Question about Buck Convertor Operation

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SUMMARY

The discussion focuses on the operation of Buck converters, specifically addressing the relationship between inductor voltage and current. The key equation discussed is v(t) = L di(t)/dt, which illustrates that a negative voltage across an inductor does not necessarily imply a reversal of current direction. Participants clarify that during the switch-off phase, the capacitor discharges, and the current Ico initially flows towards the load, with the voltage across the capacitor exhibiting ripple rather than going negative. The conversation concludes with a query about deriving the RMS current through the switch, indicating a deeper exploration of Buck converter dynamics.

PREREQUISITES
  • Understanding of Buck converter topology and operation
  • Familiarity with inductor voltage-current relationships
  • Knowledge of differential equations in electrical circuits
  • Basic concepts of capacitor behavior in DC-DC converters
NEXT STEPS
  • Study the derivation of the equation v(t) = L di(t)/dt in detail
  • Learn about ripple voltage and current in Buck converters
  • Investigate the impact of load current on capacitor discharge in Buck converters
  • Explore methods to calculate RMS current in switching devices
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Electrical engineers, power electronics designers, and students studying DC-DC converter circuits will benefit from this discussion, particularly those looking to deepen their understanding of Buck converter operation and analysis.

jendrix
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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
 
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jendrix said:
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? :smile:
 
berkeman said:
Are you familiar with the simple differential equation that relates the inductor current to the voltage across it? :smile:

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?
 
jendrix said:
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:
berkeman said:
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?
 
jendrix said:
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.
 
berkeman said:
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?
 
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).
 
berkeman said:
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?
 
  • #10
jendrix said:
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.
 
  • #11
berkeman said:
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
 
  • #12
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
 

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