This is my crcuit layout for a boost converter. Please note that I did not draw the load and its connections (for it is something taken for granted!) :A boost converter circuit

In stead of using a controller that maintains a constant output voltage irrespective of the load by changing the clocking frequency and the duty cycle, I used an Essaki oscillator for I thought it is much simpler and feasible.


  1. Can the Essaki oscillator be used to make the circuit clock?

  2. Is it correct to feed both the Essaki oscillator and the boost circuit with the same power source with the E O being connected in parallel? Does this risk of damaging the Oscillator circuit? Or should the oscillator circuit be powered by an independent power source?

Is the placement of the MOSFET Gate-Source connection correct? Will it turn on once the BC546A is made to conduct?


How about this slight modification?

[Boost converter circuit[2]


closed as unclear what you're asking by Andy aka, winny, Sparky256, laptop2d, RoyC Sep 2 '18 at 15:48

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    \$\begingroup\$ The FET's gate is connected to ground, therefore the FET will never turn on. You should re-think this circuit. Also, you should use a NE555 timer (or similar) for the oscillator circuit. \$\endgroup\$ – Jonathan S. Aug 25 '18 at 22:23
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    \$\begingroup\$ did you forget to connect the wiper on the 30k resistor? \$\endgroup\$ – jsotola Aug 25 '18 at 22:35
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    \$\begingroup\$ Did you intentionally leave the base of Q2 unconnected? \$\endgroup\$ – mkeith Aug 25 '18 at 22:53
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    \$\begingroup\$ If you want to find out if it will work, download LTSpice or one of the other free spice simulators and try it virtually. \$\endgroup\$ – Dean Franks Aug 25 '18 at 23:03
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    \$\begingroup\$ @YoussefSifi No, you are confused with how a wiper (potentiometer) and a variable resistor (rheostat) looks like. Look here to refresh your memory. A potentiometer can be converted to a rheostat by doing what jsotola implied, by connecting the middle terminal to one of its sides, which you should do if you are using a potentiometer. \$\endgroup\$ – Harry Svensson Aug 25 '18 at 23:07

You would've made googling easier if you spelled the name of Leo Esaki correctly.

What you're trying to do here is to use a common BJT as an avalanche diode to build a relaxation oscillator. This will not work well, for two reasons.

First, BJTs are not built with this purpose in mind, so they won't work reliably. The behaviour you'll get will change from one BJT to the other, with voltage and temperature variations, and also with time as the junction will get damaged by repeated breakdowns.

Second, relaxation oscillators are quite bad at generating a stable frequency. They are used when either a stable frequency is not needed (noise generators, random number generators) or at high frequencies where feedback oscillators cannot be built practically (GHz and above). In your case, the target frequency is too low to justify this design choice.


The transistor has emitter-base reverse voltage avalanche breakdown and suddenly conducts, then turns off, then has breakdown again, then turns off again oscillating like this. The 30k variable resistor should be drawn as a variable resistor and changed to be at the collector to ground.

  • \$\begingroup\$ Your answer makes no sense based on the schematic provided. The MOSFET gate is grounded which kills any chance of oscillation. As it is Vout = Vin minus the Vdrop of D1. \$\endgroup\$ – Sparky256 Aug 26 '18 at 4:51
  • \$\begingroup\$ @Sparky256 Do you think that oscillation happens if the mosfet gate is connected only to Q2's collector? \$\endgroup\$ – Youssef Sifi Aug 27 '18 at 7:09
  • \$\begingroup\$ Audioguru, long comment list with no answers are a sign of confusion over an irrational design as we see it. To avoid down votes you should skip posting a 'what-if' answer and move on, as we did. \$\endgroup\$ – Sparky256 Aug 27 '18 at 22:18

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