0
\$\begingroup\$

Question, as fast as possible: Would this circuit regulate poorly-regulated +12VDC effectively?

I am an electronics level zero, so please bear with me. Please let me know if you would me to clarify anything.

Full Question:

I'm trying to design this smart lighting system and I need a power source. Unfortunately, due to the wire run lengths, I can not rely on receiving well-regulated +3.3v directly, especially at high currents.

I want to keep the device as small and low-profile as possible, limiting my options.

I designed this circuit to regulate voltage effectively. Would this work?

I am alright with up to 0.2v of ripple. All resisters are 1k ohms, and the capacitor is 100uF.

Circuit Schematic

Also, if I messed up something which would make this circuit blow up, please let me know. This one of the first circuits I've designed from scratch.

\$\endgroup\$
  • \$\begingroup\$ How much current do you need from it? \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 23 '16 at 1:06
  • \$\begingroup\$ About 1-3 amps at most. \$\endgroup\$ – yash101 Jun 23 '16 at 1:07
  • 1
    \$\begingroup\$ 3A is not going to be a small circuit with a linear regulator. Grab a switching regulator off eBay. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 23 '16 at 1:08
  • \$\begingroup\$ How is this a linear regulator? Doesn't the MOSFET switch on when the voltage is low, and off when high? \$\endgroup\$ – yash101 Jun 23 '16 at 1:09
  • \$\begingroup\$ You don't have an inductor, and you don't have hysteresis. Performance will be in the toilet. And that's if you're lucky. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 23 '16 at 1:10
1
\$\begingroup\$

I can do a little bit math for you. If you have Vin = 12VDC, Vout = 3.3VDC and draws I = 3A at 3.3V, then you will have (12 - 3.3) * 3 or 26.1W wasted power with a linear regulator. You will get totaly 9.9W, so with other words a efficiency of 37%.

The most efficiency buck-converter (Switched-mode) will only waste under 10% of the input power. Here is a simple example:

Since you only need 3A, then you set the Rsense to 0.01 Ohm, and to get 3.3VDC out of it, you need to change Rselect to 1.66K.

\$\endgroup\$
  • \$\begingroup\$ How is the is a linear regulator? Doesn't the transistor turn on when the voltage is low, and off when it is high? If not, how can I do that? \$\endgroup\$ – yash101 Jun 23 '16 at 1:56
  • \$\begingroup\$ You are partly right. The transistor, or mosfet allows current to flow through it to ground when the voltage is to high. But the power that flows through the transistor or mosfet is equal to 26.1W, if you input 12V and draws 3A. But that doesn't make it to a switched mode converter. A switched mode convert uses high frequency to «chop» up the voltage. If you input 12V to a switched mode converter and want's 3V out, it turns the power of for 3 times X and on for 1 times X. \$\endgroup\$ – BufferOverflow Jun 23 '16 at 2:07
  • \$\begingroup\$ Wouldn't the MOSFET turn on whenever the voltage is low? Also, the MOSFET does not conduct directly to ground. It has the load in between. The capacitor smooths the voltage. Would it work (better) if I used a window comparator instead? I'm alright with a bit of ripple. \$\endgroup\$ – yash101 Jun 23 '16 at 2:41
  • \$\begingroup\$ The mosfet will only allow current to flow through if the voltage is over 3.3V. Go for a buck converter, easier to build, you don't need huge heat-sinks, it can be over 90% efficiency, it takes less space. It's only one downside with it, the parts is more expensive, or two downsides. It output some riple vs a linear that has almost no riples. \$\endgroup\$ – BufferOverflow Jun 23 '16 at 2:56
  • \$\begingroup\$ Alright. Thanks! I guess it's pointless to complicate this more than necessary! :). \$\endgroup\$ – yash101 Jun 23 '16 at 3:24

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.