A simplified boost converter circuit can be seen in Figure 1. the input voltage of this converter is always smaller than the output voltage. The boost converter works in the following way; the current flows through inductor L1 and the S1 when S1 is closed, charging L1 but no power will deliver to the load. When S1 is open, the voltage across the load equals the dc input voltage and the charge stored in the L1 is delivering current to the load.
Figure 1. Simplified Boost Circuit
You must define or calculate the following parameters when selecting an inductor for a Boost Converter; (the values at right side of the equal signs are given for calculation purpose only). This design example focuses on the “discontinuous mode”.
Since the operating frequency is 500 KHz,
Find the output power, P out
P out = (V out + V diode drop) (I output)
P out = (36 + 0.7) (1.5) = 55 W
Find the maximum current input, I input max
Find the FET volt drop, V fet volt drop
V fet volt drop = I in (max) (R on (FET))
V fet volt drop = (5.48) (0.05) = 0.274 V
Find the maximum duty ratio, D max
Find the minimum duty ratio, D min
Find the minimum load resistance, R min (for maximum load condition)
Find the maximum inductance, L max
L max = 0.74 uHy
Find the peak current, I pk
Assuming the efficiency is 93%, I pk = 15.7 Amps.
Find the RMS current, I rms
From this point on you can finish the design or use TT Electronics (TTE) Inductor Datasheet for choosing an optimum inductance and size required. Lower/higher inductance can be chosen and the current ratings will be close to the calculation but the performance should be verified using the test bench.
For this example, a good suggestion is TTE p/n: HA72L-06241R5LFTR for the automotive grade, -55 ºC to +155 ºC (AEC-Q200 Certified) and has following electrical parameters:
Table 1. Electrical specs (Transferred from the datasheet).
|TTE p/n||L @ 0 Amp +/- 20%||DCR typical||DCR max||I rms Temp rise to 40 ºC||I dc (sat) 30% Roll off||Typical picture|
Figure 2. DC bias and Temperature rise curve.
Figure 2. shows the inductance roll off is about 100 = 30% at 15.7 Amps. Since the calculated RMS current is 7.52 A, this is well below 9.0 A shown in table 1 above. So, the expected temperature rise* would be far less than 40 ºC. All off the performances must be tested on the test bench as usual.
* In this example, Irms was given by the inductor manufacturer that can be employed as a reference point (40 ºC rise) to establish the temperature rise. The temperature rise would be above, below or equal to the reference point. For an accurate temperature rise calculation, we must know the following parameters, but not limited to:
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