Transistor Amplifier Supply For A LED
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Transistor Amplifier Supply For A LED

Transistor Amplifier Supply For A LED

Figure 122.5 is a transistor amplifier circuit designed to supply a Light Emmitting Diode (LED). The output signal from the microcomputer acts as the switch for the LED. The circuit values are as follows:
Microcomputer: output resistance = RB = 1 kΩ; Von = 5 V; Voff = 0 V; current, I = 5 mA.
Transistor: VCC = 5 V; offset voltage, Vγ = VBE = 0.7;
Current gain, β = IC/IB = 95; VCEsaturation =0.2 V.
LED: offset voltage, VγLED = 1.4 V; ILED > 15 mA; Pmax = 100 mW.

(a)Determine collector resistance RC such that the transistor is in the saturation region when the microcomputer outputs 5 V.
(b) Determine the power dissipated by the LED.

The strings: S7P5A51 (Physical Change).

The math:
Pj Problem of Interest is of type change (physical change). Transistors are primarily used for signal amplification and switching. Both are change problems.

Transistor Amplifier Supply For A LED

(a)i Computer is off = Voff = 0
So, Base current, IB = 0
So, BJT is in the cuttoff region

(a)ii Computer is on = Von = 5 V
At saturation and assumimg the large signal model we have:
VCC = RCIC + VγLED + VCEsaturation
So, RC = (5 - 1.4 - 0.2)/IC = 3.4/IC
ILEDmin = 15 mA. So lets double this minimum current required to light LED
So, for 30 mA, RC = (3.4/30)103 = 113 Ω

Verifying BJT is in saturation region:
At saturation, β is significantly diminished.
IB = (Von - VBE)/RB
So, IB = (5 - 0.7)/1000 = 4.3 mA
β = 30/4.3 = 6.7.
So, since 6.7 << 95, BJT is operating plausibly in saturation region.

(b) Power disspated by LED = PLED = VγLED(IC) = 42 mW
So, power dissipation of LED is less than power max of LED
So, design is ok.


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