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DRV590GQCR Datasheet(PDF) 10 Page - Texas Instruments

Part # DRV590GQCR
Description  1.2-A HIGH-EFFICIENCY PWM POWER DRIVER
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Manufacturer  TI [Texas Instruments]
Direct Link  http://www.ti.com
Logo TI - Texas Instruments

DRV590GQCR Datasheet(HTML) 10 Page - Texas Instruments

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DRV590
SLOS365A – AUGUST 2001 – REVISED AUGUST 2002
10
www.ti.com
APPLICATION INFORMATION
LC filter in the frequency domain (continued)
If L = 10
µH and C = 10 µF, the resonant frequency is 9.2 kHz, which corresponds to –69 dB of attenuation at
the 500-kHz switching frequency. For VDD = 5 V, the amount of ripple voltage at the TEC element will be
approximately 1.7 mV.
The average TEC element has a resistance of 1.5
Ω, so the ripple current through the TEC is approximately
1.13 mA. At the 1-A maximum output current of the DRV590, this 1.13 mA corresponds to 0.113% ripple current,
causing less than 0.0001% reduction of the maximum temperature differential of the TEC element (see
equation 1).
LC filter in the time domain
The ripple current of an inductor can be calculated using equation 4.
DI
L +
V
DD *
V
TEC
DTs
L
D = duty cycle (0.5 worst case)
Ts = 1/fs = 1/500 kHz
For VDD = 5 V, VTEC = 2.5 V, and L = 10 µH, the inductor ripple current is 250 mA. To calculate how much of
that ripple current will flow through the TEC element, however, the properties of the filter capacitor must be
considered.
For relatively small capacitors (less than 10
µF) with very low equivalent series resistance (ESR, less than
10 m
Ω), such as ceramic capacitors, equation 5 may be used to estimate the ripple voltage on the capacitor
due to the change in charge.
DV
C +
p2
2
(1–D)
fo
fs
2
V
TEC
fo +
1
2
p L
3C
D = duty cycle
fs = 500 kHz
For L = 10
µH and C = 10 µF, the cutoff frequency fo = 9.2 kHz. For a worst case duty cycle of 0.5 and VTEC
= 2.5, the ripple voltage on the capacitors is 2 mV. The ripple current may be simply calculated by dividing the
ripple voltage by the TEC resistance of 1.5
Ω, resulting in a ripple current through the TEC element of 1.33 mA.
Note that this is similar to the value calculated using the frequency domain approach.
For larger capacitors (greater than 10
µF) with relatively high ESR (greater than 100 mΩ), such as electrolytic
capacitors, the ESR drop dominates over the charging-discharging of the capacitor. Equation 6 can be used
to estimate the ripple voltage.
DV
C + D
I
L
R
ESR
L = inductor ripple current
RESR = filter capacitor ESR
For a 100-
µF electrolytic capacitor, an ESR of 0.1 Ω is common. If the 10-µH inductor is used, delivering 250 mA
of ripple current to the capacitor (as calculated above), then the ripple voltage is 25 mV. This is over ten times
that of the 10-
µF ceramic capacitor, as ceramic capacitors typically have negligible ESR.
(4)
(5)
(6)


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