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I'm reviewing this schematic, I wonder what's this configuration marked with a red box.

enter image description here

what is the purpose of that configuration?

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  • \$\begingroup\$ Interesting... I would suggest putting it into a simulator and try out what it does :) \$\endgroup\$ – Dzarda Nov 3 '13 at 18:55
  • \$\begingroup\$ Isn't it filtering for the transformer taps? \$\endgroup\$ – Passerby Nov 3 '13 at 19:59
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    \$\begingroup\$ Looks like those two transistors are the main drivers for the primary of the power transformer (to their right). Each is driven from a small base winding probably so that they oscillate on their own. Basically, those two transistors make a push-pull power oscillator. \$\endgroup\$ – Olin Lathrop Nov 3 '13 at 21:20
  • \$\begingroup\$ No @OlinLathrop the primary winding of the T2 transformer is connected Q4 and Q5. The schematic does represents there only half of T2 (only the secondary side). \$\endgroup\$ – Standard Sandun Nov 4 '13 at 2:10
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A bit guessing because I don't know what the circuit actually does, it appears to me an SMPS. Not sure of switching frequency, currents and some voltages. Eg. what voltage is across C5/C6?

enter image description here

It appears to me that the transistors (Q1/Q2) are 180 degrees shifted in phase and at any time one of the two is conducting.

A more or less constant voltage will develop across the relatively large capacitor (C10/C11), charged during the positive half cycles on the transformer's winding. Charged by the transformer, the base resistor (R10/R14) and the transistor BE-diode (Q1/Q2).

The voltage across the capacitor is limited by the diode (D3/D4) in series with the resistor (R11/R12), soft clipping the voltage. The capacitor can only 'slowly' discharge through the parallel resistor (R11/R12) in series with the diode (D3/D4), which forms a path 10 times higher resistance than the path when the capacitor is charged. This causes the capacitor to charge with a more or less constant voltage.

After an initial couple cycles the capacitor is charged and the transistor (Q1/Q2) will only start conducting if the transformerś output voltage exceeds the (more or less constant) capacitor voltage.

Then when the voltage swings negative the base voltage is pushed down and the transistor will quickly switch off.

The capacitor voltage will introduce a slight delay so the transistor will only start conducting when the transformer voltage is well in its positive half cycle, to introduce a short dead time between the two transistors being switched on. This would mean the following series of events:

  • Neither of the transistors is conducting for a brief moment;
  • Q1 pulls T1 up for a moment;
  • Neither of the transistors is conducting for a brief moment;
  • Q2 pulls T1 down for a moment;
  • etc.
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  • \$\begingroup\$ This is a guess that sounds very logical. \$\endgroup\$ – abdullah kahraman Nov 8 '13 at 9:39
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in the circuit T2(1), T2(2),T2(3), are the single transformer witch is driven by TL494 PWM IC. a small portion of the main Tr is driving the Q1 base,opposite is driving Q2. 2 series capacitors after the bridge rectifier give a voltage devider, whitch swings voltage +- by switching Q1,Q2 alternatively. the small loop between T1,T2 is for synchronization and stability of the drive. AVGR

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