$ – Justme Apr 12 at 9:03
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    \$\begingroup\$ Shouldn't that be "Your question has a 8+ year old Annnnnnnnnnnnnnnnnnnnnnnnswer here :-) ? \$\endgroup\$ – Russell McMahon Apr 12 at 13:56
  • \$\begingroup\$ @Tony If the TI part isn't available where you are, search for "7805 equivalent switching". \$\endgroup\$ – Andrew Morton Apr 12 at 14:15
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Your problem is that the input voltage is too high for your output voltage.
The LM7805 as a LINEAR voltage regulator turns that voltage difference multiplied by the output current into watts of heat.
Even a relatively small current of 200mA in this case gives off about 2W of heat or even more.
If you don't have a sufficient heatsink, the 7805 overheats and reduces output voltage to protect itself from burning out.
You could cut the amount of heat generated by giving it a lower voltage at the input. Normally, you need at least 2V higher input than the output for a 7805, but you shouldn't go too high; usually 7.5VAC to 9VAC is used as a transformer output for this 5V regulator.

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