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They are push-pull (alternatively driven) drivers to the main switching stage.
Pin 12 of the TL494 is the positive supply feeding the IC and the center of the transformer through the unmarked resistor and series diode. The bias for the two transistors comes from the positive supply via the 3K3 resistors and is pulled to ground to turn them off through the internal transistors in the TL494 (C1,E1 and C2,E2).
I suspect the two diodes in the emitters of the transistor are to ensure they turn completely off as the TL494 output saturation voltage may be too high if they were grounded. By lifting the emitters of the transistors above 0V it makes the base voltage appear lower.
I am suspicious that some of the diodes may not be 1N4148 as it would be pushing them close to their limits.
The very fact that the two NPNs are pulled to ground confuses me.
Suppose these outputs are totem-pole. then for push pull we can say Transistor 1 is turned on for a maximum of 45% then 5% dead time then Transistor 2 turn on for 45% and then 5% off time and this goes on.. Bipolar drive
So here according to the above circuit the transistor T1 turns on for 55% and Turns off for 45% ???? same with T2 ???
The schematic in post #5 makes more sense. As pointed out, the biasing for the two transistors should be identical but one has an extra resistor for some reason. I suspect the resistor with no value is actually a link as adding resistance there will degrade the push-pull performance.
The 3K3 resistors in both schematics turn the transistors on and the TL494 diverts the bias current to ground to turn them off. It really doesn't make any difference whether they default to on and are turned off or if they are off until turned on as long as the drive waveform adapts accordingly. This is not a totem pole configuration where the ratio of top/bottom waveform changes the average in the middle, the transistors are turned on alternately, the duty cycle of both transistors is changed simultaneously to control the output power.
I'm not sure what you mean. The 2SC945 are drivers in a push-pull configuration. They drive a single center-tapped transformer and it is the output of that transformer that drives the main switching stage.
It may be possible to replace the transistors with MOSFETs but it wouldn't make the circuit any better, it would still have to be push-pull.
Maybe you are confusing the kind of PSU with a PWM half or full bridge circuit where the top and bottom transistors in a total pole are driven with different waveforms. In this PSU the output is controlled by the pulse width sent to both transistors although only one at a time. If you look at the TL494 data sheet you will see that both outputs are from internal transistors at the same potential, there is no 'high side' and 'low side' output as such.
To add to previous posts, the first circuit apparently has a fifth winding on the gate drive transformer which is coupled to the output of the final push pull stage. There's a slim chance this could be some sort of neat trick (the schematic doesn't indicate turns ratios or even winding polarity) but I'm 99% sure the schematic is just rubbish.
The second schematic is much more reasonable. But ATX supplies are notorious for using lots of convoluted analog circuits in order to keep costs as low as absolutely possible. Such as using an extremely basic controller like the 494.
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