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LT3080

3080fb

The second technique for reducing power dissipation,

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

LT3080. This resistor provides a parallel path for current

ﬂow, reducing the current ﬂowing through the LT3080.

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) = 1A and

IOUT(MIN) = 0.7A. Also, assuming that RP carries no more

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

Calculating RP yields:

P =

5.5V – 3.2V

0.63A

= 3.65Ω

(5% Standard value = 3.6Ω)

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

1A = 2.3W. However the LT3080 supplies only:

1A –

5.5V – 3.2V

3.6Ω

= 0.36A

Therefore, the LT3080’s power dissipation is only:

PDIS = (5.5V – 3.2V) • 0.36A = 0.83W

RP dissipates 1.47W of power. As with the ﬁrst technique,

choose appropriate wattage resistors to handle and dis-

sipate the power properly. With this conﬁguration, the

LT3080 supplies only 0.36A. Therefore, load current can

increase by 0.64A to 1.64A while keeping the LT3080 in

its normal operating range.

Figure 9. Reducing Power Dissipation Using a Parallel Resistor

–

LT3080

VCONTROL

OUT

VOUT

VIN

3080 F09

SET

RSET

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