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LT3507

3507f

The LDO pass transistor power is:

PNPN = IOUTLDO(VC – VOUTLDO)

where VC is the collector voltage on the NPN pass tran-

sistor.

Example: An LT3507 design requirements are:

VIN = 8V, f= 500kHz

V1 = 2.5V at I1 = 1.6A

V2 = 3.3V at I2 = 0.8A (used for boost, bias and V4)

V3 = 1.2V at I3 = 1A

V4 = 3V at I4 = 0.2A (from 3.3V output)

TA = 50°C, TJMAX = 125°C

θJA = 34°C/W

Schottky VF = 0.45V and Inductor ESR = 0.05Ω

PDISS(MAX) =

125

°C– 50°C

°C/W

= 2.2W

PSWDC1=

0.18

1.6A

()22.5V

= 0.14W

PSWAC1= 17ns 1.6A

() 8V

() 500k

()=0.11W

PBST1=

2.5V 3.3V

() 1.6A

+ 0.02A

= 0.06W

Similarly, PSWDC2 = 0.09W, PSWAC2 = 0.07W, PBST2 =

0.06W, PSWDC3 = 0.03W, PSWAC3 = 0.07W and PBST2 =

0.03W. Remember, the total current from channel 2 is I2

+ I4 since the LDO pass transistor draws from V2. Ignore

bias and boost currents.

PQ = 8V 3.5mA

()+3.3V 7.5mA

()=0.05W

PLDO = 8V

0.2A

100

= 0.02W

APPLICATIONS INFORMATION

The total dissipation on the LT3507 is the sum of all these

and is equal to 0.73W. Note that this is less than half of

PDISS(MAX). Next, the power dissipation of the external

components are:

PDIODE1 =

0.45V 8V – 2.5V – 0.45

()1.6A

= 0.46W

PIND1 = 1.6A

()

2 0.05Ω = 0.13W

Similarly, PDIODE2 = 0.24W, PIND2 = 0.05W, PDIODE3 =

0.36W and PIND3 = 0.05W. And ﬁnally:

PNPN = 0.2A(3.3V – 3V) = 0.06W

Thus the total power dissipated by the LT3507 and external

components is 2.08W. The thermal analysis will use these

power dissipations to calculate the internal component

temperatures. Make sure that none of the components

exceed their rated temperature limits.

RELATED LINEAR TECHNOLOGY PUBLICATIONS

Application Notes 19, 35, 44, 76 and 88 contain more

detailed descriptions and design information for buck

regulators and other switching regulators. The LT1375

data sheet has a more extensive discussion of output

ripple, loop compensation, and stability testing. Design

Note 318 shows how to generate a dual polarity output

supply using a buck regulator.

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