1
\$\begingroup\$

I have a 12 V, 5 A dual power source with ground in the middle. A 7805 regulator with two 10 μF capacitors is connected to it, outputting 5 V.

I have a 5 V mini water pump connected to that 5 V line and to the ground between the dual source. When I start my circuit, the pump starts OK but then slows down to a halt, the 5 V voltage drops to 1 V and the regulator gets really hot. Why is this happening?

The problem happens even when the pump is connected directly to 5 V instead of going to the output of the logic gates. Proteus did not give me any problems simulating what I'm trying to build:

enter image description here

EDIT: Thank you guys for the input, and sorry if the circuit isn't exactly well-designed, to say the least. It's for my analog logic and digital systems intro class. What I gather from everyone is that I should have another regulator that only has the pumps and relays with inverted diodes connected to it, and activate the relays with the logic output. This should avoid damaging my logic circuit due to either the pump taking too much current or the relays reverse-inducing voltage when deactivated. If anything about this plan sounds wrong, please tell me before I burn up more opamps lmao. Again, thanks!

\$\endgroup\$
10
  • 2
    \$\begingroup\$ Dropping 12V to 5V with a linear regulator at any reasonable current will dissipate a lot of heat. \$\endgroup\$
    – Tom Carpenter
    Commented Jan 12 at 0:08
  • 2
    \$\begingroup\$ How much current does the pump take and how big a heat-sink do you have on the regulator? \$\endgroup\$
    – Kevin White
    Commented Jan 12 at 0:33
  • 4
    \$\begingroup\$ Too much wrong to guess the fault. Driving TTL logic with +/- 12V powered op amps is one, Driving a LED with no series resistor is another. \$\endgroup\$
    – glen_geek
    Commented Jan 12 at 0:55
  • 1
    \$\begingroup\$ Linear regulators work by turning the excess voltage to heat. So more input voltage relative to output voltage means more heat. Motors use a lot of current which means even more heat. 7V to burn off will overheat an unheatsinked regulator even at currents considered to be low. You can reply with @username. \$\endgroup\$
    – DKNguyen
    Commented Jan 12 at 1:52
  • 2
    \$\begingroup\$ "How do I reply to someone in particular here?" Use @username with no spaces in username. They'll get a notification in their SE inbox. (It doesn't send out emails.) Tip: turn off the grid to improve lgibility when taking screenshots. \$\endgroup\$
    – Transistor
    Commented Jan 12 at 4:02

2 Answers 2

Reset to default
1
\$\begingroup\$

You are seriously overloading the 5 V reg.
First heatsink it better to confirm this when you notice that the pump motor runs longer, while the pump is running find out the pump current draw. Now look at the input to the reg and calc your power wasted. Think about your housing space and come to a conclusion about realistic heatsinking.
If things are too bad then consider PWM for the pump which will not waste much power.

\$\endgroup\$
1
\$\begingroup\$

If the 7805 overheats, the pump uses too much current.

In proteus it works because it does not care if you misuse a part, or simulated the pump current correctly.

To drop 12V to 5V, there is 7V over the regulator. At 0.1A to 0.2A, the regulator has to dissipate 0.7W to 1.4W as heat.

That is a lot, and the pump may take much more than 0.1A or 0.2A when it is starting up.

\$\endgroup\$

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