Boost Converter Output Voltage When Duty Cycle is 0.9

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mengghee

Joined: 29 Nov 2005
Posts: 165
Helped: 3
Location: United Kingdom

11 Feb 2006 18:03 Boost Converter Output Voltage When Duty Cycle is 0.9

I have attached a schematic diagram of my circuit and also a graph plotted on Duty Cycle Vs Vout. In the graph, it shows that when my duty cycle increases to 0.9, the output voltage is lower than the output voltage when the duty ratio is 0.8. how can that be possible ? can anybody tell me why that happens ? and if that happens, how should i prevent it ? thank you.

thanks,
mengghee

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eecs4ever

Joined: 31 Jan 2006
Posts: 181
Helped: 26
Location: Analog Environment

12 Feb 2006 9:38 Re: Boost Converter Output Voltage When Duty Cycle is 0.9

Here is how this boost converter works:

Phase 1:
When V1 is High, M1 turns on, pulling the voltage at node 2 low.
This reverse biases the diode and isolates the two halves of the circuit.
During this phase, the current going through the inductor increases, and additional
energy is stored in the inductor in the form of magnetic flux.

The output voltage / current is supplied by the Capacitor during this phase.

Phase 2:

V1 goes low. M2 turns off. node 2 increases in voltage, and the diode D2 becomes foward biased. The energy from the 5 volt Vdc and the energy stored in the inductor is injected to the right half of the circuit, causing the voltage to rise beyond 5 V.

Now to answer your question, if the Duty Cycle is too close to 1. Then there will be very little time for the Inductor to inject its energy to the right half of the circuit. and Vout will drop.

Thats my best guess.

Davood Amerion

Joined: 01 Mar 2005
Posts: 589
Helped: 90
Location: Persia

12 Feb 2006 21:15 Re: Boost Converter Output Voltage When Duty Cycle is 0.9

Dear Mengghee
In boost convertor as opposed to step down convertor output voltage is not proporational to duty cycle.
This config usualy used maximum 50% duty cycle.
why?
1- because output voltage is not depend on duty cycle.
it is better to say that output power is proporational to duty cycle.

2- in phase 2 (transistor off) inductor must enough time to disharge its energy to output circuit via D2.
this time is not fix , and depend on output voltage. for greater voltage this time will reduced.

for example, you can check with (say) 200 ohm, for R8. you will see Vo/DC curve improved significantly.

Regards,
Davood Amerion.