The power dissipation within an IC is a very important parameter because excessive temperature rise within a semiconductor junction can ruin the device. In linear series voltage regulators, the device that handles the most current is the series control element. The power dissipated in the control element is the product of the voltage across the unit times the current through the unit. The load is the current, ILOAD, through the unit as shown in Figure 9-2c and d. The voltage across the series element is VREG; therefore, the power dissipation in the series element is:
PD = VREG × ILOAD
VREG is: VREG = VDC – VS – VOUT
VREG is: VREG = VDC – VS – VOUT
VS is usually very small compared to VOUT, therefore, the series element power dissipation can be expressed as: PD = (VDC – VOUT) ILOAD
Example 3. Power Dissipation
A voltage regulator has an input voltage of +12V and is regulating a +5V supply line. The load current
is 100 mA. What is the power dissipation in the control element?
Solution:
PD = (VDC – VOUT)ILOAD
PD = (+12 – +5)V × 0.1A
PD = 7 × 0.1 = 0.7 watts
In many IC voltage regulators, especially the low-drop-out regulators, the VDC is restricted to specified
values so that the VREG is not too great across the series element at the rated load current. This prevents
exceeding the rated power dissipation of the device.
A voltage regulator has an input voltage of +12V and is regulating a +5V supply line. The load current
is 100 mA. What is the power dissipation in the control element?
Solution:
PD = (VDC – VOUT)ILOAD
PD = (+12 – +5)V × 0.1A
PD = 7 × 0.1 = 0.7 watts
In many IC voltage regulators, especially the low-drop-out regulators, the VDC is restricted to specified
values so that the VREG is not too great across the series element at the rated load current. This prevents
exceeding the rated power dissipation of the device.
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