How to Determine RE for a Sinusoidal LED Current Offset of 27 mA?

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To determine the emitter resistor (RE) for a sinusoidal LED current offset of 27 mA, two methods can be used: keeping RE constant and varying the offset voltage (VOFF), or keeping VOFF constant and varying RE. The correct formula for calculating the emitter current (IE) is IE = (V_OFF - V_BE) / RE. After calculations, an emitter resistor value of approximately 139.46Ω was suggested. It is crucial to ensure that the sinusoidal collector current does not clip by adjusting the amplitude of the input signal if necessary.
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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.
 
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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.
 
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How about providing a diagram of your circuit.
 
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?
 
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Yes.

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

PS I would just say Ie = (V_OFF - 0.7)/R_E.
 
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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.
 
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Thanks a lot man. I know it was simple, but got stucked on it for some reason.
 
tadm123 said:
Thanks a lot man. I know it was simple, but got stucked on it for some reason.

Big 10-4, tadm.
 
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