Buck convertor

[oodes]

Hi,
I am simulating a design for a buck converter on LTSpice. I have done all the calculations using the formula. My input voltage is 12V and my desired output is 9V . I am using a 75KHz switching frequency. My output ripple is 3%. The output i am getting is around 9.3V but i need an exact 9V . I know if i reduce my on time for the switch i can get my Vout down to the exact 9V
Is this the correct way to get my desired Vout reduced ? and what is the reason is it overshooting in the first place? Are there any other design parameters that affect my Vout?
Thanks in advance

 

[MrAl]

Hi,

There are lots of reasons for this but you'll have to supply more information like the schematic.

 

[alec_t]

Simulation is never going to give you an exact desired real output, because component models are only an approximation of reality. IMO 9.3V is a pretty good approximation of 9V.

 

[ronsimpson]

Simulation is never going to give you an exact desired real output, because component models are only an approximation of reality. IMO 9.3V is a pretty good approximation of 9V.

And the real parts will not give you 9.00 volts either.
If the IC has a reference that is 2.5% and the feedback resistors are 5% (or even 1%) then 9.3 volts is about right.

 

[oodes]

having some trouble uploading the schematic from LTSpice but my component values are 5uF Cap , 37uH Inductor. Outputting up to 500mA of current, so i inserted an 18ohm resistor, standard buck converter design. Just need to know what component to tune to give the desired 9V output.

 

[oodes]

I agree 9.3V is a pretty good approx and i know simulation is never gonna be the real component responses, i just need to know what component of my circuit do i need to adjust and the reason why.
Its an assignment question for buck design here it is
"If the output values produced do not fully correspond to the calculated values indicate possible reasons for this and make adjustments to your design to compensate"
Thanks in advance

 

[ronsimpson]

I ran your circuit. I don't see 9.3 volts. I see ripple from 8.93 to 9.17 volts.
If you want to change the output voltage, change the duty cycle.
---OR---
Pick a different diode. You are loosing voltage across the diode.
---OR---
You can set V1 or S1 to a different internal resistance. (cheating!)

 

[crutschow]

If you increase the value of C1 (say to 49μF) to reduce the output ripple voltage you will see that the average output voltage is very close to 9.0V.

 

[oodes]

I ran your circuit. I don't see 9.3 volts. I see ripple from 8.93 to 9.17 volts.
If you want to change the output voltage, change the duty cycle.
---OR---
Pick a different diode. You are loosing voltage across the diode.
---OR---
You can set V1 or S1 to a different internal resistance. (cheating!)

Sorry that was my amended circuit to reduce my Vo to 9V , i had adjusted the duty cycle on this already. If u reset the ton to 10uS u will see the 9.3V average between the ripple. I had read alright that the formulas for component calculation ignore the voltage loss across the diode, but can u explain why this pushes up the Vout above the desired 9V value. I know i'm being pedantic

 

[oodes]

The original design View attachment 62912 . I am just looking for a theoretical reason why my voltage goes above the 9V calculated with the 0.75 Duty Cycle. If i adjust the duty cycle my vout reduces to the desired 9V ( between the ripple). I just need the reason for this happening , and if i make the adjustment to the duty cycle i need a justification for doing so.
Thanks for the help guys

 

[MrAl]

Hello there oodes,


Well, theoretically with a perfect 0.75 duty cycle and components that are perfect with no losses and a little output averaging filter, you'd see an output voltage that is exactly 9.000000 volts. That's not 9.000001 volts nor is it 8.999999 volts, but exactly 9 volts plus or minus nothing.

Now if you figure in the error in floating point math and the errors dont average out, you might see a very small difference of 1uV or something like that, probably more typically 1e-12 volts plus or minus.

With imperfect components though, you'll see lots of different kinds of variations. The list is quite extensive...

1. The diode forward voltage will pull the output down slightly, not up. That's because it represents a power loss.
2. The diode reverse leakage, which will affect it less in most cases.
3. The inductor series resistance will decrease the output also.
4. The switch series resistance will decrease the output.
5. The capacitor ESR will decrease the output, but it plays in combination with the switch resistance too.
6. The capacitor value itself. An increase in value can cause more peak drop across the switch resistance and input source resistance.
7. The pulse timing accuracy which comes in at least two forms: a. The pulse setting itself, and b. the rise and fall times.
8. The output resistance, which causes a drop in the inductor ESR and the switch ESR so decreases the output.
9. The input voltage source series resistance, which of course causes a loss.

So you see every single component causes a drop in the output voltage from the theoretical with perfect components.
If you improve the diode choice you'll get closer results, or better yet get rid of it entirely by going to a synchronous buck (another switch).
Probably a big factor is the rise and fall times of the pulse source. You can't set them to zero you have to choose some
reasonable value like 1ns. If you choose 0 then LT spice will choose for you which will cause problems.
I'd also set the values for the ESR for all the components too rather than leave them blank or set them to zero. Choose some small value like 0.001 or even lower.

To average the output without appreciable loss, connect a 10k resistor in series with a 0.1uf cap and take the output voltage reading from across the cap and increase the simulation time somewhat until the voltage levels off. Once the voltage levels off the loss is extremely small.

It is entirely possible to set the values of the components to see very near theoretical results in the simulation.

 

[oodes]

Hello there oodes,


Well, theoretically with a perfect 0.75 duty cycle and components that are perfect with no losses and a little output averaging filter, you'd see an output voltage that is exactly 9.000000 volts. That's not 9.000001 volts nor is it 8.999999 volts, but exactly 9 volts plus or minus nothing.

Now if you figure in the error in floating point math and the errors dont average out, you might see a very small difference of 1uV or something like that, probably more typically 1e-12 volts plus or minus.

With imperfect components though, you'll see lots of different kinds of variations. The list is quite extensive...

1. The diode forward voltage will pull the output down slightly, not up. That's because it represents a power loss.
2. The diode reverse leakage, which will affect it less in most cases.
3. The inductor series resistance will decrease the output also.
4. The switch series resistance will decrease the output.
5. The capacitor ESR will decrease the output, but it plays in combination with the switch resistance too.
6. The capacitor value itself. An increase in value can cause more peak drop across the switch resistance and input source resistance.
7. The pulse timing accuracy which comes in at least two forms: a. The pulse setting itself, and b. the rise and fall times.
8. The output resistance, which causes a drop in the inductor ESR and the switch ESR so decreases the output.
9. The input voltage source series resistance, which of course causes a loss.

So you see every single component causes a drop in the output voltage from the theoretical with perfect components.
If you improve the diode choice you'll get closer results, or better yet get rid of it entirely by going to a synchronous buck (another switch).
Probably a big factor is the rise and fall times of the pulse source. You can't set them to zero you have to choose some
reasonable value like 1ns. If you choose 0 then LT spice will choose for you which will cause problems.
I'd also set the values for the ESR for all the components too rather than leave them blank or set them to zero. Choose some small value like 0.001 or even lower.

To average the output without appreciable loss, connect a 10k resistor in series with a 0.1uf cap and take the output voltage reading from across the cap and increase the simulation time somewhat until the voltage levels off. Once the voltage levels off the loss is extremely small.

It is entirely possible to set the values of the components to see very near theoretical results in the simulation.

BINGO!! Now that's the perfect answer i was looking for!! Thanks a million man!!

 

[MrAl]

BINGO!! Now that's the perfect answer i was looking for!! Thanks a million man!!

Hi,

Hey you're welcome. Now try a few changes and see if you like the results better

 

[oodes]

I ran it again, and the rise and fall time of 1nS made all the difference as i was leaving this field blank . When i used these parameters i got a voltage slightly lower than 9V, and factoring in all the losses u mentioned with the components it made sense. Thanks again ,

Hi,

Hey you're welcome. Now try a few changes and see if you like the results better