This paper discusses how to select transformers for isobuck converters and why we use the simplified formulas in the data books of max17682 and max17681. It also introduces the basic principle of isolated buck DC-DC converter.

The selection of transformer seems to be the most critical step in the design of isolated buck DC-DC converter. This paper will discuss the working principle of isolated buck DC-DC converter, which parameters should be focused on when selecting transformer, how these parameters affect the selection of transformer and how the selected transformer affects the circuit parameters.

**How does isolated buck DC-DC converter work?**

Just like the buck converter, the figure below shows us the ISO buck topology. Obviously, we can get an isolated buck converter by using transformer instead of inductor in buck circuit. The secondary side of the transformer has an independent ground.

Figure1Iso-BuckTopology

During the on time (ton), the high side switch (QHS) is on and the low side switch (QLS) is off. The transformer excitation inductance lpri is charged. The arrow in the figure shows the direction of the current. The primary current increases linearly. The slope of the current versus time depends on (VIN vpri) and lpri. During this time interval, the diode D1 on the secondary side is reversely biased, and the load current flows from the cout to the load.

Figure2OnPeriodEquivalentCircuit

During the off time (toff), QHS is off and QLS is on. The primary inductor discharges. The primary side current flows from QLS to the primary side ground, D1 is positively biased, and the secondary side current flows from the secondary side coil to the output capacitor and load. During this time, the output capacitor is charged（ Turning QHS off and QLS on does not change the direction of the current, it changes the slope of the current. The positive current decreases to 0A, then the negative current increases.)

Figure3OffPeriodEquivalentCircuit

## Which parameters will affect the selection of transformer?

When designing power supply, some parameters and specifications should be defined. These parameters will determine the selection of peripheral devices (such as input capacitance, sampling resistance, etc.), especially the transformer.

Input voltage range

Output voltage

Max duty cycle

Switching frequency

Output voltage ripple

Output current

Output power

We usually assign the maximum duty cycle (d) from 0.4 to 0.6. Minimum input voltage (VIN)_ Min) and the maximum duty cycle will determine the primary side output voltage (vpri). The primary output voltage and the output voltage (VOUT) determine the transformer turns ratio.

Output current (IOUT) and output power (pout) may be the most critical parameters affecting the selection of transformer. The output current will determine at least how thick copper wire should be used. The output power will determine which transformer frame should be used. The permeability of the skeleton indicates how much energy it can store and thus how much energy it can output. Usually, we use the DC load current multiplied by a factor to specify the ripple current of the inductor (transformer). The duty cycle and switching frequency will tell us the ton time. With the help of VIN, vpri and ripple current, we can determine the primary side inductance. The coefficients should not be too large or too small. Larger coefficient will lead to larger ripple current. Large ripple current may exceed the H-bridge current limit, which may damage MOSFET, and large ripple current will cause large ripple voltage on output capacitor due to ESR and ESL of output capacitor. On the contrary, when we want a minimum ripple current, our calculation will get an inductor (transformer) with a large inductance value, which will be a wire breaker with a large skeleton and a large number of internal windings, and the large inductance will limit the loop bandwidth and reduce the dynamic response index.

**Select transformer**

Obviously, energy is only transmitted to the secondary side during the off time. Therefore, the turn ratio can be determined by the following formula:

Where VD is the forward bias voltage of the secondary diode. For vpri, we usually allocate the maximum duty cycle in the range of 0.4 to 0.6. Vpri can be calculated by the following formula:

Where D is the maximum duty cycle, Vin_ Min is the minimum input voltage. From the above discussion, we can calculate the turn ratio. As a non isolated buck converter, the ripple currents on both sides of the inductor are equal. Easily, we can calculate the required inductance value according to the following inductance characteristic formula at ton time:

Where f is the switching frequency, Δ I is the ripple current. As we said, the ripple current is the DC load current multiplied by a factor:

K is the coefficient. But in isolated buck converter topology, it’s a transformer, not an inductor. What should we do? We know that the current on both sides of the transformer is inversely proportional to the number of turns

Among them, IPRI_ Toff is the equivalent current converted from the secondary current to the primary current in toff time. It should be noted that the secondary coil only outputs current in toff time. Maybe we can add the two currents together as the equivalent inductor current.

Where ileq is the equivalent inductive current. If the transformer has three more windings,

Is that right? Let’s take a look at the simulation results based on max17682. The following screenshot is a typical circuit of max17682 drawn in simple. The current probe has been placed on both sides of the transformer, called IPRI and isec1 respectively.

Figure4MAX17682TypicalCircuiTInSIMPLIS

The following screen capture shows the transient simulation results of the two probes. I added the sum of these two current waveforms according to the above equation.

Figure5MAX17682Typicalcircuitsimulatecurrentwaveform

Obviously, the result of current addition (cyan) is triangular wave, just like the inductor in non isolated buck converter. Therefore, the primary side of the transformer can be easily calculated Δ I：

Usually, we divide the load current ripple into 0.2 times of the DC output current. Therefore, K can be assigned as the turn ratio multiplied by 0.2. At the same time, the primary peak current should be less than the switch limit. The peak current ipk is as follows:

Then the primary inductance of the required transformer can be easily calculated

By using these parameters, such as turn ratio, primary inductance, output power, output current and isolation voltage, we can decide which inductor to use.

## Why can Maxim’s simplified formula be used?

When using the max17682 data book for the first time, you may be suspicious. The manual shows us a formula for the following screen capture. It seems even wrong in the unit of measure. What do we think of this equation? How do we understand it?

Figure6MAX17682datasheetshot

According to the above discussion, equation (10) can be rewritten as the following equation in toff time.

We assume that D is 0.6 if and only if Δ I is 0.4A polynomial (1-D) and Δ If I can be reduced, then equation (11) is the same as equation in the data table. Obviously, the formula in the data book has selected the primary ripple current. If we assign 0.6 to D, the primary ripple current is 0.4A. In quantity, the toff duty cycle is equal to the primary ripple current.