The push-pull half-bridge converter

Let us study a schematic diagram of a two-stroke half bridge converter, of an international naming «half bridge» (rice. 1).

File0052Fig.1. The push-pull half-bridge converter

While the gates of the transistors are not received voltage, they are closed. The voltage at the middle point of the capacitive divider, performed on the capacitors C1 and C2 of the same capacity, is half of the DC voltage, supply converter.

Will provide from the master oscillator to the gate trigger voltage VT2 of transistor. By chopping + Uent, capacitor C1, TV1 transformer winding, transistor VT2, -in theent potechet current. In the secondary winding of the transformer there TV1 voltage, which will be rectified diode VD1 assembly and smoothed capacitor C3. Transistor VT1 all this time has been closed.

Will provide a blocking voltage to the gate of the transistor VT2 and is based on the gate voltage of the transistor VT1. Current flows through the circuit + Uent,transistor VT1, TV1 transformer winding, capacitor C3, -in theent. In the secondary winding of the transformer will TV1 voltage of opposite polarity relative to the previous cycle, which straighten VD1 diode assembly and smooth out the capacitor C3. Then a DC voltage from the capacitor C3 is applied to the load. VT2 transistor during a second closed cycle.

How visible, current flows through the load during both cycles. output voltage ripple frequency of twice the frequency conversion, that allows the use of capacitor C3 the smoothing filter with a small rated capacity. Private hysteresis loop of the magnetic core of the transformer half-bridge converter close to the limit of the hysteresis loop.

As long as the load is not connected to the SMPS, each capacitor of the capacitive voltage divider is applied half of the DC voltage, supply converter. If the capacity of the capacitor voltage divider is not big enough, then at maximum load during each half cycle the capacitors are substantially discharged, and the voltage across them exceeds half the supply voltage converter.

Voltage, applied to the primary winding of the pulse transformer of the half bridge converter, It can be calculated by the formula:

File0053where UP - constant pressure, supply converter;

in theus - saturation voltage of a transistor.

The capacity of each capacitor of the voltage divider can be calculated by the following formula:

copy File0053Where C - the capacitance, F;

Iperv.maks - the amplitude of the total current through the primary winding of a transformer;

F – frequency conversion, Hz;

ΔUwith - the change in voltage on the capacitor for the time duration of the total current passing therethrough pulse Iperv.maks.

The magnitude of the voltage applied to the capacitor of the variable component shall not exceed the maximum allowable reference value for a component of the brand and type of. Important to remember, the rated capacity of many capacitors at high frequencies and at low ambient temperatures is substantially reduced.

Half-bridge converters are widely used in power output from a few watts up to several kilowatts.

The virtue of the half bridge inverter is a low reverse voltage, attached to each transistor in the off state, approximately equal to the DC voltage converter power supply.

This allows you to use half-bridge converters with a high supply voltage. Half-bridge converters can be incorporated without load, and thus will not be dangerous damage to parts. Pulse frequency is twice the frequency conversion.

If the voltage divider capacitors are strictly identical, switching transistors are identical to each other, magntoprovoda and hysteresis loop of the material contains no defects, it can be assumed, that the magnetization of the core of the pulse transformer offline. Such a pattern can only be ideally. So, For example, real half-bridge converter, the capacitors in the voltage divider is always different from each other and, Consequently, nesymmetrychno peremahnychyvanye transformer. However, the degree of asymmetry is usually much less than the, than in the magnetic single-ended converters transformers. One of the simplest ways to reduce podmagnichivanmya core half bridge inverter is the inclusion of a non-polar capacitor between the pulse transformer and the midpoint of a capacitive voltage divider.

The disadvantages include the presence of two capacitors in a voltage divider, destruction of the components of SMPS overcurrent in the load in the absence of the protection system, lower efficiency, than attainable in the bridge converter.

A source: Power sources. Moskatov E.A.


  1. “Will provide from the master oscillator to the gate trigger voltage VT2 of transistor. The chain Ui +, capacitor C1, TV1 transformer winding, transistor VT2, -Uvh potechet current.”

    Maybe I do not understand, or maybe I'm a little experimenting with chemistry, but, if current flows through C1, then the capacitor breakdown and it needs to be changed. no current flows through the run capacitor never (or at a constant voltage, or at a variable).

    Operation of the circuit is as follows:
    1. VT1 opened, VT2 is closed. Current flows Ui +, VT1, TV1, C2. When this charge accumulates on the capacitor plates, and does not pass through it. In this case a consumer of energy C2.
    2. VT1 is closed, VT2 open. Current flows C2, TV1, VT2, --Uvh. In this case the energy source is C2.

    C1 then need to form a capacitive voltage divider.

    • 1) we have alternating current, not stress
      2) alternating current flows through the capacitor, this is a completely correct description of what is happening

      • Respected, read the theory. DC and AC voltage, not current. In foreign publications, the current is alternating and direct, this is fundamentally wrong, their stars are holes in the sky, not planets emitting light. Maybe your land is flat? Current is, like water, turned on the tap, how much did you drink, drank so much. If it is an LED bulb, then it consumes a certain amount of current stably. At 220V, 7Watt, it turns out 0.03A. And if it's a radio, then it consumes current unstable, the program is clearly configured, one meaning, the program is a little lost, there was interference, different current, ie. the radio consumes current dynamically. Western DC / AC construction was hammered into your head. Although if you figure it out, this is just an abbreviation and in translation it will mean, that the constant voltage current and the alternating voltage current. I hope, Did you learn english, and according to the rules of translation, must be translated from the end, main word current. Learn electrical engineering and good luck to you.

  2. C1 and C2 are divisors of Uin. and energy sources, respectively C2 to VT2, and C1 of VT1 until diluted. When C1 supplies VT1 current flows through the circuit and charges C2. When C2 supplies VT2, the + -Uvh. charges C1.

  3. “Will provide a blocking voltage to the gate of the transistor VT2 and is based on the gate voltage of the transistor VT1. Current flows through the circuit Uin +,transistor VT1, TV1 transformer winding, capacitor C3, -Uvh.”
    Maybe I misunderstood something, but the current flows through C3 in the secondary circuit of the transformer, and it is galvanically isolated, ie. the current from the primary winding cannot flow through the secondary – this is nonsense. In the primary winding, the current flows through the capacitor C2. ie. it will be right: “Will provide a blocking voltage to the gate of the transistor VT2 and is based on the gate voltage of the transistor VT1. Current flows through the circuit Uin +,transistor VT1, TV1 transformer winding, capacitor C2, -Uvh.”

  4. Except for a typo, which I pointed to above, cool article! Everything is simple and clear. On other sites it is written like this, that you can't figure it out without half a liter, although half a liter is unlikely to help there. I would like to see an addition to the article, at least a couple of paragraphs, why is all this needed? Why a half-bridge converter, if there is a pavement. Saving on two diodes? Yes, of course. But with half-bridge, two capacitors are added, which are also not free. A plus, as far as I understand, capacitors hold half the charge on the plates, and the transformer won't start working, until the voltage across the capacitors exceeds 50%. ie. whether or not we use half of the potential power of the transformer, or install a less powerful and cheaper transformer. So is it?

  5. No current flows through the capacitor. There is a charge current and a capacitor discharge current. In this case, in the initial state, capacitors C1 and C2 are charged to half Uin each. When T1 opens, the discharge current of the capacitor C1 flows through the circuit: +C1, VТ1, TV1, -C1. When VT2 opens, the discharge current of the capacitor C2 flows through the circuit: +C2, TV1, VT2, --Uvh. Since the impulses are short, and capacities C1 and C2 are large, then they do not have time to discharge, and in the intervals between pulses are recharged.

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