# Energy in Boost converter

#### anhnha

Joined Apr 19, 2012
881
I am having a hard time understanding a point while reading about boost converter.
1. First switch is ON, the inductor is charged and energy is accumulated with an amount of E1.
2. Next switch is OFF, the inductor will discharge its energy to the output. After a specific of time diode is OFF (for some reason). Let's call the energy is transferred to output is E2 and E2 < E1.
So now there is a situation like this: both switch and diode are OFF and inductor still has energy , E1 - E2.
What is the voltage at the end of inductor that is NOT connected with Vin now?
Assuming that all components are ideal.

#### GopherT

Joined Nov 23, 2012
8,012
In an inductor, current lags voltage. So, in a well designed inductor, the switch will be closed until the inductor us has charged to an optimal amount, then the switch opens and there is a nice big voltage kick that dumps the energy in the inductor through the diode an in to the cap & load). Hopefully, there is enough excess energy in the cap to supply power until the following closed/open cycle.

#### Papabravo

Joined Feb 24, 2006
14,884
At the moment the switch is opened there is a large [positive] voltage transient, proportional to L*di/dt, which forward biases D1. D1 conducts and the energy of the transient charges the capacitor and is dissipated in the load. The transient voltage from the inductor drops to a lower level approaching Vin and the diode stops conducting. If the switch continues to be open the voltage drop across the inductor will be 0 volts. On the other side, as the load continues to discharge the capacitor, the controller will decide at some point to turn the switch back on. So the cycle continues. The energy difference E1-E2 just disappears as a switching loss. Ohm's law enforces the idea that if no current can flow the voltage drop must be 0 volts. Conversely if the voltage drop is 0 volts, then no current can flow.

#### Roderick Young

Joined Feb 22, 2015
408
If all components are ideal, then the diode would turn on instantly. But let's say it doesn't for some reason, and all components are ideal. Then the voltage across the inductor would zoom to infinity until the diode conducts. Back to the real world, where there is an actual resistance to the inductor, inductance in the wiring, and capacitance all around, we might see a voltage overshoot for a fraction of a microsecond, quickly settling to a slowly sagging voltage as the capacitor is charged, and perhaps a bit of ringing on the waveform, too.

#### dl324

Joined Mar 30, 2015
12,271
An excerpt from an OnSemi application note on switching regulators describing theory and showing non-ideal waveforms.

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#### anhnha

Joined Apr 19, 2012
881
If all components are ideal, then the diode would turn on instantly. But let's say it doesn't for some reason, and all components are ideal. Then the voltage across the inductor would zoom to infinity until the diode conducts. Back to the real world, where there is an actual resistance to the inductor, inductance in the wiring, and capacitance all around, we might see a voltage overshoot for a fraction of a microsecond, quickly settling to a slowly sagging voltage as the capacitor is charged, and perhaps a bit of ringing on the waveform, too.
This is what I was confused. I will read more about that.
Thank you all
I see that there is PWM is used. Is its purpose to avoid this situation?

#### Papabravo

Joined Feb 24, 2006
14,884
The PWM is designed to monitor the output voltage and make small corrections to the duty cycle to keep the output voltage within some small range. The actual range depends on the controller chip.