# Power dissipation of 2 diodes (1 is a Zener diode)

• Engineering
• Edy56
In summary, the Zener diode is not conducting because it requires 5.6 volts and sicillium-oxide one always does. To determine whether the Zener diode is conducting, check that the voltage on the diode is 5.6 volts and if it is not, calculate the currents and node voltages.
Edy56
Homework Statement
The circuit in Figure 4 uses two diodes, D1 and D2. Write which diodes are in question. Determine the value of the power dissipated on diode D1. It is known: R1= 330 Ω, R2= 1 kΩ, V1= 9 V, VZ= 5, 6 V
Relevant Equations
None
Are my equations correct? And my process of how I should solved this problem?

Have you checked whether the zener diode is conducting?

Gordianus said:
Have you checked whether the zener diode is conducting?
Zener Is because it requires 5.6 volts and sicillium one always does.

Thus, if you feel confident, compute the currents and node voltages.

Edy56 said:
Zener Is because it requires 5.6 volts and sicillium one always does.
5v6 is the Zener breakdown voltage only for that 5v6 Zener diode, which is not always the operating voltage. It could be less.
The resistors R1 and R2, with D2, form a potential divider. It is possible that the Zener diode, D1, has a voltage below 5.6 volts, and so does not conduct. Check that first.

Once you have determined the actual voltage on D1, you can solve for the current and power in D2.

Solve for the current through R1, subtract the current that flows through R2 and D2. What current remains flows through D1.

scottdave
It's hard to read your handwriting. Is ##V_1 = 5V## or is ##V_1 = gV##?

From the sketch above it looks to me like the Zener is reverse biased so the voltage after the R1 is constant 5.6V. And so you may calculate I.
Subtract 0.7 volts - the voltage drop constant across D2 - from 5.6 V and then calculate I2

berkeman said:
It's hard to read your handwriting. Is ##V_1 = 5V## or is ##V_1 = gV##?
I'd guess it's ##V_1 = 9~\rm V##.

Edy56 said:
Are my equations correct? And my process of how I should solved this problem?
I'd say no. You've written things like ##V_1 = R_1## and ##V_2=5.6~\rm V## and ##V_2=0.7~\rm V## (hence implying 5.6=0.7!). Plus you don't appear to indicate exactly what you mean by ##V_2## as far as I can see.

ETA: Oh, I see! Some of those 2s are actually Zs. There's a reason we ask you to type up your work. Your handwriting isn't as easy to read as you might think.

It would be more helpful if you explain your thought process behind your work instead of spewing a bunch of equations and asking "are they right?" That way the helpers can see where you're getting stuck and possibly help you identify what you're unsure about.

berkeman

## What is power dissipation in diodes?

Power dissipation in diodes refers to the amount of electrical power converted into heat within the diode due to the current flowing through it and the voltage drop across it. It is calculated using the formula P = V * I, where P is the power dissipation, V is the voltage drop across the diode, and I is the current flowing through it.

## How do I calculate the power dissipation of a Zener diode?

To calculate the power dissipation of a Zener diode, you need to know the Zener voltage (Vz) and the current flowing through the diode (Iz). The power dissipation (Pz) is given by the formula Pz = Vz * Iz. Ensure that the Zener diode is operating within its specified current range to avoid damage.

## What are the factors affecting power dissipation in diodes?

Several factors affect power dissipation in diodes, including the forward voltage drop (for regular diodes), the Zener voltage (for Zener diodes), the current flowing through the diode, the ambient temperature, and the diode's thermal resistance. Higher current and voltage drop lead to greater power dissipation, which can cause the diode to heat up.

## How can I minimize power dissipation in diodes?

To minimize power dissipation in diodes, you can use diodes with lower forward voltage drops, ensure proper heat sinking and cooling, operate the diodes within their specified current ratings, and use efficient circuit design to reduce the current flowing through the diodes. For Zener diodes, choose an appropriate Zener voltage and limit the current using resistors or current-limiting circuits.

## What happens if the power dissipation exceeds the diode's maximum rating?

If the power dissipation exceeds the diode's maximum rating, the diode can overheat, leading to thermal runaway, permanent damage, or failure of the diode. This can result in the diode becoming non-functional, short-circuiting, or even causing damage to other components in the circuit. It is crucial to ensure that the diode operates within its specified power dissipation limits.

Replies
1
Views
1K
Replies
45
Views
8K
Replies
5
Views
3K
Replies
4
Views
2K
Replies
12
Views
1K
Replies
6
Views
2K
Replies
15
Views
6K
Replies
2
Views
1K
Replies
33
Views
3K
Replies
2
Views
2K