Inductors/Transformers for Switching Power
3.2 Selecting the Inductor Value
cores are so small that EMI is seldom a problem, provided the core is a toroid or similar design with a closed magnetic path.
3.Ί.14 Self-Resonant Frequency
All inductors have some distributed capacitance that combines with the in
ductance to form a resonant circuit. The frequency of this self-resonance should be between 5 and 10 times the switching frequency (but not at an exact multiple of the switching frequency!). As the inductance value is set by circuit requirements, the SFR is determined by the distributed capacitance (a higher capacitance produces a lower SRF).
When the SRF is low, the normal linear ramp of the inductor current (see Fig.
3-3) is preceded by a sudden jump of current when the switching transistor turns on.
This results in so-called switching losses that lower the regulator's overall effi
ciency. As a result, distributed capacitance should be kept at a minimum so that the SRF will be high and will not seriously affect the inductor current. Distributed ca
pacitance can be lowered when the toroid is wound, either by overlapping the ends of the winding somewhat or by leaving a gap between the winding ends (rather than ending the winding at one full layer).
where
W ~ vLtON/L
The total power that can be put in (or taken out of) the inductor is the energy (EL) per cycle times the number of cycles per second:
P = F f
rL ' V o ' where fQ is the switching frequency.
This equation assumes that tON and f0 are interdependent. For PFM regulators, the clock is typically a 50% duty-cycle square wave, so:
t0 N=l/(2f0).
The inductor power (as a function of input voltage, frequency, duty cycle = 50%, and inductance) is:
or in terms of tON:
PL = VL2/(8f0L)
PL = VL W (4 L) ·
In step-up and inverting converters, the charging voltage for the inductor (VL) is usually the same as the input voltage (VIN) if switch losses are ignored.
In step-down converters, VL = VIN - VOUT (again ignoring losses) because the inductor is connected between the input and output voltage when charging.
3.2.2 Inductor Selection for Step-Up Regulators
The first step in selecting an inductor for a step-up regulator is to choose
νουτ' viN(min)' viN(max)' a n d W · Remember that in step-up design, VIN must be less than Vo u r If there is any possibility where VIN could equal (or exceed) VOUT, the inverting regulator of Section 3.2.4 should be considered.
^iN(min)and ^iN(max) c o v e r m e input-voltage range (such as voltages at the be
ginning and end of normal battery life). The output power, or POUT, is VOUT x IOUT. However, the converter must also make up for losses in the inductor, switch transis
tor, and diode. These losses typically add 10 to 25% to the required power.
In step-up design, power is supplied both through the inductor and directly from the input voltage. This is because one end of the inductor remains connected to VIN (see Fig. 1-5), as the inductor supplies current to the output. Therefore,
"OUT = "L + ( MN x *ουτ'·
The power PL needed from the inductor is then:
* L = (^ουτ ~ ^IN + * D) lour
VD x IOUT accounts for losses in the diode. Schottky diodes (such as the 1N5817) minimize this loss (which can be significant in low-voltage circuits). Table 3-3 lists some typical diodes for switching regulators. Specific diodes for specific
Table 3-3. Some typical diodes for switching regulators (Maxim Seminar Applications Book, 1989, p. 94)
Rectifiers for DC-DC Converter Circuits
Part Number 1N914 1N4148 1N4935 1N5817 1N5818 1N5620 1N5821 1N5822 1N5823 1N5824 1N5825 UDS62G UDS640 UES1001 UES1002 U ES 1003
Uvo(amps) 0.05 0.05 1 1 1 3 3 3 5 5 5 6 6 1 1 1
VF (volts) 1.0 1.0 1.2 0.45 0.55 0.475
0.5 0.525
0.36 0.37 0.38 0.48 0.48 0.895 0.895 0.895
Prv (volts) 75 75 200
20 30 20 30 40 20 30 40 20 40 100 50 150
Type Silicon Silicon Silicon Schottky Schottky Schottky Schottky Schottky Schottky Schottky Schottky Schottky Schottky Silicon Silicon Silicon
Pkg Glass Glass Plastic Plastic Plastic Plastic Plastic Plastic Metal Metal Metal TO-220 TO-220 Bead Bead Bead
regulator ICs are given in Chapter 5. In high-voltage circuits (where V0UT = 10 V and up) or if efficiency is not critical, signal diodes (such as 1N4148) can be used (provided that the diode reverse-voltage rating is not exceeded and that the recovery time is not too slow).
In ideal systems (minimum switch losses), either of the following equations can be used to find inductor power:
PL = V/(8 fOL) or in terms of tON:
PL = VtON/(4 L)·
By substituting this last equation for PL, the value of the inductor can be found by:
L = VIN /[8f0I0UT (V0UT + VD - VIN)J,
where fQ is the switching frequency and is a square wave with a 50% duty cycle.
As discussed in Chapter 1, the clock or switching frequency can be adjusted on some ICs (such as the MAX630 and MAX4193). In other cases, the regulator IC switching frequency is fixed and cannot be changed (such as on the MAX631/32/33, where the rate is preset at 50 kHz).
When using the equation to find L for a step-up design, always use the mini
mum expected value for VIN. This will ensure that there is adequate output power under all input conditions.
Besides inductance value, the selected inductor must also be rated for the peak current involved, or IPK. This is found by:
W = νΐΝ*ΌΝ^ = VIN/(2fQL),
where tON is the inductor charging time (for one clock cycle), which is equivalent to one-half of one f0 clock period. For example, for an fQ of 50 kHz, one clock period is 1/50,000 = 0.00002, or 20 μ8, and one-half clock period is 10 μ8.
The equation for IpK assumes that the switch-on resistance, or RON, is insignif
icant when compared to the inductor resistance and that the inductor voltage VL is approximately equal to VIN (they are never exactly the same). When using the equa
tion to find IpK for a step-up design, always use the maximum expected value for VIN. This will ensure that the maximum current (under all operating conditions) will be considered.
Once the IpK is found, compare the value with the inductor current rating and the switch rating. Of course, IpK must be less than both ratings.
If IpK exceeds the peak-current rating of the internal transistor switch (which