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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 .

As per the datasheet of UCC28C40 PWM controller specifications, the VDD should be 18V.

But as you can see from the schematic, Vx=Vin-Vaux . Vin is 100 to 120 Vdc ,Vaux should be around 20 V. Also,there is zener diode D12 which has breakdown voltage of 18V. So as per my calculations Vx comes out to be 100-18=82V. But for ucc28c40 to work VDD must be 18V.

So can anybody tell me where am I getting wrong in doing calculations? what is the right value obtained at Vx?

NOTE:I have tried to simplify the concerned circuit by neglecting other components. Here, Vaux is the voltage obtained from auxiliary winding.

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    \$\begingroup\$ Have you seen D12 yet? \$\endgroup\$ Sep 22, 2016 at 7:35
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    \$\begingroup\$ Your calculation of Vx is wrong, your calculation assumes zener D12 is in series with Rstartup1 and 2 and that the input voltage is dropped by a certain amount. Look at the schematic again, ignore Vaux for a moment and look what happens when for example Vx is 20 V ? What will D12 do ? What will Rstartup1 and 2 prevent from happening ? So what will the voltage on Vx be ? \$\endgroup\$ Sep 22, 2016 at 7:47
  • \$\begingroup\$ @FakeMoustache do you want to say that in the simplified diagram shown above, Vx=V_D12? \$\endgroup\$ Sep 22, 2016 at 8:44
  • \$\begingroup\$ It has to be. D12 won't let it get any higher. \$\endgroup\$ Sep 22, 2016 at 8:49
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    \$\begingroup\$ Yes it is needed, only a small amount of current can be delivered through Rstartup1,2. This current is only to charge Cvdd (680uF) to start the chip. Once started it needs more current and Rstartup1,2 cannot deliver that. You could give them a lower value but this would make them dissipate a lot of heat (since they have to drop about 80 V). So having the separate winding to feed the chip is much more efficient. \$\endgroup\$ Sep 22, 2016 at 9:51

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The idea here is that the chip gets (limited) power trough Rstartup initially, but when it starts working, the separate winding of the transformer (Vaux) provides (more) power for the chip. In any case, the zener diode is there to limit the voltage to 18V.

EDIT in response to the comment:

If Vx becomes higher than 18V above ground, the zener diode starts to conduct and the current trough Rstartup would cause a voltage drop on it and would cause Vx to go back down to 18V.

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  • \$\begingroup\$ I have tried to simplify the concerned circuit by neglecting other components. So now could you explain how could Vx be 18V? with the help of Kirchoff's voltage law or something else ? \$\endgroup\$ Sep 22, 2016 at 8:32
  • \$\begingroup\$ @user3559780 I have edited my answer \$\endgroup\$
    – Pentium100
    Sep 22, 2016 at 8:55
  • \$\begingroup\$ then is voltage given from auxiliary winding really needed? can't we obtain supply voltage (i.e. 18V) to UCC28C40 through the Vin directly ? i.e. put zener diode in between VDD terminal of UCC28C40 and Rstartup2 ( setting the Rstartup1 and Rstartup2 values accordingly)? \$\endgroup\$ Sep 22, 2016 at 9:51
  • \$\begingroup\$ @user3559780 This is done for efficiency. You can adjust the Rstatrup values, install a bigger zener diode (or use a transistor to boost the current), but you will find that those parts will dissipate more power than now, making the power supply run hotter and possibly fail efficiency regulations (because that power will be consumed whether the load is plugged in or not). \$\endgroup\$
    – Pentium100
    Sep 22, 2016 at 11:16

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