DC-DC Converter Design Problem

jegues · Oct 30, 2012

Hello all,

See figure attached for requirements regarding a design problem I am working on, in addition to my work thus far.

I've managed to come up with a number of equations for my component values but they all contain parameters that have varying ranges.

Is it possible to write enough equations to correctly solve for everything simultaneously or do I need to select a particular parameter as a starting point are work outwards from there?

I need to preform a worst case design so it must meet the design criteria in the worst case scenarios. How can I identify what these worst case scenarios are?

Thanks again!

BradtheRad · Oct 30, 2012

I had to spend hours playing with an animated simulator (Falstad's), before I got a notion what is happening in these switched coil converters.

Is it possible to write enough equations to correctly solve for everything simultaneously or do I need to select a particular parameter as a starting point are work outwards from there?

There are so many factors at work that it is hard to solve them all simultaneously.

These factors include:

* Coil henry value
* Coil's maximum current for inductive saturation, coil's maximum current for safe temperature
* operating frequency
* load resistance
* capacitor value
* supply voltage
etc.

Therefore when starting a design, you try one parameter and see what develops based on that. If you encounter something that won't work, then you may need to go back to the first parameter and change it.

Example of things that might not work: coil's henry value is too high or too low, coil is too expensive, coil is too bulky, etc.

The coil is the center of action. Looking at your equations, your next step was going to be to find the coil's henry value. This tends to be defined by the given operating frequency.

I made a user-interactive simulation of a buck converter which is oversimplified but should aid understanding.

It will demonstrate coil action as you press and release the On switch.

Click the link below. It will open falstad.com website, load my schematic, and run it on your computer. (Click Allow to load the Java applet.)

https://tinyurl.com/8zk6pjd

You will need to change some values. Right-click on a component to bring up an edit window.

This simulator won't do everything you need, but it will get you part of the way there.

I need to preform a worst case design so it must meet the design criteria in the worst case scenarios. How can I identify what these worst case scenarios are?

Worst case scenarios include:

Will my design start up reliably, or does it need prodding?

What if I short circuit the output wires? Will components be destroyed?

What if I do not attach a load? Will anything bad happen?

Could my design ever expose the load to overvoltage? What about at power-up?

What if the load is momentarily disconnected? Will my design continue to operate?
What if the load resistance is suddenly changed?

What if the supply voltage drops? Will anything bad happen?

What if oscillations stall during switch-On cycle? Will anything bad happen?
If they stall during switch-Off cycle?

Etc.

There are IC's made to drive this type of supply. They may have a few safeguards built in, to prevent disasters.

jegues · Oct 30, 2012

BradtheRad said:
I had to spend hours playing with an animated simulator (Falstad's), before I got a notion what is happening in these switched coil converters.

There are so many factors at work that it is hard to solve them all simultaneously.

These factors include:

* Coil henry value
* Coil's maximum current for inductive saturation, coil's maximum current for safe temperature
* operating frequency
* load resistance
* capacitor value
* supply voltage
etc.

Therefore when starting a design, you try one parameter and see what develops based on that. If you encounter something that won't work, then you may need to go back to the first parameter and change it.

Example of things that might not work: coil's henry value is too high or too low, coil is too expensive, coil is too bulky, etc.

The coil is the center of action. Looking at your equations, your next step was going to be to find the coil's henry value. This tends to be defined by the given operating frequency.

I made a user-interactive simulation of a buck converter which is oversimplified but should aid understanding.

It will demonstrate coil action as you press and release the On switch.

Click the link below. It will open falstad.com website, load my schematic, and run it on your computer. (Click Allow to load the Java applet.)

https://tinyurl.com/8zk6pjd

You will need to change some values. Right-click on a component to bring up an edit window.

This simulator won't do everything you need, but it will get you part of the way there.

Worst case scenarios include:

Will my design start up reliably, or does it need prodding?

What if I short circuit the output wires? Will components be destroyed?

What if I do not attach a load? Will anything bad happen?

Could my design ever expose the load to overvoltage? What about at power-up?

What if the load is momentarily disconnected? Will my design continue to operate?
What if the load resistance is suddenly changed?

What if the supply voltage drops? Will anything bad happen?

What if oscillations stall during switch-On cycle? Will anything bad happen?
If they stall during switch-Off cycle?

Etc.

There are IC's made to drive this type of supply. They may have a few safeguards built in, to prevent disasters.

Hello BradtheRad,

First off thank you very much for all your advice!

I tried to kick things off as you've said by fixing a few parameters and seeing how things unfold.

Attached to this post is my first rough attempt at poking through the design.

I chose to fix both the input and output voltages in the middle of their respective ranges and to fix myself at the lowest switching frequency (i.e. 20kHz).

I started by extracting the resistor value R from the power constraint.

I found that,

\[R \approx 5\Omega\]

From here I was able to compute a minimum L such that the converter will indeed operate in continuous conduction mode. (CCM)

I found that,

\[L_{min} \approx 95uH\]

I was able to obtain an average source current of,

\[I_{in} \approx 1.11A\]

but I was confused how I can relate this back to the source current ripple constraint. (In particular I don't know the equation for ΔIin)

To ensure CCM I selected,

\[L = 1.25L_{min} \approx 118.73uH\]

Which gave me an inductor current ripple of,

ΔiL ≈ 7.11A

I know I am probably violating a bunch of the design constraints at the moment or have poorly selected my component values but I wanted to start with something put the pen to paper so I can begin refining and retuning certain aspects of my design.

This rough attempt leaves me with the following questions,

Could you recommend a better starting point in determining some of the components values?
Should I have fixed my input/output voltages and frequency as I did? Or should I have proceeded in another manner, say by first determining the Inductor value and then determining the resistor value from there? (I.e. taken an alternative path)

Hopefully this rough attempt starts up the discussion, so I can work towards a more accurate and elegant solution that falls within the design requirements.

I look forward to your (or anyone elses) comments!

Cheers!

mtwieg · Oct 30, 2012

The input ripple current spec is bizarre. It's virtually impossible to fulfill without having an input filter, which could possibly take the form of a very large input bypass capacitor or a CLC filter.

jegues · Oct 30, 2012

I decided to further experiment, solving the outcomes of L, C, ΔiL, IL and R given a selected frequency, input voltage and output voltage.

Of course to obtain the values of the outcomes in a simple manner I made various assumptions.

These assumptions are stated below,

Assumptions:

Diode and Transistor voltage drops are 0
Minimum output current = Output current, In other words Imin = 0 (We are operating on the edge of CCM)
The value for the output current was always selected such that the output power was 100W. **NOTE: We are neglecting the power dissipated across the series resistance of the source (i.e. Prin = IL²D²rin) because for all values of IL and D I found this to be small in comparison to the 100W.

Attached below is a figure of the table of outcomes.

Are my assumptions "valid enough" that these results have actual relevant meaning and/or insight?

Can someone help me interpret my results? Am I moving in the right direction?

jegues · Oct 31, 2012

After goofing around some more with BradtheRad's Falstad circuit, I tried simulating one of the rows of my tables in the previous post.

For, f = 20kHz, Vin = 100V, Vout = 25V I obtained a L = 118uH, C=200uF and R = 6.25Ω.

Throwing these into the falstad simulation with a duty cycle of 0.25 and a 20kHz switching frequency I obtained these results. (See figure attached)

As we can see I am obtaining ~19.7V at the output, which is close to the desired 20V.

We are also observing a current of 3.1A through the load resistor, giving an output power of 61.07W which is about a third less than our desired 100W.

How can I improve on my design?

Cheers

BradtheRad · Oct 31, 2012

Since I too am learning how switched-coil supplies work, I made simulations to see if a configuration can be made that will live up to your specs.

To obtain 100W at 20V, you are pushing 5 amps through a load resistance of 4 ohms.

Your 7 amp ripple value is not too far off. It roughly matches my simulation. It creates a waveform that brings out maximum coil action.

However such a large swing puts stress on several components. It sends large current surges through the smoothing capacitor. Design articles suggest you reduce this stress by designing for a certain amount of CCM.

Your amp ripple spec is for 2 percent or less. Typical values used are 10 percent. However it may be possible to meet a 2 percent spec.

Here are screenshots of scope traces from my simulations, showing different ripple amounts.

Not shown are my layouts.

Some component values were adjusted to alter the operating frequency and make the waveforms more obvious.

- - - Updated - - -

jegues said:
After goofing around some more with BradtheRad's Falstad circuit, I tried simulating one of the rows of my tables in the previous post.

For, f = 20kHz, Vin = 100V, Vout = 25V I obtained a L = 118uH, C=200uF and R = 6.25Ω.

Throwing these into the falstad simulation with a duty cycle of 0.25 and a 20kHz switching frequency I obtained these results. (See figure attached)

As we can see I am obtaining ~19.7V at the output, which is close to the desired 20V.

We are also observing a current of 3.1A through the load resistor, giving an output power of 61.07W which is about a third less than our desired 100W.

How can I improve on my design?

Cheers

I see you put up your post while I was composing mine.

Good work with the simulator. (It appears Falstad's will not accept clock frequencies greater than 25 kHz. I suppose this is why you found it necessary to raise the coil value from your previous 95 uH figure, so the simulation could achieve your desired performance specs?)

Since you want greater output voltage, all you need to do is give proportionally more time for coil current to build. This can be done by increasing duty cycle, and/or changing operating frequency.

jegues · Oct 31, 2012

BradtheRad said:
I see you put up your post while I was composing mine.

Good work with the simulator. (It appears Falstad's will not accept clock frequencies greater than 25 kHz. I suppose this is why you found it necessary to raise the coil value from your previous 95 uH figure, so the simulation could achieve your desired performance specs?)

Since you want greater output voltage, all you need to do is give proportionally more time for coil current to build. This can be done by increasing duty cycle, and/or changing operating frequency.

Hi Brad,

If you read and view the attachement in my previous post, (post #5) in this thread you will see how I chose the values that I did.

As mentioned previously, I made a table of values for L, C and R given a Vin, Vout, and f. (Contingent on a simple set of assumptions of course)

For my figure within the Falstad Simulator I injected a clock frequency of 20kHz with a duty cycle of 0.25.

How do I figure out my source ripple? Does it relate to the ripple in the inductor current?

As you can see from your graphs, the source current spikes from 0 to the minimum inductor current, and then climbs to peak current in the inductor and then falls back to 0. (It looks like a square with a triangle sitting on it)

So is the peak-peak ripple simply Imax then?

I'm sure I could throw a large cap after the source and before the switch to not allow the source to drop its current back to 0 every DT.

Do you know how to obtain the equation for source ripple if I had added a cap right after it?

Hopefully we both gain some insight on this topic and the design at hand.

Thanks again!

BradtheRad · Oct 31, 2012

jegues said:
Hi Brad,

If you read and view the attachement in my previous post, (post #5) in this thread you will see how I chose the values that I did.

As mentioned previously, I made a table of values for L, C and R given a Vin, Vout, and f. (Contingent on a simple set of assumptions of course)

As you could see, by starting with a few specs, the rest derives from that.

Your results were pretty consistent for coil and capacitor values which are key components. It shows your approach is sound.

I noticed one figure you had for 'average source current' of 1.11 A, which turned out to be questionable.

For my figure within the Falstad Simulator I injected a clock frequency of 20kHz with a duty cycle of 0.25.

In real life you will usually obtain (or construct) a coil which is close to what you're looking for.

After you see how things are running, you will adjust frequency and/or duty cycle to get the desired output level. Or your control module will adjust it.

How do I figure out my source ripple? Does it relate to the ripple in the inductor current?

All I can picture is that source ripple refers to supply ripple. Because incoming supply current is either on or off with a single converter.

You can interleave two or more converters, which has the supply providing some current at all times.

As you can see from your graphs, the source current spikes from 0 to the minimum inductor current, and then climbs to peak current in the inductor and then falls back to 0. (It looks like a square with a triangle sitting on it)

So is the peak-peak ripple simply Imax then?

That's right, as per coil current, when talking about non-CCM.

I'm sure I could throw a large cap after the source and before the switch to not allow the source to drop its current back to 0 every DT.

Do you know how to obtain the equation for source ripple if I had added a cap right after it?

I'm not sure this would gain you anything, although it might stabilize supply voltage somewhat.

jegues · Nov 2, 2012

BradtheRad said:
As you could see, by starting with a few specs, the rest derives from that.

Your results were pretty consistent for coil and capacitor values which are key components. It shows your approach is sound.

I noticed one figure you had for 'average source current' of 1.11 A, which turned out to be questionable.

In real life you will usually obtain (or construct) a coil which is close to what you're looking for.

After you see how things are running, you will adjust frequency and/or duty cycle to get the desired output level. Or your control module will adjust it.

All I can picture is that source ripple refers to supply ripple. Because incoming supply current is either on or off with a single converter.

You can interleave two or more converters, which has the supply providing some current at all times.

That's right, as per coil current, when talking about non-CCM.

I'm not sure this would gain you anything, although it might stabilize supply voltage somewhat.

Hello again BradtheRad,

Thank you again for another informative reply. How would you recommend one suppress the source ripple then? Would a second inductor in series with the source resistance do the trick? Of course, a diode would be put in place to keep the inductor current flowing when the switch opens.

Apparently the strategy for this design is to identify the design requirements in terms of the duty cycle parameter, D. Once this has been achieved, one is to obtain a worst case D. We have also been told that as the designers we are allowed to simply select a frequency within our range, but not an input voltage. The input voltage will lie in the specified range, but it is not to be chosen by the designer.

Nonetheless, from the given range of our input/output voltages, we can easily determine a lower and upper bound on the range for D. That being said, the worst case D will either fall somewhere in this range, or at one of the bounds. (the lower bound if the worst case D is lower than our lower bound, or at the upper bound if the worst case D is larger than the upper bound)

We don't need to actually implement the control system that would be required to obtain a constant 100W of power over the range of input/output voltages, but we must show that we are indeed designing for the worst case D.

Later this weekend I am going to attempt to describe the constraints of the design in terms of D and attempt to identify the worst case D. Of course, I will post my findings or problems as they arise.

It seems as though this is a minimization problem.

If you have any tips/suggestions to further simplify my attempt at identifying and obtaining the worst case D, or perhaps another strategy to identify the worst case design, I would love to hear about it.

I'm interested to hear what your thoughts are.

Cheers!

BradtheRad · Nov 2, 2012

jegues said:
Hello again BradtheRad,

Thank you again for another informative reply. How would you recommend one suppress the source ripple then? Would a second inductor in series with the source resistance do the trick?

'Source current ripple' only has a few dozen hits searching on the internet.

It could be that source ripple refers to the supply V drop during the switch-On cycle. I don't see how it could refer to current. Switching current on and off is the key principle of this type of power converter.

Here are links to discussions other websites which mention it.

To reduce supply V ripple, you might install a smoothing capacitor across the leads to the converter, or install an inductor similar to the kind that reduce RF in a wire.

Apparently the strategy for this design is to identify the design requirements in terms of the duty cycle parameter, D. Once this has been achieved, one is to obtain a worst case D. We have also been told that as the designers we are allowed to simply select a frequency within our range, but not an input voltage. The input voltage will lie in the specified range, but it is not to be chosen by the designer.

Nonetheless, from the given range of our input/output voltages, we can easily determine a lower and upper bound on the range for D. That being said, the worst case D will either fall somewhere in this range, or at one of the bounds. (the lower bound if the worst case D is lower than our lower bound, or at the upper bound if the worst case D is larger than the upper bound)

We don't need to actually implement the control system that would be required to obtain a constant 100W of power over the range of input/output voltages, but we must show that we are indeed designing for the worst case D.

Later this weekend I am going to attempt to describe the constraints of the design in terms of D and attempt to identify the worst case D. Of course, I will post my findings or problems as they arise.

It seems as though this is a minimization problem.

If you have any tips/suggestions to further simplify my attempt at identifying and obtaining the worst case D, or perhaps another strategy to identify the worst case design, I would love to hear about it.

I'm interested to hear what your thoughts are.

Cheers!

This will cover the above plus other points.

1.

Yes, duty cycle (D) is a chief factor in the basic equation for converting V_in to V_out.
At least in CCM that is. Because DCM has a 'dead' spot hence it requires a different equation.

You need to select values which will let you vary the duty cycle between maybe 5 and 95 percent, to handle the range of possibilities as to supply voltage and load.

If your control circuit allow the duty cycle to get close to zero or 100, then it is liable to latch totally off or totally on. This can be disastrous to components, including valuable devices attached at the output.

2.

So far you went on the assumption that switch-On resistance is zero through the transistor. It's okay to do this with a high supply V. However when supply V is low, or when the supply has high internal resistance (or when some other component contributes resistance, including wires), then it can become impossible to obtain sufficient current flow into the coil, no matter what duty cycle you try.

3.

The amount of hysteresis is another factor which will determine the shape of the waveform. The narrower your hysteresis, the higher your waveform rises above zero volts.

Also notice that the amount of hysteresis impacts two major coil specs: (1) saturation current, and (2) safe power capacity. Referring to the images in my post #7. One coil must handle 9 A without saturating. The other (with narrower hysteresis) can be rated for lower saturation current, making it less costly.

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