Circuit no.2 and no.3

lochmaster draw up

circuit no.3
components:

d1 = 9v1 zener diode IzRm 5.6 mA

d2, d3, d4 = 1N4001 type diode

c1, c2 = 0.1 uf capacitors

r2, r3, r4 = 1000 ohms .25w

r5 = 390 ohms .25w

r6 = 10k ohms .25w

r7 = 330 0ohms .25w

r8 = 660 ohms .25w

led d1 (red), d1(yellow), d2(green) =1.8v 9.5 mA

ic1 = operational amplifer lm324

working :

r2,r3,r4 = 1k ohm resistors
r5 = 2.2v/0.00645A = 341.24 ohms, use of 390 ohms

r6 = given resistance of 10k ohms

r7 = 0.23v/0.000847A = 271.34ohms, use of 330 ohms

r8 = 0.4v/0.000847A =472.25 ohms

Explantion:

when testing circuit no.3 , the board made a zapping sound, later testing the board finding that the op- amp short circuited, after looking careful it was because of a soldering problem conecting two tracks of the op- amp together.

circuit no 2

Components:

3X resistors,
R1=800 ohm

R2=330 ohm

R3=1k ohm
2x capactors
2x diodes

1x zener diode
1x led 1.8v 40ma

1 LM317 transistor

working
vref=1.25............vout=5v

vout =vref(1+r3/r2)

5=1.25(1+r3/r2)

5/1.25(1+r3/r2)

4=1+r3/r2

4-1=r3/r2
3=r3/r2

990/330

3-1 ratio

3=990 ohms ...R2=330 ohms

NOTE:
as the resistor does not exict im using a 1k ohm resistor

R1 resistor before the 1.8v LED: R =v/i 5-1.8v /0.004 = 800 ohms using 1k ohm

Explantion:

starting with D13 the supply voltage enters. In parallel is a zener diode, with a capactor, these two are then earthed to ground ,
R2 330 ohm and R3 990 ohm help the voltage regulator LM317 convert the voltage to a lower level of 5v .R1 1k ohm is in series with the LED 1.8v also being ground to earth.

actual circuit

Reflection and results:

Vout=Vref

v=1.25v

VD across the output leg of the regulator to earth was 5.03v

accpectable as the supply to output is 5v

VD across the LED is 1.92v

also alright as led has a spec of 1.8v

R1 VD is 3.11v

also fine as 3.11v+1.92v=5.03v

circuit no 1

Circuit board no.1

lochmaster 3.0 draw up

The circuit board consists of:
given readings:

12volt supply
2x 5 volt supply

LED :1.8v, 20ma

vce:0.04 volts bjt

WORKING:
R14;R15:
12v-1.8v-0.04=10.16v

R=V/I :10.16V/0.02A =508 ohms

R13;R16:

Beta=ic/ib.......ic=2omA, o.02A........ib= ?......beta=10o

Ib=ic/beta...o.o2A/1oo=o.o0o2A,

R=V/I 5-o.7 V/o.o0o2A...... subtract 0.7 from the supply voltage
4.3v/0.2ma= 21500 ohms

NOTE:The resistor im going to use is 560 ohms as 508 resistor does not exist going up to the closet resistance level. As the resistor R15 is in a parallel circuit with R14 the calculations are the same.

Components :

3 voltage inputs, v in 1 being 12volts, v in 2 & v in 3 being 5 volts, the circuit ends being grounded to earth.
Also consists of two different coloured leds both being 1.8v ,there are also two NPN type transistors.
4x resistors, two being 560 ohms R14 & R 15, the other two being 1k ohm resistors R13 & R16.

NOTE: After checking data sheet ib, 0.2mA is not enough current to open the transistor.Data sheet recomands transistor needs a 5mA current.
R13;R!6:5v -o.7v/5mA=...... 4.3v/0.o0o5A= 860 ohms same problem with R14;R15 i'll change the resistor to a 1 k ohm resistor as there is no resistor at 860 ohms.

Actual circuit board:

Explantion:

R14 at 560 ohms is in parallel with R15 also 560 ohms. R14 is connected to D1 the green LED 1,8 v in series to T1 the collector leg( NPN transistor),base being postive is linked to R13 1 k ohm resistor and then to 5volt supply,

From my 12 volt supply R15 (in parallel with R14), connected to D2 through the collector of T2, base is linked to R16 1 k ohm resistor.

Both T1 and T2 are grounded to earth through their emitter .


Reflection and results:

-With the given 12.07 volt supply and two 5 volt inputs, procceding with a voltage drop across R14=10.11v and voltage across LED=1.96v giving a VTof 12.07 in the circuit.

-the voltage drop from the base to the end of R13=4.88v, this must mean .13volts is lost in the circuit somewhere between the two components.

the voltage readings for R15 should be close if not the same as R14 as the two circuits are simlar

the same situation for R 16 should be the same as R13 as the two circuits are simlar

Experimant no.7 and no.8 ( transistor as a switch )

Components: 1x small signal NPN transistor, 2x resistors




working :

vce2=21ma/0.8ma 26.25 beta
vce3-14ma/0.58ma=28beta
vce4=5ma/0.2ma=25beta

beta amplificationgain differnece = ic/ib


notes":
-Using a multi meter i did a voltage drop test across be, base and emitter.

this reading at .80volts indicates the amount of voltage flowing between the two points.

-the reading 0.06v was the voltage reading across the ce, collector and emitter of the transistor






the three regions of a working transistor, active, saturated, and cutoff.

active region is the main operation zone for amplifying small siganls digtal and analog . one problem with the active region is the lose in power dissipation mailto:T@=25%20oc
there is a current gain in this region also BETA, this is the current ratio between the base and collector

saturated and cutoff regions act like a switch across the emitter and collector junctions ,when the bjt is fully saturated it is when cloised and voltage reading will be emitent, and cutoff like an open circuit having no flow .
p-v*i
3v *0.13ma=0.39watts

experiment no.8

components :circuit on bread board ,using 470R for Rc and a bc547 npn transistor.

using 3 differnet resistors changing Rb doing this to find beta difference between the resistors.

RB 270k ohm Vbe 750 vce 1296 ib 15.6 mA ic 3.18 mA

RB 330K ohm vbe 753 vce 899 ib 19.6 mA ic 5.18 mA

RB220k ohm vbe 753 vce 578 ib 13mA ic 3.35 mA

1) 3.18/15.6= 4.90 beta ratio

2) 5.18/19.6=3.75 beta

3) 3.35/13=3.88 beta

Experiment no.6 ( transistor check )

Using a multi-meter, set to diode test, ( as transistors are made using at least two diodes)

We can work out the PNP and NPN type, and determining the collector, base and emiter, we find the voltage across the three listed above .

NOTE:

Positive leads on the base indicate that it is a p type transistor, negative leads on the base indicate that the transistor is N type . When taking the reading the lowest reading is normally the collector and the highest reading the emitter.

Experiment no.5 observation pictures

Picture no.1

100uF, 1k ohm resistor, 1v,

0.5seconds

Picture no.2

100uF, 0.1 ohm resistor, 1v,

0.05seconds

Picture no.3

100uF, 0.47k ohm resistor, 1v,

0.235seconds

Picture no.4

330 uF, 1k ohm resistor, 1v,

1.65seconds

NOTE:

The less resistance in the circuit the faster the current travels in time (ms) to complete the circuit,

Experiment no.5 ( capacitor )

Components:

1x resistor, 1x capacitor, 1x push button swicth (bridged wire)

Capacitor circuit:

Time taken to charge capacitor

Working:

no.1)
100Exp-6 x 1000 x 5
=0.5s x1000

= 500ms

no2)

100Exp-6 x 100 x 5

=0.05s x1000

= 50ms

no.3)

100Exp-6 x 470 x 5

=0.235 x1000

= 235ms

no.4)
330Exp-6 x 1000 x 5

= 1,65 x 1000

=1650ms

Experiment no.4 ( 5,1 zener diode )

Components: 1x resistor, 1x 5v1 400mW zener diode, 1x diode 1N4007

NOTE:

Readings are as such due to the zener diode being in reverse bias and the second diode in forward bias

Resulting in the 10 volt circuit to have a reading of 10.07 mA, and the 15volt circuit having 15.48 mA

Experiment no.3 ( 5.1 zener diode )

Components : 2x resistors,1x 5.1 400mW zener diode (Zd)


Vz = zener voltage
Vs = voltage supply

The value of Vz : 5volts
Vs now varys from 10 -15 v
What is Vz now: 5volts


NOTE:

Component voltage dos'nt change as this voltage amount is to operate the component, Zener diode has a reverse bias of 5.1 v and a forward bias 0.7v

Changing the supply voltage wont change a diodes voltage, only changing the diodes forward or reverse bias wil change its voltage drop.


Zener diodes are good for circuits containing different voltage components.


Reversing the polarity of the zener diode:
0.846 was measured in this reversed polarity giving a forward bias, which should read between 0.6-0.7, the reading 0.8 would mean there was a leak in the diode exceding its forward bias voltage.

experiment no.2 ( diodes )

Diodes are symbolised by a white ring on the diode, closet to the cathode end indicating the negative end of the diode.

When the cathode end of the diode is connected to the earth of the power supply the diode is in forward bias
Reverse bias is when the diode is conected back to front

calculate the value of current flowing through the diode:

calculated : i=v/r = 5.02-0.7/1000 = 0.00432 = 4.32 ma

measured : 4.35 ma

To find the amount of current flowing through the diode, i

used the formula i = v/r , dividing the voltage by the resistance giving me the current .

calculate the voltage drop across the diode :

calculated : v= i x r =

measured : 0.65 v

maxium value of current flow through the diode:

Ta=1.0@75degreesceluis

maxium value of supply voltage (Vs) at which the diode is safe to run:

v=ixr= 1x1000 = 1000@75degreesceluis

Diode is replaced by LED :

calculated: 5.02 -1.8/1000 = 3.22/1000 = 0.00322 ma

measured: 3.3 ma

NOTE: LED circuit uses more volts but less amperage than the diode.

Experiment no.1 (resistors)

NOTE:

In a series circuit resistance is added together to giving a prpportional decrease in the flow of current ,in addtion a lower resistance reading.

resistance in a parallel circuit
-

Working:

Resistor 1 : 98900 ohms

Resistor 2 : 389 ohms

Both in series :
calculated : 99289 ohms

measured : 99200 ohms

Both in parallel :

calculated : 387.48 ohms

measured : 388 ohms