1. [INC 281 MULTIDISCIPLINARY WORKSHOP WEEKLY REPORT] April 20, 2012
Name-Surname Ms. Pornprom Nisaimun
Student ID. 54261518
Group Name Instinct
M 53211801 Kung 54261502
Boss 53211808 Mic 54261507
Engineering Team
Yui 53211835 Arm 54261510
Members
Ball 53211836 Garfield
Top Ice
Product Design P’Jub
Team Members P’Ploy
Weekly Progress
Since last week there is some problem about high temperature of LM317. I have to learn
more about characteristic charge condition of Lipo battery. From my learning I have known that the
basic charging procedure is by limiting the current (from 0.2C to max 1C depending on
manufacturer) until the battery reaches 4.2 V/cell and keeping this voltage until the charge current
has dropped to 10% of the capacity C. For example charge characteristic in Fig.1 which is
Panasonic’s charge curve for their 830 mAh cells.
Fig.1 Panasonic’s charge curve for their 830 mAh cells.
Refer to old charger circuit (Fig.2), it is not has current limiting when I connect the batteries
that there is low resistance into the output circuit, it represent very load. So the circuit attempt to
retain output voltage by distributes high current to batteries. By this way, there is power loss on
LM317 by Vdrop x Icharge
Ploss = (Vin – Vout) Icharge
= (12 V – 7.9 V) x 2000 mA
= 4.1 V x 2000 mA
= 8.2 W
Fig.2 old charger circuit
Department of Control System and Instrumentation Engineering
KING MONGKUT’s UNIVERSITY OF TECHNOLOGY THONBURI
2. [INC 281 MULTIDISCIPLINARY WORKSHOP WEEKLY REPORT] April 20, 2012
I have already designed new charger circuit with limiting current by adjust RV2 and
automation adjust output voltage by BC337. The details of circuit as follows:
Lipo
7.4 V, 1500 mAh
(3.7 V x 2 )
Fig. 3 Charger circuit
Voltage Current Time
Charge condition
8V 375 mA 240 minutes
Calculation
1. Charge voltage: The max charging voltage must not exceed 4.2 V per cell, for 2
series cell is 8.4 V
2. Charge Current: The max charge rate recommended is 1C (1500 mAh), but it is
cause for power loss of LM317 since Vdrop x Icharge. 375 mA is 0.25C.
3. Time:
min
4. Ploss = (Vin – Vout) Icharge
= (12 V – 8 V) x 375 mA
= 4 V x 375 mA
= 1.5 W
5. Vout =
6.75 V = (5.308 x 10-3) R2
= 1271.665411 Ω
Department of Control System and Instrumentation Engineering
KING MONGKUT’s UNIVERSITY OF TECHNOLOGY THONBURI
3. [INC 281 MULTIDISCIPLINARY WORKSHOP WEEKLY REPORT] April 20, 2012
6. RV2 = = = 1.867 Ω
7. PRV2 = ( ) Icharge
= (0.7 V) x 375 mA
= 0.2625 W
Engineering Analysis
Strong Point
- Size PCB of new circuit can put into product.
Week Point
- The new circuit have not already test.
Next Week Job
- Test new charger circuit.
- Design alarm circuit when output voltage of 7806 less than 6 V.
- Design new PCB of charger circuit.
Department of Control System and Instrumentation Engineering
KING MONGKUT’s UNIVERSITY OF TECHNOLOGY THONBURI