I am have a bluetooth speaker that is powered by a single 18650 battery. It can only handle 6V input to the PCB. I want to add one or two more 18650's. I have a bunch of different voltage regulator IC's that are transistor shaped. I was thinking of using a LM7805 regulator to cut the voltage down to 5.0V. It says that the minimum voltage input voltage is 5V and maximum is 18V. If I were to use this, would it shut off as soon as the total voltage of the batteries drops below 5V?

  • \$\begingroup\$ Also, would it just become a open circuit below 5V? \$\endgroup\$ Jan 15 '18 at 17:05
  • 2
    \$\begingroup\$ "It says that the minimum voltage input voltage is 5V" What says that? An LM7805 typically needs at least 7V in to produce 5V out. \$\endgroup\$
    – Finbarr
    Jan 15 '18 at 17:07
  • \$\begingroup\$ No, at 5V in (to a 7805) you'll be lucky if you get 3V out, unregulated. \$\endgroup\$ Jan 15 '18 at 17:43

In general, below the minimum Vin, linear regulators present themselves simply as a saturated transistor.

The output then simply follows Vin minus the saturation voltage of the pass transistor/darlington.

enter image description here

Of course there is some minimum voltage needed to turn on that transistor.

Some devices also detect and shut down when the input is too low.

However, the older 78xx series regulators however are pretty basic and will continue to pass Vin down to a diode drop or two above ground. Vin needs to be higher than Vbe of Q17 and Q16 + VSat of Q9, so about 1.8V or so. Below that the transistors will only be partially on.

Interestingly the current limiter Q15 should still work when it is in bypass mode.

Here is a quick simulation with an LT1084-5 LDO 5V regulator into a 1K load.

enter image description here

Note above about 1.9V the output follows Vin minus about 1V. Below 1.9 it's more of a resistor.

Here is the same device with a 0.1 Ohm load.

enter image description here

Note the current limit kicks in at 6A with Vin merely ~2.5V.

  • \$\begingroup\$ Awesome thanks a lot! So when you say "will continue to pass Vin down to a diode drop or two above ground." what do you mean by diode drop? Do I have to connect a diode between the emitter and collector to achieve the results I need? \$\endgroup\$ Jan 15 '18 at 17:17
  • \$\begingroup\$ @cindercloud no I mean in order for Q17 to turn on in that schematic the input needs to be higher than Vbe of Q17 + Q16 and VSat of Q9. i.e ~1.8V \$\endgroup\$
    – Trevor_G
    Jan 15 '18 at 17:22
  • \$\begingroup\$ Ok, I see what you mean. Thanks for your help. You totally solved my problem. \$\endgroup\$ Jan 15 '18 at 17:26
  • \$\begingroup\$ For this particular LDO, the situation is a little more complicated than this answer indicates. The pass transistor isn't saturated at all when the regulator is in drop-out. Input-to-output voltage is limited by the darlington pair which operates as a follower (and the saturation of the small signal transistor driving it) in drop out. At low VIN values, Q12, Q13, and Q7 can't turn on and the output will likely not turn on until VIN is at least 3 diode voltages. \$\endgroup\$
    – user49628
    Jan 15 '18 at 17:26
  • \$\begingroup\$ @user49628 yup I already changed that. Anyhow, numbers aside, the thing will still pass current way lower than 7V or whatever. \$\endgroup\$
    – Trevor_G
    Jan 15 '18 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.