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LT3085

3085fa

Without series resistor RS, power dissipation in the LT3085

equals:

TOTAL = 5V – 3.3V

()•

0.5A

⎛

⎝⎜

⎞

⎠⎟

+ 5V – 3.3V

()•0.5A

= 0.86W

If the voltage differential (VDIFF) across the NPN pass

transistor is chosen as 0.5V, then RS equals:

S =

5V – 3.3V

− 0.5V

0.5A

= 2.4Ω

Power dissipation in the LT3085 now equals:

TOTAL = 5V – 3.3V

()•

0.5A

⎛

⎝⎜

⎞

⎠⎟

+ 0.5V

()•0.5A=0.26W

The LT3085’s power dissipation is now only 30% compared

to no series resistor. RS dissipates 0.6W of power. Choose

appropriate wattage resistors to handle and dissipate the

power properly.

The second technique for reducing power dissipation,

shown in Figure 9, uses a resistor in parallel with the

LT3085. This resistor provides a parallel path for current

ﬂow, reducing the current ﬂowing through the LT3085.

This technique works well if input voltage is reasonably

constant and output load current changes are small. This

technique also increases the maximum available output

current at the expense of minimum load requirements.

As an example, assume: VIN = VCONTROL = 5V, VIN(MAX) =

5.5V, VOUT = 3.3V, VOUT(MIN) = 3.2V, IOUT(MAX) = 0.5A and

IOUT(MIN) = 0.35A. Also, assuming that RP carries no more

than 90% of IOUT(MIN) = 630mA.

Calculating RP yields:

P =

5.5V – 3.2V

315mA

= 7.30Ω

(5% Standard value = 7.Ω)

The maximum total power dissipation is (5.5V – 3.2V) •

0.5A = 1.2W. However the LT3085 supplies only:

0.5A –

5.5V – 3.2V

7.5

= 0.193A

Therefore, the LT3085’s power dissipation is only:

PDIS = (5.5V – 3.2V) • 0.193A = 0.44W

RP dissipates 0.71W of power. As with the ﬁrst technique,

choose appropriate wattage resistors to handle and dis-

sipate the power properly. With this conﬁguration, the

LT3085 supplies only 0.36A. Therefore, load current can

increase by 0.3A to 0.143A while keeping the LT3085 in

its normal operating range.

Figure 9. Reducing Power Dissipation Using a Parallel Resistor

–

LT3085

VCONTROL

OUT

VOUT

VIN

3085 F09

SET

RSET

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