Could some electronics guru please clarify the following ? The following is the basic structure of a boost converter:
DC_in -> inductor --> diode --> capacitor --> output | | | | switch | | | GND GND The diode is forward biased in the sense that when the switch is closed, the diode is reverse-biased and prevents the capacitor from discharging. The question is: How is the maximum output voltage set, i.e., how is maximum output voltage set ? Is it done by the reverse breakdown voltage of the diode, since once that is exceeded, the capacitor will try to discharge, when the switch is on.
I'm not a guru and not even a designer. So the following is what has helped me, only. I've read some of your other posts and from them I am assured you have LTspice. Have you set this up in LTspice, yet? You should. It supports a voltage-controlled switch, which you can use to try out different switch periods and duty cycles. Make sure you test a variety of cases and see if you can make predictions.
You will need to keep in mind some limitations on inductors that aren't air-core (Bmax at low-freq or eddy current losses at high-freq) when considering core area, volt-seconds, turns, and possible gapping or distributed gap materials (ferrite), if you actually try and build something and expect it to work according to some idealized LTspice schematic.
There are some excellent PDF files available on the web, if you google for them. Look for 'boost,' and 'continuous,' and 'discontinuous,' and 'switcher,' as some example keywords. And in particular, look at the Wiki that comes up. TI writes a few good things (Unitrode series comes to mind.)
If you dont add any further bits than you've drawn, there is no voltage control. Voltage control is achieved by not switching the switch on when sufficient V exists in the capacitor.
Note to get adequate V control you need a capacitor that stores far more energy than the coil does. If, for experimental reasons, you were to use a small cap and a big coil, one single switching cycle would put a large voltage increase into the cap, making regulation unachievable.
The output voltage is set by the duty cycle, D, of switch which is defined as Ton/(Ton+Toff) where Ton is the switch ON time and Toff is the switch OFF time. This follows from the fact that in steady state the average voltage across an ideal inductance is zero. During the ON time the voltage across the inductor is DC_in, and for the OFF time the voltage is DC_in-Vout (making the simplification that diode forward drop is 0), so that the average is 0=3DDC_in x Ton + (DC_in- Vout) x Toff. Solving for Vout yields Vout=3D DC_in x (Ton+Toff)/Toff =3D DC_in/(1-D).
Just google it and you'll find a formula which says that the input/ouput voltage ratio is determined by the duty cycle. This goes as long as the minimum load current is satisfied.
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Failure does not prove something is impossible, failure simply
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nico@nctdevpuntnl (punt=.)
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