0
\$\begingroup\$

I have two circuits C1 and C2. The power supply for C1 is much lower-powered than the power supply for C2. How can I have C1 switch on C2?

I'm fairly new to circuitry, but I imagine I'm looking for a single component, whose name I cannot, for the life of me, figure out by Google.

\$\endgroup\$
4
  • 3
    \$\begingroup\$ A relay is one way. \$\endgroup\$
    – Leon Heller
    Apr 7, 2014 at 11:44
  • \$\begingroup\$ Are the supply voltages for C1 and C2 the same? Why don't you just power both circuits from the supply for C2? \$\endgroup\$
    – Joe Hass
    Apr 7, 2014 at 15:11
  • \$\begingroup\$ @LeonHeller, aha, thank you! I do not have enough reputation to upvote your comment, but that is the component I was looking for. \$\endgroup\$
    – dailgez004
    Apr 7, 2014 at 15:19
  • \$\begingroup\$ If its something like 3.3V into 5V you could use a Schmitt trigger \$\endgroup\$
    – Tim M
    Apr 7, 2014 at 15:39

1 Answer 1

Reset to default
1
\$\begingroup\$

If C2 is a DC circuit, and both C1 and C2 share a common ground, putting a N channel MOSFET at the ground of C2 and driving the gate from C1 is a common setup. Drain of the MOSFET is connected to where C2 used to be grounded. Then the Source of the MOSFET is connected to ground. When the Gate of the MOSFET is driven high enough, the junction turns on an C2 is grounded and power flows.

Make sure the current and voltage capability of the MOSFET is greater than the demands of the C2 circuit, as well as turn on gate voltage low enough to be driven by C1. MOSFETs that can be driven by lower voltage circuits are commonly called "logic level" MOSFETs.

As Leon said in the comments, a relay is a easy way to drive something that is AC or much higher power without getting yourself into trouble. Sometimes with very low powered circuits, you will need a MOSFET to drive the relay. Just make sure to look at how diodes are put on coils of relays to short out the voltage spike generated when power is removed from the coil.

\$\endgroup\$

Your Answer

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

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