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I have a little problem. For a project I have to step up a 3V button battery to 12V to supply a DC motor (specifications of the problem). The hint that the professor gave to us is to make a double dc boost converter ! But the thing that is the motor is modelized by a 110 ohms resistor and my double boost converter can't reach the 12V at the output. He also told us to add at least one more transistor for the switching part of each boost converter. I also have some difficulty to design the components of my circuit. Also one thing that i noticed is that the output voltage does depend on the switching frequency (my oscillator is a multivibrator astable and can work around 20kHz to 75kHz). double boost converter

The specifications of the project :

  • CR2032 battery or 3V supply;
  • MOSFET transistors, reference ZVNL120A ;
  • Geared motor ref. 20G150.
  • coin battery type CR2032.

@Andyaka Okay i tried with a new method with 1 boost but 5 transistors in //. How can you explain this then ? The supply can be this button pile or another 3V supply like it said in the specs. enter image description here

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The hint that the professor gave to us is to make a double dc boost converter ! But the thing that is the motor is modelized by a 110 ohms resistor and my double boost converter can't reach the 12V at the output.

A CR2032 lithium battery would be lucky to supply 10 mA. If you look at a relevant data sheet current pulse tests are only performed at 6.8 mA: -

enter image description here

A cell like this will never be able to supply 12 volts to a 110 Ω load.

Regarding what \$\color{red}{\boxed{\text{you said}}}\$ your prof said about using two cascaded boost converters, I can only say that he doesn't know what he is talking about. It's far, far easier and more efficient to use a single boost converter when going from circa 3 volts to 12 volts. But, the bigger problem is that a CR2032 cannot provide the current for a 12 volt 110 Ω load.

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    \$\begingroup\$ I'm hoping the OP misunderstood the professor's comments somehow. Otherwise, he/she should start looking for a new professor! \$\endgroup\$
    – John D
    Jan 20, 2023 at 22:02
  • \$\begingroup\$ look at my other comment \$\endgroup\$
    – ktnl527
    Jan 20, 2023 at 23:31

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