0
\$\begingroup\$

I am learning electronics. Today I tried using 7805 and 7809 voltage regulators with a 12V battery.

Though my battery is of 12V, currently it has 8.76V in it.

In my first attempt, I connected a 7805 with the battery and I got ~ 8.70V in the input terminal and in the output terminal I got ~ 4.98V which is pretty obvious.

The unexpected things happened when I connected the battery with a 7809 voltage regulator.

Here in the input terminal, I got ~ 8.69V, but in the output terminal I got ~ 7.91V.

Now this is pretty unusual to me because I am using a 9V voltage regulator and if I assume that due to the internal resistance of the wire, I am losing around ~ 0.2V then for 7809 the output reading should be ~ 8.67V.

Where is the unsolved (8.67-7.91=) 0.76V? Why isn't this the case with the 7805?

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Check the datasheet for the 7809 to see what the minimum input voltage requirement is. It's probably 11 V and since you're running on a flat battery the regulator isn't regulating. \$\endgroup\$
    – Transistor
    Commented Sep 2, 2021 at 6:31
  • 1
    \$\begingroup\$ +1 what Transistor said and also check the minimum load requirement. \$\endgroup\$
    – winny
    Commented Sep 2, 2021 at 6:36
  • \$\begingroup\$ @Transistor, yes, you are correct 👏 Vin is 11V!! Thanks man. But if it is not regulating, then why this big amount of voltage drop? Why isn't with 7805? \$\endgroup\$
    – user295082
    Commented Sep 2, 2021 at 6:37
  • \$\begingroup\$ R.I.P. to your battery .. Ensure input is 2.5V higher is old tech. while new tech. FET LDO's are << 1V drop \$\endgroup\$ Commented Sep 2, 2021 at 6:49
  • \$\begingroup\$ What you expect would be achievable with an LDO (Low Dropout) linear regulator. The 7809 is a linear regulator, but is much older than LDO regs. I'm surprised you're getting more than about 6.5V, perhaps you have a very light load (or no load). \$\endgroup\$
    – user16324
    Commented Sep 2, 2021 at 11:37

1 Answer 1

4
\$\begingroup\$

Linear regulators like the 7805 and the 7809 regulate the output voltage by varying their internal resistance.

They can only lower a voltage. They cannot raise it.

They also have a minimum difference between input and output voltage. Another way of phrasing that is that they have a minimum input voltage for them to operate.

From the LM7809 datasheet:

enter image description here

The line that says "Input Voltage" is the relevant one here. It says "11.5V to 26V."

Below 11.5V, it won't regulate. It delivers something, but it isn't specified.

You need to charge your battery. With 12V in, the 7809 will produce the expected 9V (8.65V to 9.35V.)


The 7805 operates correctly because your 8.7V input is above the minimum operating voltage for the 7805.

The 7805 requires 2V of dropout. It will work properly with an input of 7V or higher.

\$\endgroup\$
2
  • \$\begingroup\$ Yeah I think that's the reason. But then in case of 7809, why this big amount of voltage drop? And why for 7805 the voltage drop is negligible? \$\endgroup\$
    – user295082
    Commented Sep 2, 2021 at 6:48
  • 1
    \$\begingroup\$ The thing you need to look at is the difference between input and output voltage. The 7809 powered from 8.7V is not regulating. It is just passing its input to its output with a drop. The 7806 operating from 8.7V is regulating. It is dropping 8.7V to 5V. That's 3.7V that it is dropping. \$\endgroup\$
    – JRE
    Commented Sep 2, 2021 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.