# boost technique ac-dc converter

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#### mengghee

##### Full Member level 3 Boost Converter Output ?

hiye everybody,

my classmate in uni has build a boost converter with a input voltage of 7.5v, duty cycle of 0.5 from a signal generator, frequency of 40KHz. the output voltage obtain is around 11.3v. the pulse connected to the gate of the Mosfet is not connected through a resistor. the output voltage seems to be very low. i assume the effeciency is very low. is that normal ?

regards,
mengghee

#### VVV Re: Boost Converter Output ?

Certainly the voltage seems too low, provided the duty-cycle is really 50%.
Should check that. He should also check that the input voltage AT THE BOOST INPUT is really 7.5V.
We cannot talk about the efficiency unless we measure the input and output currents.

The MOSFET will be driven by the internal resistance of the generator, which can be 50 or 600 ohm usually.
A normal MOSFET gate resistor is in the tens of ohms range, so if he is driving it with 50 ohms, that is OK.

### mengghee

Points: 2

#### v_c Re: Boost Converter Output ?

If everything was ideal (no losses in the components zero voltage drops across devices) the you would expect the gain of the boost converter to be 1/(1-D) which is 2 in your case. Then the output would be 15V.

First take the suggestion of VVV -- check to make sure that the input of the converter is indeed 7.5V. Check the voltage from the input side of the boost inductor to ground. Then look at the drain-source voltage of the MOSFET when it is on and the anode-cathode voltage of the diode when it is on. You will see that they are not zero and they result in the reduction of the average output voltage. Note that the device voltage drops will have to be multiplied by either D or (1-D) to give their effect on the average output voltage.

Look at pages 25-31 of https://ece-www.colorado.edu/~pwrelect/book/slides/Ch3slides.pdf
It shows the boost converter loss, efficiency calculation.

Best regards,
v_c

### mengghee

Points: 2

#### electronrancher Boost Converter Output ?

If you can set the duty cycle to 0.5 with a signal generator, then this is only a driver stage?

It sounds like your gate driver cannot enhance the mosfet very well - you should check the voltage at the drain of the mosfet during the on time when switching - it should be very low, like 0.2v or so.

Also - do you have a current probe for your scope? This would be an excellent time to plot the inductor current.

I think that either your mosfet or your schottky diode is chosen incorrectly - you are right to suspect that 11.5v out is too low. What kind of load is attached at the output?

### mengghee

Points: 2

#### mengghee

##### Full Member level 3 Re: Boost Converter Output ?

electrorancher,

the vgs source is directly from the signal generator, i am not sure about the lab, i need to ask if they have a current probe. as this is my friend circuit ( i have yet to build mine ) he did not have a schottky diode in. and the load is 20watt 66 ohm resistor. MOSFET used is irf511 and i am sure the vgs is not connected to optimize its performance. thank you

regards,
Jeffrey

#### VVV Re: Boost Converter Output ?

The duty-cycle formula is more complicated than the simple 1-Vin/Vout used in many calculations.
The key to deducing a more exact formula is understanding that the volts*seconds applied to the inductor during the transistor on time must equal the volts*seconds during the off time.
Assuming continuous current mode operation, we can write:

(Vin-Vind-Vmos)*ton=(Vo+Vind+Vd-Vin)*toff,

where Vin is the input voltage, Vind is the voltage drop across the DC resistance of the inductor, Vo the output voltage, Vmos is the drop across the MOSFET during on time.
Dividing by T, and knowing that ton*T=DC and ton+toff=T, ir follows that:
toff/T=1-DC

Then we can re-write:

DC*(Vo+Vd-Vmos)=(1-DC)*(Vo+Vdrop+Vd-Vin)

And then DC=(Vo+Vd-Vin+Vind)/(Vo+Vd-Vmos)

This reduces to the familiar formula if:
Vd«Vo, Vind«Vo and Vmos«Vo

But the real world is not perfect, so if the duty-cycle is forced to 50% (without feedback to adjust it), taking only the diode drop into consideration, we find that the output voltage would be:
Vo=2*Vin-Vd=14.4V

Taking other factors into consideration helps explain why your friend only obtains 11.5V output (again, this is way lower than I would expect for properly chosen components and correct operation).
The first things to check are the input voltage, gate drive, and inductor DC resistance.

Points: 2