Turns ratio

Nominal turn ratio (the error must not exceed + 0.25%)... 

All such CBCTs have to be identical in turns ratio and magnetizing characteristics to avoid circulating currents among themselves. To order a CBCT the following information will be essential ... 

The turns ratio is typieally 1 1, but if other turns ratios are used, the output voltage should not eause an avalanehe failure of the power switeh. 

Round upwards to the nearest integer all fractional turns in the results. Next multiply this result by the desired turns ratio to determine the number of... 

Sinee the input eontrol IC bias voltage is about 15 V, we need to make the turns ratio about 1 1. So the seeondary turns will be 11 turns. 

The gate drive transformer is a very simple 1 1 turns ratio forward-mode transformer. There are not extraordinary demands being plaeed on the transformer sinee it is a very low power, ae-eoupled (bipolar flux), 300 kHz transformer. 

The seeondary winding s induetanee eontrols how quiekly the eore would empty itself of the stored energy for the diseontinuous-mode of operation. Sinee the input and output voltages are very elose in magnitude, a 1 1 turns ratio would be possible. This would result in a 3 qS off-time within a eomparable PWM system. Tet us use a 1 1 turns ratio it eould be bifilar wound to promote the tightest eoupling. 

Figure B-10. Figure B-10 shows the presence of a transformer. For the buck converter, the designer can assume that the turns ratio is 1 1.

The MC34262 has a high-side driver clamp of 16VDC, so in order to keep the high-side driver dissipation to a minimum, the peak voltage of the rectified auxiliary voltage should be around 16 V. Determine the turns ratio needed for this from... 

Capacitor RMS current depends only on the power, turns ratio, and inductance... 

This is also true because any trace inductance here gets multiplied by the square of the turns ratio, and reflects into the primary side, as discussed previously. This greatly increases the dissipation in the primary-side RCD/zener clamp and severely degrades the converter efficiency. We have to really struggle to minimize secondary-side inductances, especially for low output voltage rails, that is, those with higher turns ratios. 

So, is that not true for a Flyback Actually it is, because IL = /0R/(1 - D). But what has happened here is that by changing the turns ratio, we changed 70R, too, because 70R = I0/n. So by decreasing the turns ratio n, we have actually increased 70R, and therefore also 7L. 

But, if we keep the turns ratio fixed, the Flyback certainly follows the known behavior of the Buck-Boost and Boost with respect to changes in D. 

What happens if Vsw < VD In fact that is the situation in most commercial Flybacks. But note that to do a proper comparison, you have to reflect the diode drop to the primary side. And for that we have to multiply the diode drop by the turns ratio (see the equivalent Buck-Boost models of a Flyback section in my book, Switching Power Supply Design Optimization). So, for example, if the turns ratio is 20 and the diode drop is 0.6V, the effective VD we need to compare with Vsw for our time-sharing analysis is 0.6 x 20 = 12V. And that is usually greater than the (average) drop across the switch. Therefore, we tend to say that in a Flyback, decreasing D (increasing input) will worsen the total conduction loss and decrease the efficiency. But of course that never happens, because as we increase the... 

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