Boost converter not working

  • #1
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Main Question or Discussion Point

i built a boost converter, but it is supposed to boost 4 volts to 8 volts. the multimeter i am using says 0.36 volts.
here is the circuit i am using.
My Snapshot_7.jpg


there is a resistor between the arduino uno and the transistor, i forgot to include that. each resistor i rated 1000 ohms. the capacitor is rated 120
microfarad. the inductor has an unknown inductance. i removed it from a laptop recharger. the arduino uno switches the transistor on for 415
microseconds then switches it off for 415 microseconds. then switches it back on , then switches it back of and on and on. the transistor is a tip47g. the diode is sb3100. i removed it from an old multimeter.
 

Answers and Replies

  • #2
33,829
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The Arduino should have a second connection to the circuit.
415 microseconds might be too short. You could measure the current flowing through the transistor.
 
  • #3
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415 microseconds then switches it off for 415 microseconds, that means the switching frequency = 1204 Hz.

As far as I know, very few boost converters operate at such low frequencies, and unless the inductance is large, the inductor may saturate.

Today, many boost converters using in consumer products have switching frequencies ranging from a few hundred kilohertz to a few megahertz.
 
  • #4
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the arduino uno does have a connection to the negative terminal. i forgot to draw that in the schematic. so the
switching frequency should be higher?
 
  • #5
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If I don't have the equipment to measure the inductor's inductance and saturation current, and lazy to do any calculations..etc, then one simple way I can try is to increase the switching frequency to see if the output voltage is increasing
 
  • #6
NascentOxygen
Staff Emeritus
Science Advisor
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If you remove drive to the transistor (or if you remove the transistor from the circuit), the voltmeter should read around 3 to 4 volts. If it doesn't, something is amiss.
 
  • #7
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Check if the transistor still working. At low frequencies it might be cooked easily.
Try to measure the current at the battery.
Also, how big is the load?

Try it again with the transistor mounted on a heatsink, and at 10-15kHz frequency. You can use a small speaker to check if it is working.
Even higher frequency would be better, but I doubt that the Arduino would be able to drive the transistor at that speed.
 
  • #8
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i tried building a boost converter it is not working. i keep getting a reading of zero.
19a91825-b490-4ae4-ac65-e09947e4ccc5.jpeg

the capacitor has a capacitance of 120 microfarad. the base of the transistor is connected to a 450 ohm resistor. the inductor has an inductance of 10 microhenries. the arduino uno switches the transistor on and off. the voltmeter is connected to 1000 ohm resistor. the battery provides 3.98 volts roughly. the duty cycle is %50. the voltage reading is zero. why is it not working? how do i get it to work?
 
  • #9
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i think the current is .27 amps. i am trying to get it to boost the voltage to any higher voltage.
 
  • #10
berkeman
Mentor
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Your circuit schematic diagram is incorrect. No wonder it's not working. Can you post a link to the boost circuit diagram that you are using as your starting point?
 
  • #12
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could you show me a schematic of a correct boost converter circuit?
 
  • #13
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by the way i did not draw the schematic accurately.
19a91825-b490-4ae4-ac65-e09947e4ccc5 (1).jpeg


i think it is accurate this time. sorry i was under a lot of stress and i was tired. so i guess i got careless. i double checked this time.
 
  • #14
phyzguy
Science Advisor
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What does the voltmeter read if you don't switch the transistor and connect the base of the transistor to ground?
 
  • #15
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i will check. why.
 
  • #16
phyzguy
Science Advisor
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i will check. why.
Well, what would you expect to get with the transistor turned off?
 
  • #17
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if i connect the base to the ground i get a voltage reading of zero at the resistor.
 
  • #18
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by the way i did not draw the schematic accurately.
I think it's still not right. You want to connect the diode to the wire between the inductor and the collector of the transistor, so the current from the inductor will be forced to go through the diode when the transistor switches off.
This will probably damage the transistor as well. If you provide no path to ground for the current through the inductor you will get a large voltage on the collector. Make sure the transistor still blocks current if the base is grounded.
 
  • #19
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can you show me a schematic of a correct circuit?
 
  • #20
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Can I suggest Draw.io....
 
  • #21
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i don't know what Draw.io is.
 
  • #22
phyzguy
Science Advisor
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if i connect the base to the ground i get a voltage reading of zero at the resistor.
Do you see why this is not correct? If the transistor is off, the battery voltage, minus some drop for the diode, should appear across the output resistor. Perhaps you have fried the transistor? What if you take the transistor out of the circuit completely? Does the battery voltage then make it to the output resistor? You need to try to think of these kinds of debug steps on your own. It's much faster than coming here and just telling us, "it doesn't work."
 
  • #23
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can you show me a schematic of a correct circuit?
You only needed to tie the diode between the collector and the inductor.
https://www.electronics-tutorials.ws/power/switch-mode-power-supply.html
look at "The boost Switching Regulator"

Do not switch off the current through the inductor without the diode and output capacitor present. Your transistor probably has a short between collector and emittor now, wich explains the 0 Volt on the output you are seeing.
 
  • #24
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i was thinking when the transistor is on, current flows through the transistor, then through the inductor. the inductor generates a magnetic field. the diode is reverse biased. when the transistor is switched off, current does not flow through the transistor, and the inductor's magnetic field collapses. the inductor produces a voltage. this voltage is of opposite polarity to the battery voltage. the battery voltage is in series with the inductor voltage. this produces a higher voltage. the higher voltage forward biases the diode.
 
  • #25
berkeman
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when the transistor is switched off, current does not flow through the transistor, and the inductor's magnetic field collapses. the inductor produces a voltage. this voltage is of opposite polarity to the battery voltage.
No, it is in series with the battery voltage, so it adds to it. This is a boost converter topology, not an inverter topology converter.

The problem highlighted by @willem2 is that if you ran the circuit without the output diode connected correctly, there is nothing to clamp the collector voltage of the switching transistor when it is turned off and the inductor voltage "kicks" up. This can exceed BVCEO for the transistor, causing it to fail (either open or short).

It's important to use good learning resources when studying and building your first DC-DC converters. Do you understand the voltage and current waveform plots at the Wikipedia page?

1574468700307.png


Have you been studying any other DC-DC converter tutorial pages?
 
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