0
2.2kviews
Pushpull SMPS Converter
1 Answer
0
70views

SMPS with push-pull configuration is shown in Fig. It uses two power MOSFETs $M 1$ and $M 2$ and a transformer with mid-taps on both primary and secondary sides. As in flyback converter, an uncontrolled rectifier feeds push-pull SMPS. Inductor L and capacitor C are the filter components.

enter image description here

When $M 1$ is turned on, $V_{s}$ is applied to lower half of transformer primary, $i . e . v_{1}=V_{s}$ . As a result, voltage $v_{2}=\frac{V_{s}}{N_{t}} N_{2}$ is induced in both the secondary windings. Voltage $v_{2}$ in the upper half secondary forward biases diode D1 , therefore load voltage $V_{0}$ is given by $V_{0}=\frac{V_{s}}{N_{1}} N_{2}=\alpha V_{s}$ .

When $\mathrm{M} 2$ is turned on, $v_{1}=-V_{s}$ is applied to upper half of primary winding. Consequently $v_{2}=-\frac{V_{s}}{N_{1}} N_{2}$ is induced in both the transformer secondaries. As $v_{2}$ is negative, diode D2 gets forward biased and $V_{0}=\alpha V_{s}$ as before.

This shows that voltage on primary swings from $+V_{s}$ with $M 1$ on to $-V_{s}$ with $M 2$ on. Power MOSFETs $M 1$ and $M 2$ operate with duty cycle of $0.5 .$ When $M 1$ is off, the voltage across $M 1$ terminals is $V_{o c}=2 V_{s}$ . As both $M 1$ and $M 2$ are subjected to open-circuit voltage of $2 V_{s},$ this configuration is suitable for low-voltage applications only.

Please log in to add an answer.