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I am referring below schematic of transformer to generate power supplies in the system. Transformer used is step up transformer (Input-5V, Output: +/- 12V, +/- 48)

Transformer But I am not getting how transistor pairs are operating in this condition and why they have used?

( PS- From my basic understanding, parallel transistors are used for increasing current capacity, but in what condition they are switched ON, that I am not able to figure out)

Left part where 5V is written is basically following circuit:

TRF

Please someone guide. Thank you in advance.

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    \$\begingroup\$ Looks like some kind of self-oscillating push-pull to me. \$\endgroup\$
    – winny
    Jun 27, 2016 at 11:31
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    \$\begingroup\$ This is suppose to be self oscillating. Make sure the phasing of the 1,3 and 4,6 connections is correct. If they're backwards (easy to get wrong) it won't run. Experimentally reverse the connections to 4 and 6 and see what happens. \$\endgroup\$
    – Neil_UK
    Jun 27, 2016 at 11:52
  • \$\begingroup\$ What will happen if we connect unidirectional TVS Suppression Diode or zener diode at D37 ( Cathode up and anode down, with reference to above image )? \$\endgroup\$ Jun 28, 2016 at 5:45

1 Answer 1

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It is basically a multivibrator circuit. It rely on inherent imperfections on the circuit elements to aid its startup. This type of circuit will NOT stimulate properly in circuit simulators unless deliberate imbalance is introduced in at least one of the circuit elements.

What is meant by imbalance is the minor difference in one of the transistor gain, a couple of turns more in one of the coils of the center tapped winding or minor difference in resistance due to their tolerance.

Having said these, when the power is applied the collector current in top transistor row and bottom transistor row will race and one of these rows will dominate. During this time, the induced base current will be as such where it will increase this domination and cause the transistors in that row to go into saturation. This saturation cause the positive feedback base current to those transistors come to an end. At this point the other group of transistors will start to take over the domination. Now the magnetic field in the core switches direction, due to the way the connections are made. The growing magnetic field now induces base current to provide positive feedback to these transistors until the collector current saturates.

This ultimately makes this circuit self oscillate back and forth between saturation and cut-off of the two groups alternatively.

The speed of this oscillation, or the frequency of the oscillator is determined by the transformer impedances (RL value), the RC value at pin 5 of the base drive and also, a bit determined by the base resistance of the two groups of transistors.

This chopper circuit output voltage depends on the primary to secondary windings ratio. The control is open loop, thus voltage regulation is poor.

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    \$\begingroup\$ it's also known as a Resonant Royer Oscillator circuit \$\endgroup\$
    – Brian Onn
    Jun 27, 2016 at 13:09
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    \$\begingroup\$ Check your 3rd paragraph, and then edit it. The induced base current will increase the domination, this is positive feedback. That side stays on until the transformer saturates, which kills the +ve feedback, the on side starts to come out of conduction, then the +ve feedback forces it out of conduction faster and switches the off side on. It is the 'keeping the on side on until something hits an endstop' that times the oscillation. \$\endgroup\$
    – Neil_UK
    Jun 27, 2016 at 13:29
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    \$\begingroup\$ Oh.. Yes, what you are saying is absolutely true... The way I described it wrong because there is no positive feedback is mentioned. Oscillator will not work without the positive feedback. I will edit the answer to suit. \$\endgroup\$ Jun 27, 2016 at 16:30
  • \$\begingroup\$ Thank you all for answers. In schematic which winding is input ( amongst 1, 2, 3, 4, 5 ) and where we get 5V input? What is the reference point ( Gnd ) in windings? \$\endgroup\$ Jun 28, 2016 at 4:13
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    \$\begingroup\$ If you do that, when top transistors switches on the entire 5volt supply will be dumped across that diode and fail it immediately. \$\endgroup\$ Jun 28, 2016 at 6:33

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