Schematic for 5v regulator

I have been trying to assemble a simple 5V regulator circuit using the L7805CV and have had no luck at all getting the output to 5V. I have tried many circuits and multiple L7805CV regulators, thinking that maybe it was the regulator itself that wasn't doing its job.

So far the closest I've come to 5V is 6V, which won't work for what I'm trying to power.

First, I tried the simple circuit with a 10uF cap from input to ground pin and 1uF cap from ground pin to output, and was getting 3.8V output with an input of 9V. I was getting 11V out with a 12V input.

I tried another one that was similar but with a 1K resistor connected to an LED on the output, which gave me 6V. I thought maybe the capacitors were the cause so I tried going with 100uF from input to ground and a 10uF from ground to output and another 100nF on the output for filtering and still no luck. I basically played with cap values all the way up to 2200uF and didn't get anything close to what I needed. I either got really low (about 3V) or really high (close to 12V) on the output and also tried different input values of 6V, 9V, 12V, and even 18V and nothing worked.

I used a heat-sink on the l7805CV and am stumped how this is happening. I eventually went back to one of the first designs that had the output of 6V and tried using a diode on the output for its voltage drop and was getting almost the exact reading on the output (actually a few mV higher) and am stumped at this point. After a while I got bored and tried out a few different transistors I had around to see how they would affect the circuit and was surprised to find almost exact results.

What am I doing wrong? How can I successfully achieve 5V on my output?

  • 1
    \$\begingroup\$ Welcome to EE, first rule is you need to show us a schematic of what you've done. The most obvious question I have here without one is whether you've been testing these circuits with an output load. Linear regulators won't work without a load on the output, because they're basically an emitter follower, and if the emitter can't put current into a resistance, you won't get a meaninful voltage. They are very simple to use normally, so you should be getting the correct results. All you need is an input voltage, an output load and the other pin grounded. That's it. \$\endgroup\$
    – Ian Bland
    Aug 27, 2017 at 23:54
  • \$\begingroup\$ You are clearly doing something wrong. How much current are you trying to get? Check the pinout of the regulator. Check your meter on a known voltage source. (Maybe replace the battery too). No idea what you mean when you say you tried a few different transistors to see how they would affect the circuit. \$\endgroup\$
    – John D
    Aug 27, 2017 at 23:55
  • \$\begingroup\$ I didnt know that Ian and I didnt try connecting it when I saw the undesired voltage. I am sorry about not posting any schematics and will do so next time. They were just generic ones that can be found by google searching 5v regulator circuit. Should I expect a higher or lower voltage under no load? I can just add a resistor as a load correct? \$\endgroup\$
    – MadRob
    Aug 28, 2017 at 0:00
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    \$\begingroup\$ With no load, I might expect the output of a linear regulator like the 78xx family to be a little higher than advertised. It should take very little load (5 - 10 mA should do) to get the voltage down to the specified voltage. I'm sure I've measured something near the correct voltage with no load other than a DVM, so you may be doing something wrong. Wrong pinout? Ground pin not grounded? \$\endgroup\$ Aug 28, 2017 at 0:25
  • 1
    \$\begingroup\$ You can edit your question to include a schematic. \$\endgroup\$ Aug 28, 2017 at 0:27

3 Answers 3


The usual reason for an unexpectedly high output voltage on a properly wired (as in pinout correct) 7805 regulator is a flakey ground connection. An open on the GND pin will cause the output to rise to a volt of two of the input. 100 ohms resistance will cause the output to rise by about half a volt.

Solderless breadboards are prime offenders, and this can expose the circuitry connected to the regulator to damaging voltage, so it's best to do that part with solder.

  • \$\begingroup\$ i actually had no idea breadboards had this flaw and will definitely be mindful of that from now on. Thank you very much for sharing that with me. \$\endgroup\$
    – MadRob
    Aug 30, 2017 at 4:32

Here is a L7805CV datasheet.

Note that on page 7, for the 5V version, Vout is specified with loads of 5 mA to 1A and there is a footnote that says "a Minimum load current for regulation is 5 mA".

To meet this minimum current requirement a load of no more than
R = V/I = 5V/5mA = 1 K is required.

That said, "usually" a 7805 will "more or less just work" if connected correctly. Caps as shown help but usually are not essential to make it work. Even Vout is liable to be "about right" without a load. (These older style regulators power themselves between Vin and ground so load current is less important.)(The LM317 family power themselves from Vin to Vout so a minimum Iout is needed for correct operation).


  1. Try it with a 5mA or higher load. (1 K or lower).

  2. Look at the data sheet(s) and ENSURE that you have the pinout correct. ||

  • 1
    \$\begingroup\$ In particular the internal feedback divider of around 5K provides an adequate load for non-pathological situations. \$\endgroup\$ Aug 28, 2017 at 5:25
  • \$\begingroup\$ @SpehroPefhany Ithink that that is essentially what I adumbrated BUT I'm not sure what pathological situations are wrt voltage regulators. \$\endgroup\$
    – Russell McMahon
    Aug 29, 2017 at 9:51
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    \$\begingroup\$ I'm thinking temperature at the high extreme Tj(max) or even the thermal cutoff temp (175C?) so maximum leakage yet no load, which would be a bit of an unusual scenario, but conceivable. \$\endgroup\$ Aug 29, 2017 at 11:01
  • \$\begingroup\$ @RussellMcMahon Thank you very much for all the information you provided me with. It really gave me a different insight to how I go about reading and applying information from a datasheet. I actually came across the exact datasheet (which makes me happy because to me that means im on the right track) and i actually did try the circuit you are referring to and i did double check the pinout , however i also ignored the minimum load requirement as well. funny you mention the lm317 because I actually tried it with that one also. Anyway, is this the best way to regulate 5v? \$\endgroup\$
    – MadRob
    Aug 30, 2017 at 4:26
  • \$\begingroup\$ best way depends on specs for input V,I, Load, ripple, efficiency cost, time to build or buy, easy to learn, etc... otherwise undefined what is best, so be clear next time with all these paramters in your mind and search google images for answers \$\endgroup\$ Aug 30, 2017 at 15:28

From looking at the internal schematic and example test circuits, I'm pretty sure it works with input at 6V min with no load and 7V min at 1A . 7.5 V is ideal. , but 9V drops too much power across regulator.

It must use a good heatsink to keep cool. It will shutdown when too hot. If it is too hot to hold, its too hot. Expect the heatsink to be at least 1 sqin per watt. For a rectified input, the input cap must be large enough to ensure the Vin min does not sag below minimum so you must measure AC pp and DC voltage. My guess 7.5Vdc avg in with 1Vpp ripple @ 1A needs C=80ms/5 ohms = 1.6mF not 100uF.

If you raise Vin more than Vmin, it must dissipate the drop voltage x load current, so 9V is not good.

Recheck your connections and shorting the input will damage the part, so reverse power diodes are suggested between in and out.

There is no need to put series diodes on output, but if you only have 9V, use diodes on input to run cooler to reduce the drop power across series regulator to just above Vmin at 1A.

  • \$\begingroup\$ thank you for sharing your knowledge with me. I had assumed that 9v and even 12v were perfectly fine to use as inputs because i saw so many circuits designed that way. And i only tried the diode on the output for its voltage drop (which didnt occur) and i also attenpted to lower the jnput voltage with the same method on the input. Was i wrong to think that? Forgive me for my ignorance but when you say reverse power diode between in and out you mean a diode like: Input |<-- output or pin1 |<---pin 3 or left pin |<---- right pin right? \$\endgroup\$
    – MadRob
    Aug 30, 2017 at 4:56
  • \$\begingroup\$ correct . , datasheet advises this to protect output in case of input short to ground only. \$\endgroup\$ Aug 30, 2017 at 15:23
  • \$\begingroup\$ "old" bipolar LDO's have a dropout from input to output of about 2V for bias and emitter follower, modern ones use MOSFET's which permit much lower dropout or minimum Vin-out needed to regulate. Too much and you dissipate deltaV*I \$\endgroup\$ Aug 30, 2017 at 15:27

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