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I have an AC power source that when rectified with 470 µF capacitor gives 38 V DC with no load. I have a 7805 (TO-220) which the data sheet says has a maximum input of 35 VDC. I'm trying to power an ESP8266 module that when connected to a bench power supply indicates that it's using less than 100 mA and maybe 100 mA when it powers a relay. There might be current spikes that the power supply meters aren't fast enough to show.

When feeding the 38 V directly into the 7805 I do get 5 V DC, but when powering the 8266 something happens and the output falls to about 2.5 V. I think the regulator is shutting down. When I put a resistor in series with the input voltage (50 to 120 ohms, small light bulbs actually) the module does operate properly until I activate the relay.

If instead of using the AC source I use a bench power supply to feed DC into the bridge I can crank the power supply up to the max of 30 V and the module works fine, the 5 V DC is stable.

Do I need to look at another way to generate 5 V DC from this AC source, maybe a switching power supply module of some sort, or can I do something other than using a huge wattage resistor to get the input voltage under control?

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    \$\begingroup\$ Exactly what is this AC power source? \$\endgroup\$ Feb 1, 2022 at 4:32
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    \$\begingroup\$ Peggy, I agree with the recommendation that you use a transformer to lower the voltage. It's efficient and, given you only need 5 V DC as the output, it's nonsense to linearly dissipate so much overhead voltage and well worth the effort to use a transformer to bring it into a better range of what you need. You could just add a load resistor to your capacitor bank. It doesn't take that much to lower the peak unloaded voltage. But while simple, I really do think the transformer will pay off much better. It also nips in the bud noise spikes on the AC supply line. Lots of good to the idea. \$\endgroup\$
    – jonk
    Feb 1, 2022 at 5:02
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    \$\begingroup\$ Do you have to use a 5V relay and use the regulated 5V for it? Why not a 12V or 24V relay? \$\endgroup\$
    – Justme
    Feb 1, 2022 at 6:09
  • \$\begingroup\$ Doesn't esp8266 work with 3.3V? \$\endgroup\$ Feb 1, 2022 at 8:26
  • \$\begingroup\$ @Tryingtogetsome OP probably has a dev board with a 5V input for USB and a 3.3V reg on the board. \$\endgroup\$
    – J...
    Feb 1, 2022 at 15:33

6 Answers 6

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First of all these old LDOs require a bit of load to operate correctly, but that's not the problem here. Nor is the AC supply.

The main issue is that you simply cannot use a 7805 to go from 38V to 5V, it will melt through the floor even if you use TO220 with massive heatsink and keep current load to a minimum. The voltage span is far too big. Simply don't use such old crap regulators for this purpose, use a switch regulator.

but when powering the 8266 something happens and the output falls to about 2.5 volts. I think the regulator is shutting down.

Yes it will go into thermal shutdown, cool down, go back up again, and from there it will oscillate back and forth.

the data sheet says has a maximum input of 35 VDC

The very first thing you need to learn in electrical engineering is that when a datasheet says absolute maximum ratings it talks about stress values that the part can handle for a short amount of time. These are not normal operating conditions that you should design after.

The recommended operating conditions found below electrical characteristics say 25V. But from experience we know that 25V to 5V is too big a span too, it will get very hot. Back in the days when people were still building things like this with crappy LDOs, you would chain multiple of them and take the voltage down in steps. For example a LM317 down to 15V then a 7805. And still you need TO220 and heatsinks for it.

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    \$\begingroup\$ Also in these days of global warming, it should be made criminal to design in these kind of parts with non-existent efficiency. Particularly in series after a 230VAC -> x VDC transformer that sits in people's mains outlets just burning away 24/7. Such a huge waste of energy. \$\endgroup\$
    – Lundin
    Feb 1, 2022 at 8:07
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    \$\begingroup\$ "First of all these old LDOs require a bit of load to operate correctly..." - The 7805 is not an LDO, far from it ;) \$\endgroup\$
    – marcelm
    Feb 1, 2022 at 13:35
  • \$\begingroup\$ @marcelm Yeah I guess we should use the term linear regulator instead of LDO. They were marketed as low drop-out at some point in time. \$\endgroup\$
    – Lundin
    Feb 1, 2022 at 13:59
  • \$\begingroup\$ At "less than 100mA" average draw, an 7805 with a big heatsink should be warm but manageable, it's like 4 watts of waste heat... \$\endgroup\$ Feb 2, 2022 at 7:13
  • \$\begingroup\$ @rackandboneman I just recently hand-made a 7805 regulator board for the purpose of troubleshooting a switch regulator and supplying a RF part with a clean supply. Current draw was some 50mA and supply was 24VDC. With a single 7805 TO220 and heatsink, it still got very hot to the point where you'd burn yourself if touching the heatsink. So I did the circuit with a LM317 in series to take half of the blow, also with a heatsink. Now the board can supply 1A or so without going into thermal shutdown but still it gets hot. The datasheet saying that these can each supply 1.5A stand-alone is a joke. \$\endgroup\$
    – Lundin
    Feb 2, 2022 at 8:02
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The easiest solution is to use a transformer with a much lower AC output voltage. Also be sure you have high frequency bypass regulators on it. Check your 7805 and see how many watts it can dissipate, this is on the data sheet. While you are there check its maximum operating temperature. It will probably be in the range of about 62 degrees C per watt. Your project wattage is calculated by multiplying the voltage drop across the 7805 times the current it consumes. You will find you have a lot of heat with your current AC source probably more then enough to put it into thermal protection mode. I would suggest you get a standard wall wart rated at about 12V 2A DC, this will let you do a lot of things and it will work fine with your 7805.

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A switching power supply is probably a good choice, and more efficient — but an alternative would be to use a transformer to drop the voltage of the AC before rectification. With a 4:1 transformer you would be getting 9 or 10 volts after rectification instead of 38, so your linear regulator would be well within its specs, and it would only be turning ~50% of its input power into heat instead of ~85%.

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  • \$\begingroup\$ Someone should mention the rule of thumb that linear regulators draw the same current from their input as they are providing on their output*. The voltage difference between in/out, times that current, is the amount of heat that needs to be disposed of. So in Peggy's case we have 100mA times (38-5) volts equals 3.3 watts. This is quite a bit more than a TO-220 can vent without assistance. -- *A little more, actually, but not enough to bother about generally. \$\endgroup\$ Feb 1, 2022 at 16:57
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If the input volts is too high you can pre-regulate with transistor, resistor and zener. This scheme does waste power but it is robust for input surges when you choose your pass transistor well. You still maintain the accuracy of the existing 3 terminal regulator.

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You can use a resistor divider before the 7805 input to reduce the voltage. Resistors don't drop out. Alternatively, you can put a resistor in series with the 7805 input that will keep its input above the dropout voltage at maximum load, and a second resistor in parallel with that serial combination to keep the output above 5 volts at minimum load. Door number 2 keeps overall power dissipation the same as no resistors at all, and minimizes 7805 dissipation.

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First... mind the tolerances of your mains supply. EG, it might be 110V +-20%. That means your circuit should be able to safely deal with the output voltage of that transformer even if there is 132V at the input!

Then ... you can either use series resistors, parallel (bleeder) resistors, further Voltage regulators, a switching pre regulator, a different design of filter (eg C-L-C), extra series diodes in the rectifier, a little extra secondary winding on the transformer (in series and in reverse polarity with the main secondary), and many other methods to bring the voltage down a bit.

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