This is my first design of a asynchronous buck converter, Vin 36V, Vout 24V, Iout 1A, Frequency switch 100KHz, based on "Basic Calculation of a Buck Converter's Power Stage" - slva477b Texas Instruments. It works as intended, while 1A load connected, without load outputs 29V. From other posts I understand that for this type of buck, duty cycle MUST be adjusted properly for different loads. How can this be accomplished, if possible, without feedback? Does the circuit need a minimum load ? Thank you enter image description here

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    \$\begingroup\$ Turn off the grid, man. Turn off the grid before taking screengrabs! \$\endgroup\$
    – Transistor
    Commented Mar 25, 2018 at 13:36
  • \$\begingroup\$ First off, there is no reason why you wouldn’t want feedback. Second, not really. If you are operating in CCM all the time and the input voltage doesn’t change, duty cycle will stay relatively unchanged. \$\endgroup\$
    – winny
    Commented Mar 25, 2018 at 13:38
  • \$\begingroup\$ turned off grid. sorry. the reason for not using feedback is the buck will be used for keeping output between 24v-30v, just to prevent overheat of regulator. so, if supply fixed input of 36v, there is no need of feedback to adjust the duty, in CCM mode, right ? \$\endgroup\$
    – johnger
    Commented Mar 25, 2018 at 13:48
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    \$\begingroup\$ the buck will be used for keeping output between 24v-30v That makes no sense to me. Why drop only a few volts across the efficient buck circuit (not calling it a regulator because it lacks feedback so it is not a regulator) and then drop most of the remaining voltage across an inefficient linear regulator? It makes no sense to me. More sense would make: regulate (using feedback) the output of the switching converter to 8 - 10 V. Then drop only 3 - 5 V across the 7805. \$\endgroup\$ Commented Mar 25, 2018 at 14:05
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    \$\begingroup\$ Using a proper buck controller will not only be cheaper, but you also get soft-start, under voltage protection, over current protection, thermal shutdown, etc. Unless this is an academic exercise and you want to learn, I'd stick to a proper IC, which would also be mucho more efficient. It's also fine to have one switching regulator per voltage output needed, rather than linear ones. \$\endgroup\$
    – Andrés
    Commented Mar 25, 2018 at 15:20

1 Answer 1


Using a buck converter as a pre-regulator to drop the input voltage for a linear regulator is a valid approach and way simpler than a full regulated buck converter. An unregulated buck converter avoids a lot of the headaches you'd have with a regulated one.

You can avoid the output voltage rise without load by replacing D4 with another N-MOSFET. This FET should be on when Q3 is off and vice-versa. Make sure to include some dead-time so you don't have shoot-through between the FETs.

With those two FETs (that form a half-bridge), energy can flow back from the output capacitor to the input of the buck converter when the voltage at the output is too high, allowing the converter to always operate in CCM mode and giving you the desired output voltage even with no load. As a bonus, it will also be more efficient because the voltage drop across D4 will be avoided.

As Bimpelrekkie already stated in the comments above, you should probably drop the voltage to about 8V for the 7805 and use a second identical buck converter to create a higher-voltage rail.

  • \$\begingroup\$ This has solved most of my doubts for now, next step is building the circuit. Have followed your advice, used a 2nd Nmos, adjusted a little bit the duty and voila, 24Vout with or without load, and increased efficiency. I will build this circuit to have a realistic view of this, for sure I miss some other things which could better the circuit. \$\endgroup\$
    – johnger
    Commented Mar 25, 2018 at 16:01
  • \$\begingroup\$ @johnger The circuit looks good to me. It doesn't have dead-time control, but you might get away without it because your gate drivers have slower rise times than fall times. If you do get shoot-through and therefore need dead-time control, you'll notice because your FETs will get really hot (and might die) even with no load. \$\endgroup\$ Commented Mar 25, 2018 at 17:40

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