# Current source for boost converter

I'm designing a boost converter to boost 3.3 V to ~25-30 V. I have successfully simulated it in SPICE, but the current draw for converter is enormous - about 8-9 A. How can I handle this nuisance? This is the boost converter circuit I used.

I used $R_L = 10 \Omega$ so it doesn't go into discontinuous mode.

I don't need actually any output current, because all I need is to hold a 25-30V potential over a Numitron tube.

I managed to google some versions of this converter featuring MOSFET driver, but failed to understand how they work. Could you please light it up a bit for me?

• This question is meaningless without specifying the output current. – Brian Drummond Jul 8 '14 at 10:44
• What value for R did you use in your simulation? – Andy aka Jul 8 '14 at 11:37
• I used R=10 Ohms, because other way it was going into discontinous mode. – Mike Spark Jul 8 '14 at 11:44
• Is the basic idea of that circuit to switch faster than the capacitor discharges through the load resistor and consequentially force more charge on the capacitor plates and thereby raise the voltage? – sherrellbc Jul 8 '14 at 12:51

Neglecting the losses of any kind thus assuming a 100% efficiency, one can thing about a boost convert as a voltage translator with constant power. I mean that Pin = Pout.

$$P_{in} = V_{in}I_{in}$$ $$P_{out} = V_{out}I_{out}$$

thus

$$V_{in}I_{in} = V_{out}I_{out}$$

of rewritten differently:

$$\frac{V_{in}}{V_{out}} = \frac{I_{out}}{I_{in}}$$

thus if you boost 3.3V to 30V and see a power consumption of 9A, then

$$\frac{3.3V}{30V} = \frac{I_{out}}{9A}$$

$$I_{out} = 1A$$

If you designed your boost convert to supply roughly 1A then it is working fine. Because we assumed a 100% efficiency, the real available output current would be less than 1A.

Usually you do the computation backward, you know your output current requirement and you can compute what would be the input current.

Late answer to a year-old question...

If you need to boost a voltage from 3.3 V to 25-30 V but need very little current, consider a voltage multiplier (such as a Villard cascade or a Dickson charge pump) instead of a boost mode switching supply.

• This doesn't answer OPs question though. It would be better as a comment. – Funkyguy Sep 30 '15 at 18:35