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| converter schematic |
This buck converter is driven by 555 timer IC and converts 12V dc from a battery to 5V dc. The 555 IC (operated in astable mode ) generates pulses of voltages of 42% duty cycle. This pulses turns the transistor on and off and drives the inductor L (buck converter circuit), and as a result generates 5V dc output. For this configuration,
Ra =20K
Rb =27K
C =1nF
Q = TIP41C or any of the same specs
Cout = 2200uF (higher the better)
I think the diode D is backwards.
ReplyDeletethank you sir/maam Anonymous
ReplyDeletefor your comment and letting simple-electronics know about the diode orientation problem. The converter schematic above is already modified. If you find additional problem kindly post a comment.
thnx.
wat is the value of inuctor L
ReplyDeleteThe inductor is probably between 600 uH and 1mH.
DeleteThese are expensive.
Why?
Well the TIP41C can carry 6 amps so the coil wire will have to be sizable. This means thicker wire which means a physically larger coil.
To use a smaller, cheaper inductor change the components to these values:
Ra = 2K
Rb= 2K7
C= 1nF
Q= MOSFET FQP33N10 or similar
L= 60uH
Can I use any values between 600uH and 1mH if I implement the above circuit???? Btw, your Rb is only 2K or 27K? Thank you
ReplyDeleteHello, I drew that circuit with the multisim stimulation software but the output voltage is varied but not steady in 5V. So,I can i solved that problem. I already changed some components values but the problem is still there.
ReplyDeleteits output gives only 10.78V instead of giving 5V
ReplyDeleteis there any changes to get 5V..??
Without a load there is nothing to pull the output voltage down. there is nothing sensing the output voltage. This is not a true buck converter. The output voltage will depend on input voltage and load. This circuit is closer to a pwm driver than a buck converter. It may have good current regulation. This would probably make a good led driver but i wouldn't use it to charge usb devices.
ReplyDelete