I am currently working on a flyback design charging a capacitor in the nF range to 200V from an input voltage of around 1.5V. As the final circuit should be as small as possible, my component choices are very limited.

Furthermore, the flyback switch should be driven by a max. voltage of 2V DC. My current problem is now, that there are currently no MOSFET devices available with a Vdss high enough.

My question now: Are BJTs feasible as a switch (higher breakdown voltage) and if so do they consume any power when turned off? (I am guessing not). Is it possible to drive them from only 2V?

  • \$\begingroup\$ I already did(?) \$\endgroup\$
    – LaMontagne
    Commented Sep 23, 2014 at 7:21

3 Answers 3


I have recently finished a similar design. We are now at the pre-production phase, having already validated the first prototypes, which operate perfectly under similar constraints as the ones you describe in your question. Because of IPR limitations, I can only give you some general guidelines.

Based on the following assumptions:

  • You are trying to avoid the use of a transformer by all means (which may not be the case),
  • The power requirements of the load is in the miliwatt range,
  • You don't want to step-up the voltage supply first, but will use the 2VDC directly for powering your flyback design block.
  • You need to keep the cost and PCB space to the absolute minimum.

I suggest the following approach:

  1. Rule out the MOSFET and use a fast switching high voltage BJT (rated at least to 200V, better yet to 300V).
  2. Select the most appropriate inductor for your constraints and depending on your requirements for maximum power delivered to the load.
  3. Calculate the maximum flyback voltage, and make sure you can get at more than 200V voltage flyback "peaks". The involved parameters here are: ILpk (inductor peak current), Cts (total sum of parasitic capacitances at the flyback node) and SWITCHING SPEED of the BJT.
  4. The last one is hugely important, and rarely mentioned in the technical literature. You may have a big ILpk and very little Cts, but if your switching (off) speed is not fast enough, the flyback voltage will suffer. Reminder: Vflyback = -L * dI/dt.

The above can be implemented with:

  1. A simple (single) inductor or
  2. A 1:1 coupled inductor.

Apparently, from the design equations point of view, there seems to be no need for the 1:1 coupled inductor, as the single inductor will work exactly as well, most probably with less resonance apparatus. However, using a 1:1 coupled inductor will help avoiding EMI generation. I suggest the use the 1:1 coupled inductor in case you have to EMI certify your final product.

** However the previous assumptions, should you want to explore the flyback transformer approach, there now exist very nice miniature (micro-power) SMD flyback transformers. **

For instance, this Coilcraft model allows up to 1:100 turns ratio with 300Vrms isolation among primary and secondary coils:


  • \$\begingroup\$ Thanks a lot for your response! Finally it looks like there might be a solution for my problem. Just for my understanding, if you are speaking of an inductor you mean a coupled inductor with a 1:1 ratio? \$\endgroup\$
    – LaMontagne
    Commented Oct 23, 2013 at 20:57
  • \$\begingroup\$ It was a really nice coincidence that I came up finding your question after having finished the testing of a pretty similar design! You can either use a simple (single) inductor or a 1:1 coupled inductor. Apparently, there is no need for the 1:1 coupled inductor, the single inductor will work exactly as well, most probably with less resonance issues, however, the 1:1 coupled inductor will help you avoid EMI issues. I will jump directly to it if you are unsure of your EMI requirements or have to certify the final product. \$\endgroup\$ Commented Oct 24, 2013 at 9:11
  • \$\begingroup\$ You can also use a miniature flyback transformer from Coilcraft. I am editing now the answer to include some more information... \$\endgroup\$ Commented Oct 24, 2013 at 9:27
  • \$\begingroup\$ Really good answer +1 \$\endgroup\$
    – Andy aka
    Commented Oct 24, 2013 at 17:44

Using a transformer is my preferred option. The primary would be centre tapped and connected to the 2v rail. Transistors would pull-down on each leg of the primary in turn at a reasonable frequency of about 100kHz. This turns your primary voltage into 8v p-p\$^1\$.

If your turns ratio were 200:8 that doesn't sound too bad to me. Output would be a little short of 200volts but a little bit of resonance tuning on the secondary would soon sort that out.

You could even make a case for a smaller turns ratio with a Cockcroft Walton diode multiplier on the back end to double or triple the voltage.

Plenty of transistors would do this and several mosfets providing gate voltage threshold was below 1v.

You could of course look at an energy harvesting chip to give you 5v and then use a fly-back switcher from people like linear technology. Here's quite a useful device from linear tech: -

enter image description here

Application 1 works from a supply as low as 1.2V and produces 5V at over 200mA. This could then feed the 2nd circuit which can produce up to 350V. It should be noted perhaps that the internal switching transistor is bipolar in this chip. Also note the use of a 2-stage Cockcroft Walton diode-capacitor multiplier on the output of the 2nd circuit.

\$^1\$ Before anyone starts down-voting for saying 8Vp-p you need to analyse what happens on the primary. After one side of the transformer has been grounded it will rise up to twice the supply voltage (i.e. 4V) due to the action of the other side of the transformer being grounded by the other transistor. Hence one side has 4Vp-p and the other side has antiphase 4Vp-p. Net effect 8Vp-p.

  • \$\begingroup\$ You are absolutely right about the 8Vpp. :-) The topology you mention requires a transformer with Center Tap (CT) in the primary, which makes it a little bit more bulky/expensive, however it is my favorite for low noise switched mode converters, much lower than any other flyback topology AFAIK. \$\endgroup\$ Commented Oct 24, 2013 at 16:14

You don't need high voltage MOSFETs. For such big ratios, the transformer is a natural choice. This way, on the primary side, the voltages will be low enough to be handled with low power MOSFET, that will provide higher efficiency and easier design.

Using BJTs is possible of course and they does not consume energy when turned off. Driving BJT from 2V is easy. As a rule BJT will consume a little bit more energy than MOSFET, because it consume some constant base power, when turned ON.

But the problem with BJTs is the same - I highly doubt, that raising the voltage from 1.5V to 200V without transformer is possible. In theory yes, but in practice...

  • \$\begingroup\$ The O.P. is concerned mostly with gate voltage, if I understood it correctly. \$\endgroup\$ Commented Oct 22, 2013 at 21:05
  • \$\begingroup\$ Hm, Vdss is maximal drain-source voltage IMHO... \$\endgroup\$
    – johnfound
    Commented Oct 22, 2013 at 21:08
  • \$\begingroup\$ My problem is actually that I am limited in the transformer turns ratio as there are only certain components available which respect the space constraints. For this reason I was thinking about a BJT as the power is very low and efficiency less of a problem like this. \$\endgroup\$
    – LaMontagne
    Commented Oct 22, 2013 at 21:13
  • \$\begingroup\$ @user30784 - the maximal voltage when there is no transformer is limited by the parasitic capacitances. I never saw such big ratios to work in practice. Did you made some modeling? \$\endgroup\$
    – johnfound
    Commented Oct 22, 2013 at 21:19

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