LM2587: FLYBACK, VIN=11V, VOUT_1=12V, VOUT_2=5V

Hi Zack,

first of all, thanks for the fast and helpful answer. But there are still two more questions that came up when scrolling through the datasheet of the LM2587-5.0:

“Figure 29. Triple-Output Flyback Regulator” shows a Transformer with ratios of N1 = 1:0.8, N2 = 1:0.8 and N3 = 1:0.35. The corresponding Voltages are +12V, -12V and +5V. My first question is: Why is the Transformer in the example (T4) rated for Voltages of +18V to +36V? Why is it not recommended to use, for instance a +14.7V Battery?

My second question relates directly to your responds: If the Flyback is working in continuous mode, the correct equation should be

V2/V1 = (N2/N1) * (D/(1-D))

This means, that given the above ratios, an input voltage of for example +11.1V and a duty cycle of 26.5%, the output voltages should be +5.0V, +11.4V and -11.4V, which is an acceptable result for my purposes. So my question is: Why do I have to use equation

V2/V1 = 1/(1-D)

for discontinuous mode? Doesn’t the flyback continuously switch between both modes, depending on the conditions on the secondary side?

Many Thanks in advance
Franz Braun
:-)

Hi Franz,

Sorry for the late response. Figure 27 shows an example designed for the some applications using 24V(18~36V ) voltage bus. You are right, the equation should be V2/V1=N2/N1*D(1-D). The reason that they choose turns ration 1: 0.8 is at minimum input voltage, the duty cycle is  around 47%. Flyback has sub-harmonic oscillation problem and needs slope compensation when duty cycle is higher than 50%. So when using 11V battery, there could be stability problem.