# Transistors and Photodetectors

STATEMENT OF PURPOSE: We describe how a simple BJT works. We build simple amplifiers with this BJT. We study how to bias the BJT. We also use the BJT as a switch. Switches are useful for many applications as a switch. Switches are useful in numerous applications.

EQUIPMENT:

Circuit with NPN BJT

Speed of light board

Photo transistor

Photo darlington

Fiber optic cable of different lengths

Photo transistor and photo darlington

Oscilloscope and PC

RESULTS

The two IN of the scope are connected to the base and collector resistors of the BJT to display the voltage vs time. The shape of the graph is displayed as amplified in the Collector circuit.
Then we test the BJT as a switch (we apply a bigger signal as to enter the switching region of transistor). At 0.38 V on IN A and 0.30 V on IN B the transistor is OFF. At 1.70 V on IN A and 4.00 V on In B the transistor is ON.
Then we use the BJT as an amplifier. The transistor is positioned at mid-way between OFF and ON states. Thus the BJT is in the active mode.
For a base resistance of 10 Kohm and a base voltage of 0.4 Volts (IN A) the current is 4E-5 A.
The collector resistance is 1 Kohm and the collector voltage is 0.9 V (IN B) so Ic=9E-3 A.&
The amplification is thus Ic/Ib=9E-3/4E-5=225
We connect tha main circuit board to another circuit board through through the optic cable using the optical cable. On the first board there is a LED emitting light while on the receiving board there is a photo diode or a photo transistor and an amplifier. The IN C of the scope is connected to the OUT of the amplifier. he scope is connected at the out of the amplifier.
First we use the photo diode in series with a 100 Kohm resistor. On IN A we have a 0.65 V, on IN B we have 1.5 V and on IN C we have 0.5 V
Then we use the photo transistor as detector. With the photo transistor the signal received is bigger: IN A we have 0.6 V, IN B we have 0.9 V and IN C we have 4 V.
The we use the photo darlignton as detector. The signal received is the biggest. IN A and IN B we have less than 0.1 V , on IN C we have e have 4V+/-0.5 V
Then we use a photo logic as detector. Now the signal is square different from the initial sinusoidal signal detector. Now the signal at IN C is square different from the initial signal. On IN C we have 4 Volts. On IN C we have 4 V.
CONCLUSIONS
The sensitivity of the diode is the least, then the sensitivity of the photo transitor and of the darlington are bigger.

The photo detectors used were:

1. IF-D91 photo diode

2. IF-D92 photo transistor

3. IF-D93 photo darlington

4. IF-D95 photo logic detector

PART 3

We try to find the speed of light in the cable.

We know the length of the cable, the optical index of refraction and the delay time in the cable. From these data we find the speed of light.

The data is

Length (m)         Time delayed (ns)        Speed in fiber (m/s)                 Index of fiber
$0.30±0.05$      $2*10^{-9}±1^{-10}$           $1.5*10^8±2.5*10^7$           $2.00±0.22$
$5.00±0.05$      $30*10^{-9}±1.0^{-10}$     $1.6*10^8±2.2*10^6$           $1.88±0.13$
$10.00±0.05$    $60*10^{-9}±1.0^{-10}$     $1.6*10^8±1.7*10^6$           $1.88±0.21$
$33.00±0.05$    $160*10^{-9}±1.0^{-10}$   $2.06*10^8±1.3*10^5$        $1.46±0.12$

CONCLUSIONS

The optical detector is part of a logic gate, so the voltage provided is uniform. The average lux provided to the detector by the cable over several trials was 180 Lux. However, a minimum was detected by turning down the lux until the logic gate could no longer detect the input. This was found experimentally to be around 40-60 lux, depending on the strength of the connection. The behavior of current that runs through the resisters does follow Ohms Law. Comparing two resister, one with 100 kilo Ohms and the other one with 1 kilo Ohm, by changing the voltage we can get to change the current, according to Ohms Law.