# Boost converter inductor discharging question

1. Feb 11, 2017

### TheRedDevil18

I'm reading up on boost converters and so far I understand that it basically uses the inductor as a source together with the actual supply voltage to boost the voltage. During the discharging period, why would the current (if allowed to) drop to zero ?, wouldn't the supply or battery still be supplying current to the load ?

Source
https://en.wikipedia.org/wiki/Boost_converter

2. Feb 11, 2017

### Baluncore

The output reservoir capacitor receives a flow of charge through the inductor each time the switch opens. If the load does not draw sufficient current to pull the output voltage back down below the regulated voltage, then the regulator will prevent another switching cycle until it does. The current through the inductor may then fall to zero while waiting.
The situation can also occur when charging a battery from a low voltage supply.

3. Feb 11, 2017

### Staff: Mentor

Were you to keep the switch OPEN for a long time, then inductor current would reach a fixed value. With steady current the voltage across the inductor will be zero, meaning the voltage across the load would be equal to the voltage output by the supply: hence, no voltage boost!

By switching the coil to ground at judiciously-chosen intervals, the coil generates a boosted voltage across the load. In the arrangement shown, with no capacitor, the load voltage will be delivered as a pulsed voltage, the pulses separated by drops to zero volts.

In operation, with a fixed load, the inductor current will be found to be approximately constant, only.

4. Feb 11, 2017

### TheRedDevil18

So when the inductor is discharging the total input current is inductor current + source current ?

Why would the load voltage drop to zero ?, when the switch is closed the capacitor will try to maintain a constant voltage to the load as long as it is large enough and charges when the switch is open which creates the ripple ?

5. Feb 11, 2017

### cnh1995

Inductor current is 'equal to' source current. They are in series after the switch is opened.
Output voltage=inductor voltage+source voltage, which is more than the source voltage. Hence the name 'boost' converter.
NO is referring to the circuit diagram in the wiki article which doesn't have a capacitor.

6. Feb 11, 2017

### Staff: Mentor

In the circuit you linked to, the load current is the inductor current while ever the diode is conducting.

There is no reservoir capacitor shown in the schematic you referenced.

7. Feb 11, 2017

### TheRedDevil18

Thanks for the replies. From what I understand now is that when the load is connected (switch open), the current in the circuit drops (more resistance) but because of the inductor it prevents the instantaneous drop in current but if it where allowed to fully discharge then it would reach steady state current and because there is no di/dt then the inductor voltage is zero, correct ?

8. Feb 11, 2017

### Staff: Mentor

That would happen, yes.

9. Feb 11, 2017

Ok thanks