Design the circruit No.1

Built and test the circuit shows below:

1. Components

     In this circuit. There are 4 resistors, 2 transistors, 2 LEDs and one TTL

2. Calculation

    I chose BC547 npn transistor and find its datasheet show as below

    For collector circuit: 

    I have Vled=1.8V and Iled=20mA. When the transistor is saturated, Vce=0.6V.

    Then Vr=Vs-Vled-Vce=12V-1.8V-0.6V=9.6V

            R=Vr/I=9.6V/20mA=480R

    For base circuit:

    Ib=Ic/beta  I have Ic=20mA, if I want find maximum Ib, then Beta should be minimum, Ibmin=110

    Then Ibmax=20mA/110=0.18mA. But if Ib=0.18mA Ic=20mA, the transistor is cut off. show as below

     So i chose i chose Ib=5mA and Vbe= 1.1V from data sheet when transistor is saturated.

     Then R=V/I=(5V-1.1V0/5mA=780R.

3.Design the layout

     2 LED collector circuit connected in parallel share the 12V supply. Each LED controlled by and 5V supply connected to transistor base. When Turn on the 5V supplier the transistor switch from cut off to saturated and allow the current flow through collector to emitter, and LED glows

4. Built the circuit

    I can not find 780R and 480R resistors, so i take 1000R and 560R instead and for safety.

    I built the circuit and works fine show as below

     

5. Measuring and Explain

    I measured the voltage cross the components.

    Vs=12V. Thats the voltage supply

    Vinput=5V. Thats the voltage from the TTL

    For LED1 circuit:

    Vr14=10.01V Vled1=2.10V Vr13=4.28V Vbe=0.82V Vbc=0.77V Vce=0.04V

    The collector circuit Vce+Vr14+Vled1=12.114V=Vs.

    The base circuit Vbe+Vr13=5.1V=Vinput.

    For LED2 circuit

    Vr15=9.79V Vled2=2.13V Vr16=4.53V Vbe=0.80V Vbc=0.60V vce=0.19V

    The collector circuit Vce+Vr15+Vled2=12.11V=Vs

    The base circuit Vbe+Vr16=5.33V=Vinput

Transistor

Transistor just like combination of 2 diodes share one end.NPN transistor share the anode, and PNP transistor share the cathode.

Identify the Transistor

Here is the transistor, there are 3 wires of it. Base, collector and emitter.




identify the wire of transistor by multimeter

Touch any 2 wires by multimeter and check the reading
Transistor BC547

0.732V if middle wire is + and left wire is -

0.725V if the middle wire is + and right wire is -









Because transistor just like 2 diodes share one terminal, there are reading when the middle wire is +, left and right wires are -, there is two diodes share + terminal. So BC547 is NPN transistor,and the middle wire is base. Vbe>Vbc, so left wire is emitter and right wire is collector.

Transistor BC557

0.73V if middle wire is - and right wire is +

0.735V if middle wire is - and left wire is +










There are reading when the middle wire is -, left and right wires are +, there is two diodes share - terminal. So BC557 is PNP transistor,and the middle wire is base. Vbe>Vbc, so left wire is emitter and right wire is collector.

Diode test:
NPN
Vbe=0.732V Veb=N/A Vbc=0.725V Vcb=N/A Vce=N/A Vec=N/A
PNP
Vbe=N/A Veb=0.735V Vbc=N/A Vcb=0.730V Vce=N/A Vec=N/A


Built the circuit on breadboad and measure Vbe and Vce

I build the circuir show as below

Measure Vbe=0.791V
Vbe is the voltage between base and emitter. It means the transistor is active.

Measure Vce=0.055V
Vce is voltage between collector and emitter.
This reading means the transistor is saturated.

Saturated:
If the transistor in saturated region, that means it fully works. The Vbe is high enough to fully open active the transistor, the current flow through collector to emitter has no resistance, so Vce is very low.
Cut off
If the transistor in cut off region, Vbe is too low to active the transistor, the current can not flow through collector to emitter. Between collector and emitter, transistor like a insulator, Vce is OL.
Active:
The region between cut off and saturated is active. Vbe is in the range between cut off and saturate. The current can flow through collector to emitter but still have some resistance, Vce has some reading.

If Vce=3V,  P=VI. more Ib, the transistor dissipate more power.
i.e. Vce=3V Ib=0.2mA, then Ic=5.5mA. P=3V x 0.0055A=0.0165W
     Vce=3V Ib=0.4mA, then Ic=10mA. P=3V x 0.01A=0.03W

When Vce=2V, Ib=0.2mA, Then Ic=5mA. Beta=Ic/Ib=25
When Vce=3V, Ib=0.4mA, Then Ic=10mA. Beta=Ic/Ib=25
When Vce=4V, Ib=0.6mA, Then Ic=15mA. Beta=Ic/Ib=25
Beta the current gain is constant, can not be changed for same transistor.


Set up the following circuit on a breadboad . Use a 470R for Rc and BC547 NPN transistor.

The breadboad circuit shows below

Measure:
Rb=47KOhm
Vbe=0.712V Vce=0.091V Ib=90uA Ic=6.1mA
Rb=220KOhm
Vbe=0.688V Vce=0.549V Ib=19.3uA Ic=5.2mA
Rb=270KOhm
Vbe=0.682V Vce=0.996V Ib=15.8uA Ic=4.26mA
Rb=330KOhm
Vbe=0.675V Vce=1.372V Ib=12.9uA Ic=3.51mA
Rb=1MOhm
Vbe=0.643V Vce=2.52V Ib=4.3uA Ic=1.2mA

In this circuit. Vbb and Vcc are always 5V. V=IR. Vbb=5V. Increase Rc, the total resistance in base circuit is increased, so the current Ib decreased, and Vbe decrease. Current gain for the transistor increased when Ib decrease, if the transistor in saturated, Ic will be the maximum, and beta=Ic/Ib, when increase Ib, Beta decrease, and if the transistor in cut off. decrease Ib, beta will increase. When increase the current gain In collector circuit. Ic decreased. Vcc=5V and Rc is 470Ohm. That means, the voltage drop across Rc is decreased. Vcc=Vc+Vce. Vc decreased, so Vce increased.

Ib=90uA Beta=Ic/Ib=6.1mA/90uA=67.8
Ib=19.3uA Beta=Ic/Ib=5.2mA/19.3uA=269.4
Ib=15.8uA Beta=Ic/Ib=4.26mA/15.8uA=269.6
Ib=12.9uA Beta=Ic/Ib=3.51mA/12.9uA=272.1
Ib=4.3uA Beta=Ic/Ib=1.2mA/4.3uA=279.1

When Ib decreased, current gain beta increased slowly.
The load line shows if Vce drops to 0, transistor has the maximum Ic and if Ic drops to 0, then Vce goes to maximum

The Capacitor

The capacitor stores electric charge and also can discharge it. There are two metal plates close together and separated by an insulator. Those two plates one is positive and the other is negative. Non-electrolytic capacitor is no polarity requirements. The charge stored depends on the size or capacitance of the capacitor, and its unit is Farads.

Identify the capacitor

This is Non-electrolytic capacitor, no polarity requirements, it can take high voltage.

This is electrolytic capacitor . The terminal which is marked "-" is negative. The polarity must be corrected in circuit. It has leakage current.

 This is tantalum capacitor. It is physically smaller and lower leakage.

This is variable capacitor, it can adjust the capacitance.

EIA code

EIa code is shown the capacitance of the capacitor. It consists 3 numbers. To identify the capacitance, write down first 2 numbers, then add numbers of zero as the third number shown.
i.e: EIA code is 105. the capacitance is 5 zero after 10, it is 1,000,000F

Charge Time
Capacitors take time to charge, to calcualte the time, the formula is T=RxCx5. R is the resistance in the circuit, C is the capacitance of the capacitor.
Capacitor charged 2/3 applied voltage in first 1/5 time of charging, then the speed of charging slow down.

Exercise
 Build the circuit show as below, calculate and observe the charge time of the capacitor.

Calculation:
1.  C=100uF R=1Kohm
     Calculation:  T=5xRxC= 5 x 1000ohm x 100 x10(-6)F = 500mS

2.  C=100uF R=0.1Kohm
     Calculation:  T=5xRxC= 5 x 100ohm x 100 x 10(-6)F = 50mS

3.  C=100uF R=0.47Kohm
     Calculation:  T=5xRxC= 5 x 470ohm x 100 x 10(-6)F = 235mS

4.  C=330uF R= 1Kohm
     Calculation:  T=5xRxC= 5 x 1000ohm x 330 x 10(-6)F = 1650mS

Observation:
Bulit the circuit on breadboad:

1.  C=100uf R=1Kohm

T=500mS








2.  C=100uF R=0.1Kohm

T=50mS






3.  C=100uF R=0.47Kohm

T=250mS









4.  C=330uF R= 1Kohm

T=250ms x 7 =1750mS

Circuit 1,2,3 have same capacitor, but different resistor. R1>R3>R2, T1>T3>T2

Less resistance in the circuit, less charging time for same capacitor.

Circuit 1 and 4 have same resistor, but different capacitor. C4>C1, T4>T1

Less capacitance in the circuit, less charging time for same resistance.

Capacitor charge about 63% in first 1/5 of total charge time, and then slow down the speed of chargeing, so the graph shows a curve.

Diodes LED and Zener diodes

Diodes and LED
Diode works like a one way valve, let the current flow in one direction and LED is a diode that can emit light when it works.

To identify the cathode without a multimeter, there is a brand marked at cathode for diode.

(ref: http://www.markallen.com/teaching/images/electronics/diode.jpg)

Longer wire is anode and shorter wire is cathode for LED.

(ref: http://www.societyofrobots.com/images/electronics_led_diagram.png)

Measure the voltage drop of diode and LED
    LED:   voltage drop in forward biased direction: 1.774V  voltage drop in reverse biased direction: N/A
    Diode: voltage drop in forward biased direction: 0.565V  voltage drop in reverse biased direction: N/A


Exercise: For Vs=5V, R=1KOhm, D=1N4007 build in series circuit on a breadboard.
      The circuit show as blew:

The1000Ohm resistance, one 1N4007 diode in series circuit with 5V supplier.

Calculate the first value of current flowing through the diode:
 Id=Ir=Vd/R=(Vs-Vd)/R=(5V-0.7)/1000Ohm=4.3mA.
Measure the current flowing through the diode, put multimeter in series in to the circuit.
The reading is 4.8mA







The reading is close to my calculation, but not 100% matched, because I took the 0.7V as the knee voltage of the diode, maybe a little different to the fact one.








I measured the voltage drop across the diode.
The reading is 0.648V. It is not 0.7V.






Here is the datasheet of 1N4007 diode

The average value of the current that can flow through the diode is 1.0A at 75degree.


For the safety, the current through the diode should not be over 1A. If we have 1000ohm resistor in circuit, the maximum voltage drop across the resistor is V=IR=1A x 1000ohm=1000V.
Vs=Vr+Vd  The maximum Vs=1000V+0.7V=1000.7V fro safety.

Replace the diode by an LED and measure the current, the reading is 3.5mA.


Calculate the current:
I=Vr/R=(5v-1.8v)/1000ohm=3.2mA.


The voltage drop across the LED is not but close to 1.8V, it cause the calculation is different from the measurement.






Zener Diodes
1. Built the circuit show blew. two 100ohm resistor, Vs=12v, one 5v1 400mW zener diode

    I build the circuit on breadboad show as below:

    Measure the voltage drop across the zener diode and the reading is 4.96V when Vs=12V.

    Vary Vs from 10V to 15V, the value of Vz is vary but still close to 5V

     Vz is vary in small range and always close to 5V, that is because the zener voltage of zener diode is 5.1V,       the circuit just let 5.1V to make the current through the zener diode in reverse direction, no matter what Vs is.So, zener diodes can be used to protect the components from the high voltage damage if build in parallel circuit.

Reverse the polarity of the zener diodes and measure the Vz.

   The reading of Vz is 0.84V, that is because the knee voltage of the zener diode is 0.84V.

2.Built the circuit show as below: one 1Kohms resistor, one 5v1 400mW zener diode, one 1N4007 diode, 

   Vs=10&15V

   I build the circuit show as below:

   

   Set the Vs= 10V and measure the V1, V2, V3, V4

V1=4.63V

V2=0.64V

V3=5.28V

V4=4.94V

   The current in the circuit is I=I4=V4/R=4.94V/1000ohm=4.94mA.

   Set Vs=15V and measure the V1 V2 V3 and V4

V1=4.79V

V2=0.67V

V3=5.47V

V4=9.54V

   The current in the circuit is I=V/R=9.54V/1000ohm=9.54mA

   In this circuit, resistor zener diode and diode in series circuit. V3=V1+v2, Vs=V3+V4. The current through any component is even. If Vs vary, V1 always is zener voltage, V2 always is the knee voltage, so just V4 changes.