Question E7D13:
Which of the following calculates power dissipated by a series linear voltage regulator?  

Choices:  
A: Input voltage multiplied by input current  
B: Input voltage divided by output current  
C: Voltage difference from input to output multiplied by output current  
D: Output voltage multiplied by output current    

Correct Answer is  C: Voltage difference from input to output multiplied by output current  

To find power dissipated in a series linear regulator, remember it “burns off” the excess voltage as heat while passing (approximately) the load current.

1) Why the correct answer is correct
- In a series linear regulator, the pass device drops the difference between input and output voltage while carrying the load current.  
  $$P_{\text{diss}} = (V_{in} - V_{out}) \times I_{out}$$
- This matches “Voltage difference from input to output multiplied by output current.”

2) Why the other choices are incorrect
- A: $V_{in} \times I_{in}$ is the input power, not the regulator’s dissipation. The regulator’s dissipation is input power minus output power.
- B: $V_{in} / I_{out}$ has units of resistance, not power.
- D: $V_{out} \times I_{out}$ is the output power delivered to the load, not the heat in the regulator.

3) Useful formulas and rules of thumb
- Core dissipation formula:  
  $$P_{\text{diss}} \approx (V_{in} - V_{out}) \, I_{out}$$
- More exact (including quiescent current $I_q$):  
  $$P_{\text{diss}} = (V_{in} - V_{out}) \, I_{out} + V_{in} \, I_q$$
  with $I_q$ usually small.
- Efficiency (approximate for linear regs):  
  $$\eta \approx \frac{P_{out}}{P_{in}} \approx \frac{V_{out}}{V_{in}}$$
- Thermal check:  
  $$\Delta T \approx P_{\text{diss}} \times \theta_{JA}$$
  Ensure adequate heatsinking.
- Headroom (dropout): $V_{in}$ must exceed $V_{out}$ by at least the dropout voltage.