LM3402_14 Datasheet, PDF (26/36 Pages) Texas Instruments – 0.5A Constant Current Buck Regulator for Driving High Power LEDs

LM3402, LM3402HV
SNVS450D – SEPTEMBER 2006 – REVISED FEBRUARY 2010
www.ti.com
In expectation that more capacitance will be needed to prevent power supply interaction a 2.2 µF ceramic
capacitor rated to 100V with X7R dielectric in an 1812 case size will be used. From the Design Considerations
section, input rms current is:
I_IN-RMS = 0.35 x Sqrt(0.82 x 0.18) = 134 mA
(66)
Ripple current ratings for 1812 size ceramic capacitors are typically higher than 2A, more than enough for this
design.
RECIRCULATING DIODE
The input voltage of 60V ±5% requires Schottky diodes with a reverse voltage rating greater than 60V. Some
manufacturers provide Schottky diodes with ratings of 70, 80 or 90V; however the next highest standard voltage
rating is 100V. Selecting a 100V rated diode provides a large safety margin for the ringing of the switch node and
also makes cross-referencing of diodes from different vendors easier.
The next parameters to be determined are the forward current rating and case size. In this example the high duty
cycle (D = 49.2 / 60 = 82%) places less thermals stress on D1 and more on the internal power MOSFET of the
LM3402. The estimated average diode current is:
I_D = 0.361 x 0.18 = 65 mA
(67)
A Schottky with a forward current rating of 0.5A would be adequate, however at 100V the majority of diodes have
a minimum forward current rating of 1A. To determine the proper case size, the dissipation and temperature rise
in D1 can be calculated as shown in the Design Considerations section. V_D for a small case size such as SOD-
123F in a 100V, 1A Schottky diode at 350 mA is approximately 0.65V and the θ_JA is 88°C/W. Power dissipation
and temperature rise can be calculated as:
P_D = 0.065 x 0.65 = 42 mW T_RISE = 0.042 x 88 = 4°C
(68)
C_B AND C_F
The bootstrap capacitor C_B should always be a 10 nF ceramic capacitor with X7R dielectric. A 25V rating is
appropriate for all application circuits. The linear regulator filter capacitor C_F should always be a 100 nF ceramic
capacitor, also with X7R dielectric and a 25V rating.
EFFICIENCY
To estimate the electrical efficiency of this example the power dissipation in each current carrying element can
be calculated and summed. Electrical efficiency, η, should not be confused with the optical efficacy of the circuit,
which depends upon the LEDs themselves.
Total output power, P_O, is calculated as:
P_O = I_F x V_O = 0.361 x 49.2 = 17.76W
Conduction loss, P_C, in the internal MOSFET:
P_C = (I_F2 x R_DSON) x D = (0.361² x 1.5) x 0.82 = 160 mW
Gate charging and VCC loss, P_G, in the gate drive and linear regulator:
P_G = (I_IN-OP + f_SW x Q_G) x V_IN P_G = (600 x 10^-6 + 3 x 10⁵ x 3 x 10^-9) x 60 = 90 mW
Switching loss, P_S, in the internal MOSFET:
P_S = 0.5 x V_IN x I_F x (t_R + t_F) x f_SW P_S = 0.5 x 60 x 0.361 x 40 x 10^-9 x 3 x 10⁵ = 130 mW
AC rms current loss, P_CIN, in the input capacitor:
P_CIN = I_IN(rms)2 x ESR = (0.134)² x 0.006 = 0.1 mW (negligible)
DCR loss, P_L, in the inductor
P_L = I_F2 x DCR = 0.35² x 1.1 = 135 mW
Recirculating diode loss, P_D = 42 mW
Current Sense Resistor Loss, P_SNS = 69 mW
Electrical efficiency, η = P_O / (P_O + Sum of all loss terms) = 17.76 / (17.76 + 0.62) = 96%
Temperature Rise in the LM3402HV IC is calculated as:
T_LM3402 = (P_C + P_G + P_S) x θ_JA = (0.16 + 0.084 + 0.13) x 200 = 74.8°C
(69)
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