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I don't understand how this circuit works. I'm not sure if it's correct. It should give a DC output, but how would it do that?

enter image description here

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    \$\begingroup\$ Where did you find the circuit? \$\endgroup\$ – Andy aka Nov 25 '14 at 10:02
  • \$\begingroup\$ I give you a link... but I don't know it is right or not.. :) techtunes.com.bd/electronics/tune-id/105328 .. I simulate it.. it works.. but the voltage gradually increases.. from 0 to 9... I used all the resistance 10k while simulating... \$\endgroup\$ – Raihan Khalil Nov 25 '14 at 10:12
  • \$\begingroup\$ I turn on the simulator and keep it up to approximately 20 minutes.. Now it is showing 130 V.. :) \$\endgroup\$ – Raihan Khalil Nov 25 '14 at 10:20
  • \$\begingroup\$ The link you provide is undecipherable and not written in English (standard for this site) \$\endgroup\$ – Andy aka Nov 25 '14 at 10:37
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    \$\begingroup\$ You have the "input" and "output" labels swapped over. The circuit is an AC-output inverter, not a DC-output PSU. \$\endgroup\$ – John Honniball Nov 25 '14 at 13:29
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Edit: As drawn, this is a clever synchronous rectifier using BJTs as very-low-drop diodes. It would be suitable for a measuring instrument since the voltage drop is very low (no diode drop loss). A full-wave rectified output appears at the terminals.

If you reverse it (the way you'd normally draw it with flow from left to right), you get this:

This circuit is called a blocking oscillator. The input is the DC and it produces a kind of square wave high voltage AC at the output of the transformer.

It depends on either the transistor coming out of saturation (due to insufficient base current for the current at the collector, which increases with on-time) or the transformer saturating (due to core saturation due to the high collector current) for the turn-off.

They're not very well controlled or very efficient circuits, but back in the dark ages, I made some simple inverters that were not much more than a 'filament transformer' a couple 2N3055s and some resistors (plus a snubber) (similar to this) that could run small AC-powered measuring instruments from a 12VDC source.

enter image description here

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  • \$\begingroup\$ If I do the opposite thing?? I mean if I give 220 or more AC input in the transformer then can't I get DC output?? \$\endgroup\$ – Raihan Khalil Nov 25 '14 at 15:06
  • \$\begingroup\$ Interesting question- does it work as a synchronous rectifier? \$\endgroup\$ – Spehro Pefhany Nov 25 '14 at 15:10
  • \$\begingroup\$ IF it works as a synchronous rectifier, the transistors would be in reverse saturation when conducting. \$\endgroup\$ – Oskar Skog May 8 '17 at 9:30
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I think you (or the author) got the input and output swapped. This seems to be a circuit for generating a sort-of 220V AC output from a 5-12V DC input. The two transistors and the secondaries of the transformer (here used as primaries!) form an oscillator. A comparable circuit can be found here:

enter image description here

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  • \$\begingroup\$ may be.. then how it works... ? If you swap the input and output.. \$\endgroup\$ – Raihan Khalil Nov 25 '14 at 10:21
  • \$\begingroup\$ @RaihanKhalil Why would you want to know how this works if you are not sure what it does? \$\endgroup\$ – Andy aka Nov 25 '14 at 10:38
  • \$\begingroup\$ I don't see how this circuit could possibly oscillate. \$\endgroup\$ – Dave Tweed Nov 25 '14 at 12:09
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    \$\begingroup\$ The only thing on that site that gets remotely close to a schematic is a blurry hand drawn image which has a label "InPuT <gibbersih> 12 V D[C?]" and at the transformer "OUt 2?? V". Running the site through google translate suggests that one of the components needed is a car battery, so I think you are right in that this is meant to be some kind of 12V => 220V thing. \$\endgroup\$ – PlasmaHH Nov 25 '14 at 13:23
  • \$\begingroup\$ @PlasmaHH I think you're right; the schematic is for a 12VDC-to-220VAC inverter. \$\endgroup\$ – John Honniball Nov 25 '14 at 13:28

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