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I apologise for the attached picture, I just made sure everything is readable so you can assist.

So I'm wondering if this is a logical idea, if 12 volts is supplied, to the circuit, as the capacitor reaches the zener 5.1v + 0.65v of the transistor Vbe, it will switch off the p channel MOSFET, and thus stop the current flow from supply, until the voltage in the capacitor drops below 5v1 + 0.65v again, which in turn will allow the p channel MOSFET to be switched on again.

Is this possible? Please feel free to correct me, thank you.my buck converter sketch

BELSmith
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    Without hysteresis, such circuits will often just settle in the linear zone, with the output a steady 5.7 ish volts. – Unimportant Mar 22 '21 at 12:52
  • But that's exactly what I want – BELSmith Mar 22 '21 at 13:00
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    @BleedingEdgeLab No it's not, that will fry your transistor. What Unimportant is saying is that this is a linear regulator. – Hearth Mar 22 '21 at 13:05
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    There will be half of power waist on Q2, no advantage of SMPS. –  Mar 22 '21 at 13:06
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    The point is that this is NOT a buck regulator but rather a (poorly designed) linear regulator. Q2, however, will need to dissipate the entire load of dropping from 12 to 5V. – jwh20 Mar 22 '21 at 13:24
  • As most comments say: this will behave as a linear regulator. If you disagree then explain to us how Q1 can only be on or off and how it will **not** be "half conducting" such that there will be about 6V at the output. – Bimpelrekkie Mar 22 '21 at 13:30
  • You could answer a lot of your own questions by running a simulation of this circuit. – Elliot Alderson Mar 22 '21 at 13:34
  • No oscillator means no buck converter! – StainlessSteelRat Mar 22 '21 at 13:36
  • TI has an example schematic in their [datasheet for the LM317HV](https://www.ti.com/lit/ds/symlink/lm317hv.pdf?ts=1616403593034&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM317HV) where it's slightly abused into acting as a switching buck regulator (9.2.11 on page 17). Perhaps if you analyze that circuit you could incorporate some of its features into yours. – brhans Mar 22 '21 at 13:52
  • You get brownie points for being a dare-devil, but the results are [these](https://i.stack.imgur.com/qzfgH.png). `V(eff)` is the efficiency (`V(p)` being the dissipated power). Mandatory grain of salt. – a concerned citizen Mar 22 '21 at 17:56
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    THIS CIRCUIT WILL FUNCTION AS A BUCK CONVERTER AS IT IS SHOWN. Adding formal hysteresis will improve efficiency BUT the energy stored in the inductor and Cout provides an energy transfer after turn off which allows operation. I have built many such hysteretic converters and used them in successful commercial products. Im out at present. I may provide more detail when back in my office. – Russell McMahon Mar 23 '21 at 04:27
  • Please, I'd really like to hear more – BELSmith Mar 23 '21 at 06:19

2 Answers2

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As above, what you have is a linear regulator.

BUT

If you add hysteresis to the circuit, you now have what is called a hysteretic converter, where the output bounces back and forth rapidly between slightly above and slightly below the desired output voltage. In this case, Q1 is now switching between off and fully enhanced (on) for much lower power dissipation. So, less heat, but more noise on the output. This can be done without an inductor, but it increases the stress on the switching transistor.

In its most simple form, a hysteretic converter is a voltage comparator driving the switch transistor.

Way back, the first switching power supply I had to (constantly) maintain was a 20 A hysteretic converter built onto the backplane of a Data General NOVA 1220. An undersized heatsink and poor airflow caused the transistor to burn up about every 6 months.

AND ...

Gold Star for using reference designators.

AnalogKid
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You could sense inductor current with an upper trip point and a lower trip point .This would self run .You could regulate the output volts using a simple error amp that changes the current trip point .This uses more parts and does work .

Autistic
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