I have to step up a 4V supply (the generator is a button battery) to a 12V voltage to supply my motor.

How do I size the components of my step-up converter? To compute for example the inductance and the capacitance of my converter, I have to set the frequency.

For the oscillator, I will use an astable multivibrator. I have to use the ZVNL120A MOSFET for the transistors. Should I use two transistors for the switch of my step-up converter?

This is a quick simulation of a step-up converter with the components set quite randomly. For this simulation I set the frequency to 100kHz.

Little simulation on ltspice pour the step-up converter

  • 1
    \$\begingroup\$ How much current does you motor need? A "button battery" sounds rather small to be powering a 12V motor. Keep in mind that when you go from 4V to 12V, you'll need three times the current from the battery. If the motor needs 100mA, then your booster will draw 300mA from the battery. \$\endgroup\$
    – JRE
    Jan 14 at 19:27
  • \$\begingroup\$ What are the characteristics of you motor and battery? \$\endgroup\$ Jan 14 at 19:52
  • \$\begingroup\$ @BruceAbbott The motor is a 20G150, the rated current for the motor is 110mA and for the resistance value of the motor i found on the datasheet "winding resistance" = 1.7 Ohms, is that the right value to represent my motor ?? \$\endgroup\$
    – ktnl527
    Jan 15 at 10:25
  • \$\begingroup\$ Is this your motor? conrad.com/p/… If not then please provide a link to the datasheet of your motor. Can you measure the winding resistance? \$\endgroup\$ Jan 15 at 10:55
  • \$\begingroup\$ @BruceAbbott the motor is : fr.rs-online.com/web/p/moteurs-a-courant-continu/3989669 No the project is just by simulation :( \$\endgroup\$
    – ktnl527
    Jan 15 at 11:02

2 Answers 2


should I use 2 transistors for the switch of my step-up converter?

That isn't necessary (unless you want to use a MOSFET to replace diode D1 in order to get a few percent more efficiency).

how I can size the components of my step-up converter?

Well, you should decide what the maximum current out should be and, this should consider (usually) the stall current of your motors and the full load current. I see you have picked 500 Ω to represent the load and this represents a load power of 288 mW (i.e. 12 volts squared divided by 500 = 0.288) so, I'm guessing that you motors are quite small in power rating.

Your circuit will also be operating in what we call discontinuous conduction mode (DCM) as opposed to CCM (continuous conduction mode) with the values you have chosen. This basically mean that at the duty cycle used (0.66), the inductor current will naturally fall to zero amps towards the end of each switching cycle. That isn't a problem in itself because, the load is a motor but, for other (more precious) loads you may choose to operate in CCM. This would happen in your case if the inductor value rose to circa 190 μH.

I've developed this little web-based calculator that can show you the currents. It's not perfect (I should try to make it able to deal with smaller power levels) but it should help: -

enter image description here

Unfortunately, it barely shows the current waveforms but, it will give you something to play with (if you want to experiment with different values). It calculates the required duty cycle based on the input/output voltages, inductor and load resistance.

The web-page also shows you the formulas for deriving the info yourself.

To compute for example the inductance and the capacitance of my converter, I have to set the frequency.

The capacitance plays only the role of ripple voltage reduction in a boost converter like this and, isn't relevant in deciding the operating frequency. However, inductance is important.

Realistically, you need to decide whether you want to run in CCM (better/lower ripple voltage) or DCM. Yours will be fine in DCM and, you could choose a lower value of inductor. On the other hand, if you are running in DCM it may be important to keep the peak current drawn from your unspecified battery to the lowest possible so, choosing a higher value inductor will reduce this. The fact that the battery is a button type does raise concern on its ability to deliver anything like 0.1 A.

In my picture above, the peak inductor current (that's the peak current drawn from the input supply) is 0.196 amps but, if you went for CCM and chose a 220 μH inductor, the peak current would be 0.133 amps. Again, using a button cell is a concern.

  • \$\begingroup\$ I found on the datasheet of my motor "winding resistance = 1.7 Ohms", is that the right value which represent my motor ?? (my motor is a 20G150). I will go read your fabulous doc on the boost converter. Thanksss \$\endgroup\$
    – ktnl527
    Jan 15 at 10:52
  • \$\begingroup\$ @ktnl527 that resistance sounds highly problematic if that's the case when using a button battery/cell. What should I do when someone answers my question? \$\endgroup\$
    – Andy aka
    Jan 15 at 10:55
  • \$\begingroup\$ I just tried to change the resistance to 1.7ohms and yes the Vout can't go higher tant 2.5V :(. Yes i guess it's problematic but it was the specifications of the project :(. So it's really impossible ? if yes idk why they gave us these specs \$\endgroup\$
    – ktnl527
    Jan 15 at 11:00
  • \$\begingroup\$ Please link the data sheet of the button battery/cell you are supposed to use. \$\endgroup\$
    – Andy aka
    Jan 15 at 11:01
  • \$\begingroup\$ The 1.7 ohms will dictate the stall current for your motor by the way. The normal running current for the motor might be between 5 and 10 times lower. Even so, the equivalent resistance would become 17 ohms and, this still sounds a problem for a button cell. \$\endgroup\$
    – Andy aka
    Jan 15 at 11:18

First you must consider the characteristics of your motor. At stall it draws ~0.57 A at 12 V. This is the current it will initially try to draw on startup. Once it gets up to speed it should draw ~0.06 A with no load, or 0.11 A with rated load.

You are boosting the voltage by 3 times so the battery current will be 3 times higher than the motor current, ie. at least 0.18 A at 12 V - assuming the battery delivers 4 V and the booster is 100% efficient. In practice the booster won't be 100% efficient and the battery voltage will drop, requiring even higher battery current.

The name 'button cell' generally refers to a small single cell battery used in low power devices such as wristwatches and hearing aids. These have a maximum current rating in the region of 10's of mA, much too low for your application. Your battery will need to deliver considerably more than 0.33 A at rated motor load, and possibly higher than 1.7 A on startup.

The booster also needs to handle this current. Instead of a 1N914 you should use a Schottky diode rated for at least 0.5 A, and the inductor should be similarly rated. The FET should have RDS(on) no higher than 0.1 Ω when driven with the available Gate voltage (4 V?). These ratings assume that you can start the motor slowly (eg. using a PWM ramp to slowly raise the voltage) to limit startup current.

To regulate the output voltage you will have to either vary the PWM ratio at constant frequency, or keep the pulse width fixed and vary the frequency. If you use a fixed PWM ratio the voltage will vary depending on motor current (higher voltage at lower current).

Boost frequency depends on the inductance - the smaller it is the higher the frequency (or shorter the pulse width) must be to avoid saturating the inductor.


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