How can I limit the DC input voltage from an AC rectified source to keep it within limits for 7805?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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%.

Answer 4

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.

Answer 5

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.

Answer 6

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.