Power Dissipation

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

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.

Power Dissipation

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

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.

Voltage Regulation

From Figure 9-2c. The input voltage VDC is the voltage out of the rectifier and filter of Figure 9-1.
There is an impedance associated with the rectifier and filter.It is RS shown in Figure 9-2c. As a result, the output voltage, VOUT, is equal to:
VOUT = VDC – ILRS – VREG
= VDC – VS – VREG
= VDC – (VS + VREG)
Using this information, the regulation can be explained as follows: VDC is always considered constant. The regulator varies VREG to keep VOUT constant as IL and VS change. With an increase in IL and thus VS, VOUT would tend to decrease; however, VREG is reduced to compensate and VOUT remains constant. A decrease in IL causes a decrease in VS, but regulation compensates by increasing VREG so that VOUT remains constant. VDC was considered constant but if VDC changes, either up or down, regulation follows to compensate by increasing or decreasing VREG to keep VOUT constant.

Percent Regulation :

The load regulation of a voltage regulator can be expressed as follows (where the load is some specified
current value): % Load Regulation = (Vno load – Vload × 100) / Vload
If Vno load = 11V and, at a specified current, Vload = 10V then
% Load Regulation = (11 –10 × 100) / 10
% Load Regulation = 10%
Common percent load regulation for IC voltage regulators is from 1% to 5%.
Example 2.
Voltage RegulationTo have 1% load regulation for the above supply where VNO LOAD = 11V, at the specified load, what does the VLOAD have to be?
1% = (11 – VLOAD × 100) / VLOAD
0.01 = 11 – VLOAD
VLOAD
VLOAD (1 + 0.01) = 11
VLOAD = 11/ 1.01 = 10.89V.

Voltage Regulation

From Figure 9-2c. The input voltage VDC is the voltage out of the rectifier and filter of Figure 9-1.
There is an impedance associated with the rectifier and filter.It is RS shown in Figure 9-2c. As a result, the output voltage, VOUT, is equal to:
VOUT = VDC – ILRS – VREG
= VDC – VS – VREG
= VDC – (VS + VREG)
Using this information, the regulation can be explained as follows: VDC is always considered constant. The regulator varies VREG to keep VOUT constant as IL and VS change. With an increase in IL and thus VS, VOUT would tend to decrease; however, VREG is reduced to compensate and VOUT remains constant. A decrease in IL causes a decrease in VS, but regulation compensates by increasing VREG so that VOUT remains constant. VDC was considered constant but if VDC changes, either up or down, regulation follows to compensate by increasing or decreasing VREG to keep VOUT constant.

Percent Regulation :

The load regulation of a voltage regulator can be expressed as follows (where the load is some specified
current value): % Load Regulation = (Vno load – Vload × 100) / Vload
If Vno load = 11V and, at a specified current, Vload = 10V then
% Load Regulation = (11 –10 × 100) / 10
% Load Regulation = 10%
Common percent load regulation for IC voltage regulators is from 1% to 5%.
Example 2.
Voltage RegulationTo have 1% load regulation for the above supply where VNO LOAD = 11V, at the specified load, what does the VLOAD have to be?
1% = (11 – VLOAD × 100) / VLOAD
0.01 = 11 – VLOAD
VLOAD
VLOAD (1 + 0.01) = 11
VLOAD = 11/ 1.01 = 10.89V.