# Question about Buck Convertor Operation

Hi ,

I have a query on Buck convertors , I have attached a picture that has the circuit and associated waveforms.

http://i.imgur.com/ydQ20m8.png

When you look at the inductor it has a constant flow of current through it with a changing magnitude but the same direction , but the waveform shows the voltage through it being positive and later on negative.If the voltage across the inductor was negative wouldn't that mean the direction of current through it would also change?

*edit*

I did some further reading , is it because voltage across an inductor is linked to the change in current , so the current could fall but still be positive and this would create a negative voltage across the inductor?

Thanks

berkeman
Mentor
Hi ,

I have a query on Buck convertors , I have attached a picture that has the circuit and associated waveforms.

http://i.imgur.com/ydQ20m8.png

When you look at the inductor it has a constant flow of current through it with a changing magnitude but the same direction , but the waveform shows the voltage through it being positive and later on negative.If the voltage across the inductor was negative wouldn't that mean the direction of current through it would also change?

Thanks
Are you familiar with the simple differential equation that relates the inductor current to the voltage across it?

Are you familiar with the simple differential equation that relates the inductor current to the voltage across it?

Hi ,

I edited my original post

"I did some further reading , is it because voltage across an inductor is linked to the change in current , so the current could fall but still be positive and this would create a negative voltage across the inductor?"

Does this sound ok?

berkeman
Mentor
Hi ,

I edited my original post

"I did some further reading , is it because voltage across an inductor is linked to the change in current , so the current could fall but still be positive and this would create a negative voltage across the inductor?"

Does this sound ok?
Yes, that is correct. The equation I was asking about is $$v(t) = L \frac{di(t)}{dt}$$
Are you familiar with how to interpret such equations?

Last edited:
Yes, that is correct. The equation I was asking about it $$v(t) = L \frac{di(t)}{dt}$$
Are you familiar with how to interpret such equations?

I do but I was a little rusty on circuit , I have a further question on Ico if you could help? When the switch is off wouldn't the capacitor discharge and have the current Ico flow towards the node at Vo?

berkeman
Mentor
I do but I was a little rusty on circuit , I have a further question on Ico if you could help? When the switch is off wouldn't the capacitor discharge and have the current Ico flow towards the node at Vo?
Yes, I think that's what they are showing in the waveforms. How much it discharges depends on the load current and the inductor current.

Yes, I think that's what they are showing in the waveforms. How much it discharges depends on the load current and the inductor current.

So immediately after the switch opens which direction would Ico be flowing?

berkeman
Mentor
It would be positive for the first part of the OFF cycle (when the inductor current is enough to maintain Vo across the load, and would reverse as the inductor current fell below Vo/Rload. Co is like a reservoir storage capacitor. The ripple current in and out of Co is an important part of the design of a buck converter (the cap must be able to handle the ripple current, and the ripple voltage you get across the load depends a lot on the DCR of Co).

It would be positive for the first part of the OFF cycle (when the inductor current is enough to maintain Vo across the load, and would reverse as the inductor current fell below Vo/Rload. Co is like a reservoir storage capacitor. The ripple current in and out of Co is an important part of the design of a buck converter (the cap must be able to handle the ripple current, and the ripple voltage you get across the load depends a lot on the DCR of Co).

But if Ico changed direction then wouldn't also the voltage across the capacitor change direction too? And if so wouldn't that also change the voltage across the load if you follow the KVL rule?

berkeman
Mentor
But if Ico changed direction then wouldn't also the voltage across the capacitor change direction too? And if so wouldn't that also change the voltage across the load if you follow the KVL rule?
The voltage on Co ripples about the nominal value. It does not go negative. For example, for a 5V output DC-DC converter, it's typical for you to get about 200mVpp of ripple on the 5V output due to the switching action. So if you measure the output with an oscilloscope, you see it rippling between 4.9V and 5.1V, at the frequency of the DC-DC switching action.

The voltage on Co ripples about the nominal value. It does not go negative. For example, for a 5V output DC-DC converter, it's typical for you to get about 200mVpp of ripple on the 5V output due to the switching action. So if you measure the output with an oscilloscope, you see it rippling between 4.9V and 5.1V, at the frequency of the DC-DC switching action.

Thanks , it's becoming clear now.I was wondering , I'm trying to derive the equation for the rms current through the switch but am having some trouble.

I have to prove that it is root(k*Io^2 + delta(i)^2 /12 )

Bit stuck on how to start

This is my approach to the problem

http://i.imgur.com/R3zxz8k.jpg

When the switch is on for a period of k*Tp the current rises from Iin which is = to kIo the current when the switch is on rises by delta(i)

Does this approach look ok? I'm not having much luck

Thanks