Question about Buck Convertor Operation

  • Thread starter jendrix
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In summary, the conversation discusses the behavior of current and voltage in a buck converter circuit, specifically focusing on the inductor and capacitor. The inductor has a constant flow of current with a changing magnitude, but the voltage across it can be positive or negative, depending on the direction of current flow. This is due to the equation v(t) = L \frac{di(t)}{dt}, which relates the voltage and current in an inductor. The capacitor acts as a storage reservoir and its voltage ripples around the nominal value, but does not go negative. The conversation also touches on the derivation of the equation for the RMS current through the switch in a buck converter.
  • #1
jendrix
122
4
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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  • #2
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:
 
  • #3
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?
 
  • #4
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 [tex]v(t) = L \frac{di(t)}{dt}[/tex]
Are you familiar with how to interpret such equations?
 
Last edited:
  • #5
berkeman said:
Yes, that is correct. The equation I was asking about it [tex]v(t) = L \frac{di(t)}{dt}[/tex]
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?
 
  • #6
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.
 
  • #7
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?
 
  • #8
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).
 
  • #9
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
 

1. How does a buck converter work?

A buck converter is a type of DC-DC converter that steps down a higher input voltage to a lower output voltage. It uses an inductor and a diode to store and release energy, and a switch (usually a transistor) to control the flow of current. By switching the transistor on and off at a high frequency, the buck converter is able to regulate the output voltage to a desired level.

2. What are the advantages of using a buck converter?

One of the main advantages of a buck converter is its high efficiency. Since it steps down the input voltage instead of dissipating it as heat, it is able to deliver more power to the load. Additionally, it is a simple and compact circuit, making it suitable for a wide range of applications. It also provides a stable output voltage even with variations in the input voltage.

3. What is the difference between a buck converter and a boost converter?

A buck converter steps down the input voltage, while a boost converter steps it up. This means that a buck converter is used to regulate a lower output voltage, while a boost converter is used to generate a higher output voltage. Additionally, the components and circuitry of these two converters are different, with a boost converter using a capacitor instead of an inductor.

4. How do I choose the right buck converter for my application?

Choosing the right buck converter depends on several factors, including the input and output voltage requirements, the desired output current, and the efficiency and size constraints. It is important to select a converter with a suitable switching frequency and component ratings to ensure stable and efficient operation.

5. How can I improve the efficiency of my buck converter?

There are several ways to improve the efficiency of a buck converter, such as using high quality components, minimizing the switching losses, and reducing the parasitic resistance in the circuit. Proper design and layout techniques, such as minimizing loop inductance and using short, wide traces, can also help improve the efficiency. Additionally, choosing an appropriate control scheme and operating at a suitable switching frequency can also contribute to higher efficiency.

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