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Ecen E5 LT2 Group 3

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0% found this document useful (0 votes)
17 views6 pages

Ecen E5 LT2 Group 3

Uploaded by

MaeDumangon
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Republic of Philippines

POLYTECHNIC UNIVERSITY OF THE PHILIPPINES


COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

ECEN-E5: ADVANCED POWER SUPPLY SYSTEMS


LEARNING TASK 2

LEADER:

DUMANGON, HYACINTH V. (4-3)

MEMBERS:

BALLINAN, JANE NICOLE (4-1)

PASTOR, HILARIE MAE (4-1)

OLAYON, CRISJOHN (4-3)

CONVERTER STAGE – PART 1


DESIGN THE LINEAR CONVERTER STAGE OF A POWER SUPPLY WITH SPECIFICATION SHOWN
ON THE TABLE BELOW:

PARAMETER VALUE UNIT

INPUT VOLTAGE (RMS) 120 VRMS

INPUT FREQUENCY 50 Hz

OUTPUT VOLTAGE (AVE) 5 VAVE

OUTPUT VOLTAGE RIPPLE (PK-PK) 2 VPP

OUPUT LOAD (MAX) 1 A

STEP 1: PASS TRANSISTORS – THE CONTROLLING INSTRUMENTS

PARAMETER EQUATION WHERE VALUE

Q1 EMITTER
(I E (Q 1) ) – Q1 Emitter Current
CURRENT
I E (Q 1)=I L 1A
(I E (Q 1) ) I L – Load Current

Q1 BASE CURRENT
Refer to Simulation (I B ( Q 1)) – Q1 Base Current 20.414 mA
(I B ( Q 1))
Q1 CURRENT I E (Q 1) (I B ( Q 1)) – Q1 Base Current 47.99
GAIN I B (Q 1)=
βQ 1+1
(β Q 1 ) – Q1 Current Gain
Republic of Philippines
POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

I E (Q 1)
(β Q 1 ) β Q 1= −1
I B (Q 1 )
Q2 BASE
CURRENT
Refer to Simulation (I B ( Q 2) ) – Q2 Base Current 0.292 mA
(I B ( Q 2) )

I E (Q 1)
Q2 CURRENT I B (Q 2)=
βQ 2+1 (I B ( Q 2) ) – Q2 Base Current
GAIN
68.91
(β Q 2 ) I E (Q 1) ( β Q 2 ) – Q2 Current Gain
β Q 2= −1
I B (Q 2 )

STEP 2: OPERATIONAL AMPLIFIER – THE CONTROLLER

PARAMETER EQUATION WHERE VALUE

( AV ( X 1 )) – Q1 Emitter
Current
X1 VOLTAGE
GAIN V omax( X 1) V omax ( X 1) – X1 Maximum
AV ( X 1) = 1.51
V imax(x 1) Output Voltage
( AV ( X 1 ) )
V imax (X 1) – X1 Maximum
Voltage

X1 SOURCE
CURRENT
I S (X 1)=I B (Q 2) (I S ( X 1) ) – Source Current 0.29 mA
(I S ( X 1) )

STEP 3: FEEDBACK RESISTORS – THE MEASURING INSTRUMENTS

PARAMETER EQUATION WHERE VALUE

FEEDBACK R3 1.683 kΩ
AV ( X 1) =1+
RESISTORS R4 R3, R4 – Feedback Resistors 3.3 kΩ
(R3, R4) R3=R 4 ( A¿ ¿V ( X 1)−1)¿
Republic of Philippines
POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

RESULT

PARAMETER EQUATION WHERE VALUE

NO LOAD ¿ ¿ Refer to Simulation 5.03 V


AND FULL LOAD ¿ ¿ ¿ ¿ – average output voltage 5.5 V
OUTPUT at no load
VOLTAGES
¿ ¿ – average output voltage
at full load

LOAD REGULATION % 𝐿R = ¿
(% LR)
% LR - % Load Regulation 9.3 %

INPUT POWER
Refer to Simulation Pi – Input Power 8.5 W
(P¿¿ i)¿
η - Efficiency
Po
EFFICIENCY (η) η= x 100 % Po – Output Power 58.82 %
Pi
Pi – Input Power
Republic of Philippines
POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

Q1 and Q2

FULL LOAD
Republic of Philippines
POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

NO LOAD

POWER
Republic of Philippines
POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
COLLEGE OF ENGINEERING
Electronics Engineering (ECE) Department

PARAMETER EQUATION WHERE VALUE

𝑁𝑃 - No. of turns of transformer’s


TURNS RATIO 𝑁𝑃 𝑉𝑃 primary winding.
= 𝑁𝑆 - No. of turns of transformer’s
secondary winding.
𝑁𝑆 𝑉𝑆
(𝑅𝑀𝑆) -
RMS voltage on transformer’s 21.606:1
primary winding.
22:1
(𝑅𝑀𝑆) -
RMS voltage on transformer’s
secondary winding.
𝐍𝐏
- Turns ratio

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