Expedite Transformer Calculations for Flybacks

Jan 1, 2008 12:00 PM
By Kirby Creel, Senior Design Engineer, Data-tronics, Romoland, Calif.

Design Example

With the background provided, we can now tackle the problem of charging a capacitor to a given voltage in a stated amount of time. Designs begin with a list of known values.

Let's illustrate the process by considering the following example of a typical design problem, for which the values are listed in the table. The application is for charging a defibrillator capacitor. (Caution: The charged capacitor used in this example can provide a lethal shock.)

The first step is to determine the number of Joules required to charge the capacitor:

where U is energy in Joules, V_CAP is the capacitor voltage and C is the capacitance in Farads.

The next step is to calculate the number of charging pulses (N) in the stated time:

In the third step, calculate the energy required per charging pulse (U_P):

The fourth step is to make an estimate of the circuit efficiency and include the factor in calculating the energy that must be supplied. All calculations up to this point are based on the assumption that there were no losses in the switching transistor, diode and transformer (winding or core). Switching transistor, diode, and transformer losses are shown to a first order. Second-order losses are those for winding capacitance and leakage inductance.

Fig. 3 provides an estimation of a typical loss factor expressed as an efficiency figure. The losses must be included in the power supplied from the dc source, as shown in the next calculation. Note that Fig. 3 is an estimate and results will vary. Use the result from Eq. 4 (in this case, 200 J) to find the efficiency value. A number of variables in circuit design, layout, transformer design, components and others will affect the result. (For very high-voltage designs, even leakage current paths across the surface of the pc board and leakage within the capacitor must be considered.)

In the fifth and final step, solve for the unknowns. There are two unknowns and two equations previously presented that will provide the answer. The unknowns are the transformer primary inductance and the peak current. The 500-µJ figure calculated in the previous equation can be solved with an almost infinite number of solutions. The remainder of the design requirements and the 500-µJ calculation limits the answer to one solution (as shown next).

Eq. 1 is an inductance formula, while Eq. 3 is solved for the inductance. The right-hand side of both equations are equal to each other. The resulting expression, with only one unknown, can be solved for the peak current. Having found the peak current, the value is entered into the original equation and solved for the inductance. For a “sanity check,” both formulas are solved for the inductance. The equal results provide confidence that the calculations were performed correctly.

Calculate the inductance by substituting the value of I_PEAK in both Eq. 1 and Eq. 3 as a check:

Fig. 4 is the completed design with the calculated values included.

A similar application to charging a defibrillator capacitor is charging a photoflash capacitor. Linear Technology's LT3468 IC performs most functions in a small footprint. The IC is designed around the specific application with some other applications mentioned in the data sheet. One limitation of this device is the breakdown voltage of the switching transistor, which at 70 V at 25°C makes it better suited for lower-voltage applications.

The LT3468 data sheet^[2] provides design information, circuits and typical waveforms, and gives a designer good insight into the process, possibly sparking new avenues of application. (Note: The data sheet provides a warning related to working with high voltage. Many of the circuits discussed can provide lethal shocks when working properly.)

For further illustration of the design process, the following low-power example was built and tested. Test data, notes on circuit operation and waveform are shown. The design starts with this list of given values:

• C = 6 µF

• V = 600 Vdc
• Charge time = 10 sec
• Circuit = flyback
• Frequency = 50 kHz
• Maximum duty cycle = 45%
• Maximum on time = 9 µs
• Input = 12 Vdc

Using the technique shown previously, the inductance and peak charging current are calculated here:

• Eq. 4 = 1.08 J
• Eq. 5 = 500,000 pulses
• Eq. 6 = 2.16 µJ/pulse
• Eq. 7 = 4.32 µJ/pulse (efficiency is estimated to be 50%)

Calculate the inductance by substituting the value of I_PEAK in both Eqs. 1 and 3.