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Feb 26, 2022 at 9:59 comment added Marcus Müller I hope the answer I've linked to well explains why that's the case.
Feb 26, 2022 at 6:30 comment added Shobeh @MarcusMüller I thought the 3 V DC motor could handle the power coming from the 5 V regulator. The motor did work as imagined but the 5 V regulator did seem to be on the verge of burn out....
Feb 26, 2022 at 6:27 comment added Shobeh @StainlessSteelRat yeah. The 3 V DC motor could take power from a 5 V, 500 mA charger and run fast. This baby is hardy!
Feb 26, 2022 at 6:23 vote accept Shobeh
Feb 25, 2022 at 16:51 comment added StainlessSteelRat 9V battery supplying 5V regulator supplying 3V motor. This seems logical to you! And you are worrying about the regulator overheating. I'm surprised the 3V motor is not smoking! Consider a 3.3V switcher driven by your 9V battery. SPCLVR - Society for Prevention of Cruelity to Linear Voltage Regulators.
Feb 25, 2022 at 16:42 history closed Marcus Müller
devnull
Elliot Alderson
ocrdu
StainlessSteelRat
Duplicate of Why is the linear voltage regulator going hot although within its specification?
Feb 25, 2022 at 16:12 comment added Hearth @Shobeh You can't use a 7805 with a 6 volt input; it won't regulate right. The 7805 needs about 1.5~2 volts of difference between its input and output, if I remember right. If you want to use a lower input to output differential, you'll need a low-dropout regulator (and note that low-dropout regulators are pickier about input and output capacitance, so read the datasheet). But really, you should be using a switching regulator if you're concerned about things heating up.
S Feb 25, 2022 at 15:31 history suggested Ralph
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Feb 25, 2022 at 10:48 comment added Marcus Müller no, you'd simply need to use something else than a linear regulator when you're dropping a non-negligible amount of the supply voltage; 1V difference between in- and output might not be enough for the LM7805 to work properly. Also, using a 5V regulator with a 3V motor was ... questionable, to begin with.
Feb 25, 2022 at 10:44 comment added Shobeh @MarcusMüller Thanks for the assistance, comrade! I think I must weaken the power of 9 V battery and turn it into 6 V so the LM7805 will be usable. Thanks!
Feb 25, 2022 at 10:37 review Suggested edits
S Feb 25, 2022 at 15:31
Feb 25, 2022 at 10:37 comment added Shobeh @MarcusMüller I really need to read more...ahahhahahahah!
Feb 25, 2022 at 10:30 review Close votes
Feb 25, 2022 at 16:48
Feb 25, 2022 at 10:28 history edited ocrdu CC BY-SA 4.0
added 6 characters in body; edited title
Feb 25, 2022 at 10:28 answer added Ralph timeline score: 4
Feb 25, 2022 at 10:21 comment added Marcus Müller It is very straightforward! The answer to the question I've referred you to explains what the LM7805 does, in very basic electronics terms. The fact that you need decoupling caps stems from dynamic control theory, in the end; why might not be straightforward, but "read the datasheet and see it says you need caps" is pretty straight, I'd say :)
Feb 25, 2022 at 10:20 comment added Shobeh @MarcusMüller I thought using LM7805 is a straightforward process. Boy I was wrong....
Feb 25, 2022 at 10:16 comment added Marcus Müller not to mention that cable twisting is not an appropriate way of connecting wire to an IC, or a motor.
Feb 25, 2022 at 10:15 comment added Marcus Müller also bad, but unrelated here: the LM7805 needs decoupling caps. You must not use it without.
Feb 25, 2022 at 10:14 answer added Andy aka timeline score: 3
Feb 25, 2022 at 10:09 answer added Justme timeline score: 2
S Feb 25, 2022 at 10:05 review First questions
Feb 25, 2022 at 10:18
S Feb 25, 2022 at 10:05 history asked Shobeh CC BY-SA 4.0