# Calculate duty cycle of flyback converter MY ATTEMPT

I found a formula for the DC transfer function for the flyback converter:

$$\frac{V_{out}}{V_{in}}=\frac{N_s}{N_p}\cdot \frac{D}{1-D}$$where $$\ \frac{N_s}{N_p} \$$ must be the ratio between the number of turns of the two coil-inductors, and $$\ D\$$ is the duty cycle. And since we know $$\V_{out}\$$ and $$\V_{in}\$$, we just need to find $$\\frac{N_s}{N_p} \$$, and then we can solve the equation with respect to $$\D\$$.

However, I can't seem to find any way to find the $$\\frac{N_s}{N_p}\$$-ratio, when I only know the inductances of the coils. How can I arrive at the value for $$\\frac{N_s}{N_p}\$$, so that I can solve the original equation?

I hope someone can help me with this.

## Edit 21/02-21

Here I am, one year later going through this problem again, to see if I can solve it. However, even with Andy aka's formula $$\\frac{N_s}{N_p}=\sqrt{\frac{L_s}{L_p}} \$$ I am unable to solve it.

$$\frac{15 \text{V}}{110\text{V}} = \sqrt{\frac{16.2 \cdot 10^{-6}\text{H}}{1.28 \cdot 10^{-3}\text{H}}} \cdot \frac{D}{1-D} \Rightarrow D=45.20\%$$ But the correct answer is $$\D=46.15 \% \$$ (according to my instructor). I even tried to take into account any mutual inductance assuming zero flux leakage and 100% magnetic coupling $$M=\sqrt{L_sL_p}= \sqrt{16.2 \cdot 10^{-6}\text{H} \cdot 1.28 \cdot 10^{-3}\text{H}}=0.144 \text{mH}$$ But that didn't get my anywhere.

## Final edit

After approximately one year, I finally solved the problem. Since $$\V_{in} \$$ is in RMS, you have to multiply with $$\\sqrt{2} \$$ to get the peak voltage.

$$\frac{15 \text{V}}{110\text{V} \cdot \sqrt{2}} = \sqrt{\frac{16.2 \cdot 10^{-6}\text{H}}{1.28 \cdot 10^{-3}\text{H}}} \cdot \frac{D}{1-D} \Rightarrow D=46.15\%$$

However, I can't seem to find any way to find the $$\\frac{N2}{N1}\$$-ratio, when I only know the inductances of the coils.

This is likely to be what you are looking for: -

$$\\dfrac{N_p}{N_s} = \sqrt{\dfrac{L_p}{L_s}}\$$

The assumption for this is that both coils share the same magnetic core and are 100% coupled.

• Splendid! I get the following: $\frac{N_2}{N_1}=\sqrt{\frac{1.28 \text{mH}}{16.2 \mu \text{ H}}}=8.889$. Plugging this value into the original equation we get: $\frac{110 \text{V}}{15 \text{V}}=8.889 \cdot \frac{D}{1-D}%$. Solving for $D$ I arrive at this duty cycle: $D=45.21%$. Is my procedure correct?
– Carl
May 22 '20 at 12:27
• Your procedure assumes that the circuit is operating in CCM - they are the formulas you have used but, there's nothing in the question about the load nor the switching arrangements that suggest it is operating in CCM. The question doesn't mention that the rectified waveform is smoothed by a capacitor either. So, unless you know otherwise and didn't post that information, the assumptions may be wrong. The turns ratio is correct and 8.889 is the right number. May 22 '20 at 12:30
• I posted the entire question, there is no missing information. So I hope I have used the expected formulas.
– Carl
May 22 '20 at 12:36
• Well, you have to state your assumptions and the assumption is that the circuit is operating in CCM. If you wanted to assume that the circuit worked in DCM you'd have a different formula so maybe, if this is homework, you should provide both answers to cover all likelihoods. May 22 '20 at 12:39
• I guess the words "ideal flyback converter" could be construed to mean a type with synchronous switching and they will always operate in CCM irrespective of load. May 22 '20 at 12:59

Going back to basics, if D = 50% then D/(1-D) would be 1 and

Vout/Vin = 1/8.889 implying that Vout = 110/8.889 or 12.375V which is lower than 15V clearly D/(1-D) has to be 15/12.375 or 1.212 which implies that D is > 0.5

D is 0.548