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My circuit uses main electricity in parts of the circuit. However my AVR and other components only need 5V. So I'm using a step down transformer to generate 12V. It's then regulated to 5V.

The transformer won't convert it to dc for me will it?

It just steps it down then outputs a certain voltage depending on it's coils. So I would need to convert it to dc myself right?

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    \$\begingroup\$ Yes, you answered your question correctly for yourself. \$\endgroup\$
    – Jaroslav Cmunt
    Commented Apr 1, 2011 at 10:41
  • \$\begingroup\$ Bear in mind that transformers are rated for a.c. rms and fully loaded. Once you regulate and smooth, you get close to the peak voltage (minus diode drops), and will probably be 10% higher than that for small current loads on a small transformer (less for larger transformers, see the regulation figure). \$\endgroup\$
    – Martin
    Commented Apr 1, 2011 at 10:47
  • \$\begingroup\$ Rectify it baby... en.wikipedia.org/wiki/Rectifier \$\endgroup\$
    – kenny
    Commented Apr 1, 2011 at 10:52
  • \$\begingroup\$ Same exact application, somewhat different question: electronics.stackexchange.com/questions/515/… \$\endgroup\$
    – endolith
    Commented Apr 1, 2011 at 13:44
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    \$\begingroup\$ Comments below assume that you are talking about a transformer when you say "transformer." Some people I know (not EEs) say "transformer" when they really mean "external power supply." \$\endgroup\$
    – Solomon Slow
    Commented Aug 24, 2017 at 15:10

4 Answers 4

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You are correct. The transformer will only reduce the voltage (and increase the available current), so you need to add additional circuitry to rectify, smooth and regulate your 12\$V_{AC}\$ transformer output to 5\$V_{DC}\$.

This is the type of circuit you should be looking to build:

AC to DC PSU

  1. The transformer reduces the voltage from mains to 12\$V_{AC}\$ (RMS).
  2. The Diode Bridge (known as a bridge rectifier) will convert 12\$V_{AC}\$ to 15\$V_{DC}\$. The voltage is \$\sqrt{2}\$ times minus two diode voltage drops higher than the input voltage because the rectifier output is the peak AC voltage, not the RMS AC voltage.
  3. The first capacitor will smooth out the ripples that come from the output of the AC to DC bridge rectifier.
  4. The LM7805 regulator will maintain a constant voltage as the load varies. For example if you are switching a light bulb on and off, the current will go up and down, and if you didn't have a regulator then the voltage would drop as the bulb is switched on. The regulator keeps it at the 5\$V_{DC}\$ your microcontroller needs.
  5. The final small capacitor filters out any noise or interference on the regulated side of the circuit.
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  • \$\begingroup\$ Thanks for the schematic BG100. I see you seen Avatar. It was awesome. \$\endgroup\$
    – Ageis
    Commented Apr 1, 2011 at 12:59
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    \$\begingroup\$ @BG100, sorry but this is a common error. 12V a.c. into a bridge does not product 12V d.c. More like 17V pk (12 * 1.414) \$\endgroup\$
    – Martin
    Commented Apr 1, 2011 at 13:05
  • \$\begingroup\$ Sorry I've been forced onto this project. I'm on a steep learning curve here. So which one of these rectifiers will do the job for me. maplin.co.uk/bridge-rectifiers-19088 \$\endgroup\$
    – Ageis
    Commented Apr 1, 2011 at 13:08
  • \$\begingroup\$ Also whats PIV and RMS \$\endgroup\$
    – Ageis
    Commented Apr 1, 2011 at 13:09
  • \$\begingroup\$ @Martin: Absolutely right... I'll edit and correct. Thanks. \$\endgroup\$
    – BG100
    Commented Apr 1, 2011 at 13:27
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The transformer is the first part of the power supply, but you need more to get DC.

The varying input voltage of the transformer creates an equally varying magnetic field in the metal core. This magnetic field in turn creates an, again varying, voltage on the output. Both input and output voltage are sine waves. It's called AC (alternating current) because the sign of the output voltage changes continuously, 100 or 120 times per second, depending on the country you live in.
You rectify this AC voltage to get rid of the sign changes; one pin will always be positive with respect to the other one.

enter image description here

The one thing which remains to be done now is to flatten the curve, get rid of the bumps. This is done by a capacitor. You now have a DC (direct current) voltage which is already usable for a number of situations. This voltage, however, may still show slight variations, which may be unwanted. To get rid of those you follow the capacitor by a voltage regulator.

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If you need 12 VDC, it may be simpler to use an off-the-shelf wall wart with a 12 VDC output. Such devices do convert to DC for you. A few of them use exactly the circuit BG100 shows, but nowadays many of them use a switching regulator technique that requires more parts, but has a lower net cost.

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  • \$\begingroup\$ +1 One should definitely know the principles behind a power supply, but when it comes to building a product (or hobby project) an off the shelf wall wart is the way to go: These things don't cost a fortune, and you get a nice case, cable, connector. Plus: You can be sure to meet all requirements for mains power (distances, isolation, ...) \$\endgroup\$
    – 0x6d64
    Commented Sep 15, 2011 at 9:51
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    \$\begingroup\$ .. and it makes product regulatory approval easier, and you don't have to worry about safety yourself. \$\endgroup\$
    – pjc50
    Commented Sep 26, 2012 at 11:09
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The transformer won't convert it to dc for me will it?

Correct!

AFAIK it just steps it down then outputs a certain voltage depending on it's coils. So I would need to convert it to dc myself right?

Correct again!

Your transformer will convert \$120V_{AC}\$ to \$12V_{AC}\$. So you'll need to regulate it to \$5V_{DC}\$.

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    \$\begingroup\$ or in this case, 230V to 12V \$\endgroup\$
    – JustJeff
    Commented Sep 15, 2011 at 11:13

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