Power Dissipation in a simple circuit

[Abhinav]

I have a transistor. And a Five Volt power supply.

The five volt supply is connected to the base of the transistor.

There are two 0.1uF capacitors and one 4.7k resistor on the line that connects the 5V to the base of the transistor.
At the base of the transistor, there's a resistor which connects the base to the ground,
And the emitter is also grounded. Open drain.

How am I supposed to calculate the value of power dissipation in each element ?
Can u please let me know how to approach this problem?

 

[(*steve*)]

open drain? Do you mean open collector?

If there's no load, you've essentially got 5v across a resistor and a diode.

 

[AnalogKid]

First, post a schematic. There is no amount of text description that will tell us as much about the circuit and the components as will a schematic . Identify each component with a reference designator, like "R1" for a resistor.

ak

 

[hevans1944]

There is no amount of text description that will tell us as much about the circuit and the components as will a schematic .

However, a verbose description will improve your English language skills and can be used as an adjunct description of the circuit schematic. Although some of us here are somewhat challenged in sketching and posting schematics, please make the effort. A simple pen and ink drawing on white paper can be photographed with a cellphone digital camera and uploaded here as a JPEG image or PNG image attachment.

How am I supposed to calculate the value of power dissipation in each element ?

You multiply the voltage drop across each element by the current in that element to obtain the power dissipated in that element. So, for a 4.7 kΩ resistor in series with the base of a transistor and connected to a 5V supply with the emitter grounded, you need to know what the voltage drop from base-to-emitter is so you can subtract that value from the 5V supply. Assume the base-to-emitter forward-biased voltage is 0.7V. That leaves 5 - 0.7 = 4.3V to drop across the resistor. The current in the circuit will be 4.3V/4700Ω = .000915 A. The power dissipated in the resistor will be this current multiplied by 4.3V = 0.00393 W. The power dissipated in the base-emitter junction will be this current multiplied by 0.7V =0.000640 W. There is nothing connected to the "open drain" so no current there and no other power dissipated in the transistor. Did this explanation help you?

 

[Abhinav]

However, a verbose description will improve your English language skills and can be used as an adjunct description of the circuit schematic. Although some of us here are somewhat challenged in sketching and posting schematics, please make the effort. A simple pen and ink drawing on white paper can be photographed with a cellphone digital camera and uploaded here as a JPEG image or PNG image attachment.


You multiply the voltage drop across each element by the current in that element to obtain the power dissipated in that element. So, for a 4.7 kΩ resistor in series with the base of a transistor and connected to a 5V supply with the emitter grounded, you need to know what the voltage drop from base-to-emitter is so you can subtract that value from the 5V supply. Assume the base-to-emitter forward-biased voltage is 0.7V. That leaves 5 - 0.7 = 4.3V to drop across the resistor. The current in the circuit will be 4.3V/4700Ω = .000915 A. The power dissipated in the resistor will be this current multiplied by 4.3V = 0.00393 W. The power dissipated in the base-emitter junction will be this current multiplied by 0.7V =0.000640 W. There is nothing connected to the "open drain" so no current there and no other power dissipated in the transistor. Did this explanation help you?

Yes. Really helped. Thanks! What if there's a capacitor connected from the base to the ground? How am I calculate the power dissipated in a capacitor?

 

[AnalogKid]

What if there's a capacitor

What if you post a schematic so we can see what you are talking about. Hint - if you post this question on a third forum, they will ask for a schematic also.

ak