Post History
I have the following problem: - Using Wikipedia I find that during CCM we have the relation: - $$\frac{V_{out}}{V_{in}} = -\frac{D}{1-D} $$Solving for $D$ and inserting values gives me $$D...
#3: Post edited
by
Lorenzo Donati
·
2023-08-10T16:02:50Z (over 1 year ago)
Retagged.
#2: Post edited
by
Carl
·
2022-03-13T14:18:21Z (over 2 years ago)
- I have the following problem: -
- >
- Using Wikipedia I find that during CCM we have the relation: -
- $$\frac{V_{out}}{V_{in}} = -\frac{D}{1-D} $$
- Solving for \$D\$ and inserting values gives me
- $$D=\frac{20\text{V}}{20\text{V}-24\text{V}} = -5 $$
- But this doesn't seem to make sense. First of all, the duty cycle is negative... Second the duty cycle is **more** than 100%!
Why do I get this erroneous result?
- I have the following problem: -
- >
- Using Wikipedia I find that during CCM we have the relation: -
- $$\frac{V_{out}}{V_{in}} = -\frac{D}{1-D} $$
- Solving for \$D\$ and inserting values gives me
- $$D=\frac{20\text{V}}{20\text{V}-24\text{V}} = -5 $$
- But this doesn't seem to make sense. First of all, the duty cycle is negative... Second the duty cycle is **more** than 100%!
- Why do I get this erroneous result?
- **EDIT**
- Turns out I used the wrong formula. Using the formulas given by my instructor for the course: -
- 
- yields \$d=0.454545 \$
#1: Initial revision
by
Carl
·
2022-03-13T07:27:48Z (over 2 years ago)
Duty cycle of buck-boost converter
I have the following problem: -
>
Using Wikipedia I find that during CCM we have the relation: -
$$\frac{V_{out}}{V_{in}} = -\frac{D}{1-D} $$
Solving for \$D\$ and inserting values gives me
$$D=\frac{20\text{V}}{20\text{V}-24\text{V}} = -5 $$
But this doesn't seem to make sense. First of all, the duty cycle is negative... Second the duty cycle is **more** than 100%!
Why do I get this erroneous result?