BJT simple Transistor problem

In summary, the value of the emitter resistor for which the LED sinusoidal current will have an offset level of I=27 mA is 139.46Ω.
  • #1
tadm123
14
0

Homework Statement



We have a transistor with RB, RE and A LED being a GaAs Light-emitting Diode (LED with a forward voltage of approximately 1.5V.

Values for RB= 2K, Vs-offset=5, Vs-amplitude=4, V-frequency= 1KHz, let β=100

what is the value of the emitter resistor for which the LED sinusoidal current will
have an offset level of I=27 mA?

Use two methods to create such a current:
Method A: Keep RE constant (say 200Ω) and vary VOFF of the VSIN source.
Method B: Keep VOFF constant (say 5V) and vary RE.

In each case, once you set the DC current to 27mA, check the sinusoidal collector current
iC(t), and make sure that it doesn’t get clipped. If it does, reduce VAMPL of the VSIN
source, from 4V to a lower level, but as large as you can make it to keep the LED
currents from clipping

Homework Equations


The Attempt at a Solution



By I offset do they mean the value of I current (in collector)?

If it is then I used the formula for current in emitter

Ie= (Vin-Vbe)/ (Re- Rb/β+1)

plugging in values and solving for Re I get => Re= 139.46Ω.

I guess this is more of a PSPICe problem but I just want to make sure if by the current offset they meant the value of the current.
 
Physics news on Phys.org
  • #2
tadm123 said:
By I offset do they mean the value of I current (in collector)?
Most likely. The LED is the collector load, is it?

If it is then I used the formula for current in emitter

Ie= (Vin-Vbe)/ (Re- Rb/β+1)
Check that formula.
 
  • Like
Likes 1 person
  • #3
How about providing a diagram of your circuit.
 
  • #4
Here's a better view of the HW problem.

Problem_HW.jpg


I copied that exact same diagram on PSPICE.
Made a Bias point analysis and got these Voltage and Current values.

http://postimg.org/image/bcry7ph7l/

Now to solve the problem. I'm not really sure what to do but I used the formula on "Hint".

I use the formula for the emmiter current:
IE= (Vin- VBE)/ (RE + RB/β+1)

Do I have to just solve in Method 1 plug in values and solve for Vin, keeping RE constant?

And in Method 2 solve for RE and plug in values keeping VOFF constant?
 
Last edited by a moderator:
  • #5
Yes.

In both cases you are gunning for Ie = 27 mA.

PS I would just say Ie = (V_OFF - 0.7)/R_E.
 
Last edited:
  • Like
Likes 1 person
  • #6
tadm123 said:

Homework Statement



We have a transistor with RB, RE and A LED being a GaAs Light-emitting Diode (LED with a forward voltage of approximately 1.5V.

Values for RB= 2K, Vs-offset=5, Vs-amplitude=4, V-frequency= 1KHz, let β=100

what is the value of the emitter resistor for which the LED sinusoidal current will
have an offset level of I=27 mA?

Use two methods to create such a current:
Method A: Keep RE constant (say 200Ω) and vary VOFF of the VSIN source.
Method B: Keep VOFF constant (say 5V) and vary RE.

In each case, once you set the DC current to 27mA, check the sinusoidal collector current
iC(t), and make sure that it doesn’t get clipped. If it does, reduce VAMPL of the VSIN
source, from 4V to a lower level, but as large as you can make it to keep the LED
currents from clipping

Homework Equations





The Attempt at a Solution



By I offset do they mean the value of I current (in collector)?

If it is then I used the formula for current in emitter

Ie= (Vin-Vbe)/ (Re- Rb/β+1)

plugging in values and solving for Re I get => Re= 139.46Ω.

I guess this is more of a PSPICe problem but I just want to make sure if by the current offset they meant the value of the current.

Forget Rb/beta +1. RE = (5-0.7)/0.027 ohms.
 
  • Like
Likes 1 person
  • #7
Thanks a lot man. I know it was simple, but got stucked on it for some reason.
 
  • #8
tadm123 said:
Thanks a lot man. I know it was simple, but got stucked on it for some reason.

Big 10-4, tadm.
 

1. What is a BJT transistor?

A BJT (Bipolar Junction Transistor) is a semiconductor device that is used to amplify or switch electronic signals. It consists of three layers of doped semiconductor material, namely the collector, base, and emitter, and can be either NPN or PNP type.

2. How does a BJT transistor work?

A BJT transistor works by controlling the flow of current between the collector and emitter terminals based on the input current at the base terminal. When a small current is applied to the base, it causes a much larger current to flow from the collector to the emitter, thus amplifying the signal.

3. What are the different modes of operation for a BJT transistor?

There are three modes of operation for a BJT transistor: active, saturation, and cutoff. In the active mode, the transistor acts as an amplifier. In saturation mode, it acts as a switch, allowing maximum current to flow from the collector to the emitter. In cutoff mode, there is no current flow between the collector and emitter.

4. How do you analyze a simple BJT transistor circuit?

To analyze a simple BJT transistor circuit, you need to determine the operating mode of the transistor and the values of the resistors and currents in the circuit. Using Kirchhoff's laws and Ohm's law, you can calculate the voltage, current, and power at various points in the circuit.

5. What are some common applications of BJT transistors?

BJT transistors are widely used in electronic circuits, including amplifiers, switches, oscillators, and digital logic circuits. They are also used in audio and radio frequency amplifiers, power supplies, and motor control circuits.

Similar threads

  • Electrical Engineering
Replies
19
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Electrical Engineering
Replies
3
Views
795
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Electrical Engineering
Replies
5
Views
2K
Replies
19
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
25
Views
8K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
4K
  • Electrical Engineering
Replies
5
Views
4K
  • Electrical Engineering
Replies
4
Views
2K
Back
Top