enter image description here A step down dc to dc converter converts a high dc input voltage into a low dc voltage. The basic process is

high dc input voltage->oscillator->step-down transformer->rectification->dc-output voltage

As one can see that a circuit of DC to DC buck converter (Vin = 100-120 ,Vout = 5V,Iout = 2.5A) is shown which is obtained from Texas instrument's webench software for schematic .

In the above design consider only the transformer primary part. In the primary side ,it consists of two types of windings 1.primary and 2.auxiliary winding Auxiliary winding is used to supply power to the UCC28C40 .

Among the two ends of primary,one end is connected by dc input i.e. Vin (in between is the snubber circuit that control the effects of the leakage inductance and improve the reliability of the power supply) and the other end is from drain of the MOSFET.

I know that UCC28C40 has an internal oscillator that can convert dc voltage into ac voltage and that voltage is given to one terminal of the transformer But what about second terminal part of primary winding of the transformer (indicated with vertical red arrow )? According to my observation,its a dc voltage .So how transformer could converts the dc voltage to ac voltage?

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    \$\begingroup\$ It doesn't. The MOSFET does. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 22 '16 at 9:01
  • \$\begingroup\$ @IgnacioVazquez-Abrams yes but what about dc voltage Vin given to one end of primary winding of the transformer with snubber circuit in between it ? Does snubber circuit too converts dc to ac signal? \$\endgroup\$ – user3559780 Sep 22 '16 at 9:03
  • \$\begingroup\$ The snubber circuit absorbs the energy from the transformer when the MOSFET is off. Absorbing energy causes the voltage to drop. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 22 '16 at 9:20
  • \$\begingroup\$ @IgnacioVazquez-Abrams Snubbers just control the effects of the leakage inductance and improve the reliability of the power supply. So they directly give dc voltage to one terminal of transformer primary reducing the leakage current. So according to me, on one terminal end of primary we have 100V dc input voltage (Vin) and on the other side we have 18 V ac voltage. So net voltage would be combination of ac and dc voltage. Is it right? \$\endgroup\$ – user3559780 Sep 22 '16 at 9:21
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    \$\begingroup\$ Why must there be a voltage drop? Where does the current flow to? You have to draw a loop which includes ground or the negative terminal of the power supply. \$\endgroup\$ – pjc50 Sep 22 '16 at 11:13

As Ignacio says in the comments, M1 makes the AC. Well not by itself of course. It does so because it is switched on/off rapidly by the chip U1.

When M1 conducts, current flows from +100V, through the transformer's primary winding, through M1 to ground. The current causes the magnetic flux in T1 to increase. When U1 switches off M1, the magnetic flux decreases.

The energy from this constant change in magnetic flux is extracted from the other 2 windings on T1. Only a small part should go through Dsnub as that energy will be lost.

What happens in a mains transformer when connected to AC ? Same thing, a large change in magnetic flux, it follows the mains voltage. So from the secondary winding's view of transformer T1, there is no difference compared to using AC to make the magnetic flux change or "chop" a DC voltage to make that same flux change.

So there is no "proper" AC voltage at the primary winding, it's a "chopped" DC voltage. But that still works !

  • \$\begingroup\$ so can I say that net voltage applied at the primary of the transformer is => 100V dc voltage(Vin) + 18V ac voltage from M1 =>118 VAC ? \$\endgroup\$ – user3559780 Sep 22 '16 at 9:56
  • \$\begingroup\$ No, that voltage is either 100 V (M1 is on) or 0 V (M1 is off). I would call that an AC voltage of 100 V peak-peak. The 18 V AC is not related nor involved in this. That 18 V is just the DC voltage which feeds the chip. Note that the winding for Vaux is a separate winding and it is also a secundary winding. \$\endgroup\$ – Bimpelrekkie Sep 22 '16 at 10:04
  • \$\begingroup\$ sorry sir but please here note that I am talking about 118VAC between only 2 ends of the primary winding not the auxiliary winding. please refer upper half of the primary winding. here, one end is from Vin and other end is from M1 through snub circuit. \$\endgroup\$ – user3559780 Sep 22 '16 at 10:11
  • \$\begingroup\$ Good ! But since you included the 18V I thought you were confused by the aux winding. That's why I stated that it is not involved in this case. \$\endgroup\$ – Bimpelrekkie Sep 22 '16 at 10:14

The UCC28C40 receives supply during startup through R startup 1 and 2. Immediatly after startup VDD supply commes from the second winding on the transformer after being rectified and limited. The second winding is only for the supply of the UCC.

The winding indicated by the arrow is the primary winding controlled by the UCC.

Remark: Since the system is a flyback converter, the flux in the core is increased during the open time of the UCC and the decreasing flux during the off time of the UCC is inducing the output voltages on the other two windings


This is an example of of a flyback type power supply and the voltage across all the windings is ac not dc.

the operation is as follows when the FET M1 turns on there is no current in any of the other windings so the current ramps up in the primary storing energy in the transformer core. When we turn M1 off this current still wants to flow so the voltage at the drain of M1 rises above Vin. The voltage across all the windings reverses direction and the energy stored in core is released through the secondary and auxiliary windings. There is a small amount of energy in the leakage inductance of the primary which can not be released by the secondary so to prevent the drain voltage going excessively high and damaging the FET we need to clamp the voltage which is the function of Dsnub, Csnub, Rsnub1 and Rsnub2.


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