Author Topic: new circuit  (Read 1439 times)

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new circuit
« on: October 10, 2008, 17:11:54 pm »
sinewave booster circuit

When power is applied and the switch starts to turn on and off at a set frequency, the inductor stores energy in the form of a magnetic field during the ON time. The current in the inductor increases linearly with time until it saturates (a condition that we do not want).

During the OFF time, the inductor stops receiving energy from the source. At this point the electromagnetic field created during the ON time starts to collapse in reverse thus inducing a voltage with inverse polarity across the inductor. What happens at this point is very important. Since the voltage across the inductor is now reversed you can think of the inductor as a battery connected in series with the input voltage source. This is why there is a voltage increase at the output of the boost converter.

The output diode rectifies the voltage from the inductor and delivers it to the output capacitor. The output capacitor smoothes out the wave.

I will not discuss any complex mathematical equations used to find different values of currents and voltages in different parts of the circuit but a few things are important to note:

Output Voltage determination
Output Current determination
Size of the inductor
Efficiency

The output voltage is determined by the ratio of the input volts and the length of the pulse. Your switching frequency has a period of 10uS, the input voltage is 12V and the the ON time is 5uS, then your output volts will be:

Vout=12*(10/(10-5)) = 24V (minus output diode voltage drop)

The output current of a boost converter will be directly related to the inductor current during the ON time of the cycle. Since we have a switching frequency and an inductor, the most important factor that determines the inductor current is the inductor's impedance at that frequency.

If the frequency is too high for say a 100uH inductor it will act as a huge resistor so its current will be very low. With low inductor current low output currents will be present. On the other hand if the frequency is too low for the inductor high currents will be present through the inductor thus high output current.

Choosing the right inductor value for an operating frequency is essential. Too much current through an inductor will cause the ferrite core to saturate. When an inductor saturates it acts as a dead short causing MOSFETs to burn and sparks to fly.