A question regarding boost converter

Thread Starter

xw0927

Joined Dec 19, 2010
114
Hello everyone,

I have a question regarding boost converter.

From all the things I studied, no matter website or books.

For CCM, output voltage of boost converter depends on duty cycle only.

from equation in this website
http://www.daycounter.com/LabBook/BoostConverter/Boost-Converter-Equations.phtml

Vout/Vin = 1/(1-D)
if D = 0.56; Vout/Vin = 2.27;
let Vin = 12, Vout = 27.27V
however, I found that this is not true.
As inside my project, the duty-cycle is only 56%, it can also boost up my input voltage from 12V to 70V++

So, I would like to ask, is there any other equations could explain about this.
 

Adjuster

Joined Dec 26, 2010
2,148
Your circuit has an effective load resistance of 33kΩ, which is very large. The output voltage may therefore rise sufficiently for the coil current to fall to zero before the end of each cycle.

Your circuit may thus be working in discontinuous conduction mode, which would invalidate the sort of calculations shown in the page you linked to.

In discontinuous mode, the input power is essentially independent of the output conditions, and the output voltage rises to whatever value will achieve a power balance in the output load, subject to losses and voltage limitations of components. For this reason, operating a boost converter into an open-circuit can be dangerous, unless it is protected by some sort of voltage limitation system.
 
Last edited:

SgtWookie

Joined Jul 17, 2007
22,230
That brings up a closely related situation you are having with that circuit.
You had mentioned before that the output voltage would be around ~46v or so initially with your load, some kind of plastic goo, but then after about a half-hour, the output voltage would rise to ~50v or so.

It may be that your load falls very low in resistance, and the 555 won't go low enough in ON-time, so the inductor puts out too much current and your output voltage rises, even though the CTRL input is held near ground.

Adding a 2nd load in parallel with your "plastic goo" load is a work-around that should keep your output voltage relatively steady. Since I have not built the circuit, you will have to experiment with various resistors to find the minimum resistance necessary to maintain the output voltage. I suggest starting with 47k Ohms.
 

Thread Starter

xw0927

Joined Dec 19, 2010
114
Your circuit has an effective load resistance of 33kΩ, which is very large. The output voltage may therefore rise sufficiently for the coil current to fall to zero before the end of each cycle.

Your circuit may thus be working in discontinuous conduction mode, which would invalidate the sort of calculations shown in the page you linked to.

In discontinuous mode, the input power is essentially independent of the output conditions, and the output voltage rises to whatever value will achieve a power balance in the output load, subject to losses and voltage limitations of components. For this reason, operating a boost converter into an open-circuit can be dangerous, unless it is protected by some sort of voltage limitation system.
thanks for you explainations, but I have a few points are not clear.

I don't understand "The output voltage may therefore rise sufficiently for the coil current to fall to zero before the end of each cycle." How would it happen?

I was thinking about DCM mode also, but I couldn't find sufficient information regarding DCM ( in aspect of theory and equations)

However, I refer to
http://services.eng.uts.edu.au/~venkat/pe_html/ch07s3/ch07s3p1.htm

the result i got is also too far from calculation.
 

Thread Starter

xw0927

Joined Dec 19, 2010
114
That brings up a closely related situation you are having with that circuit.
You had mentioned before that the output voltage would be around ~46v or so initially with your load, some kind of plastic goo, but then after about a half-hour, the output voltage would rise to ~50v or so.

It may be that your load falls very low in resistance, and the 555 won't go low enough in ON-time, so the inductor puts out too much current and your output voltage rises, even though the CTRL input is held near ground.

Adding a 2nd load in parallel with your "plastic goo" load is a work-around that should keep your output voltage relatively steady. Since I have not built the circuit, you will have to experiment with various resistors to find the minimum resistance necessary to maintain the output voltage. I suggest starting with 47k Ohms.
Yup, you're right, the resistance drops.
But, the problem i face now is, the theoritical equation is too far different from what i did here...As i did a lot of modification according to your advice, and I got the desired output at the end. But, I don't know how to justify it :( ( a huge problem to a student if the result couldn't be explained) Sigh
 

Adjuster

Joined Dec 26, 2010
2,148
It seems likely that you have discontinuous conduction. If you are able to monitor the waveform, at least on a simulation, you may be able to see this directly. Otherwise, consideration of the on and off periods and the voltages across the coil should quickly put the matter beyond doubt.

The description in the article you linked to gives conditions for different conduction modes - have you read it?. Here is another article: boost converters are discussed from slide 20. http://ecee.colorado.edu/copec/book/slides/Ch5slide.pdf

When the converter has a very high load resistance, the voltage ratio calculated from a formula involving the load resistance but ignoring losses may not be accurate. The reason for this is that with only a small dissipation in the external load, dissipation in losses may be a relatively large fraction of the total power.
 
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