Ericsson Internal
ECHWANG
PRODUCT SPECIFICATION
4 (5)
Prepared (also subject responsible if other)
No.
2/1301-BMR 636 01/54 Uen
31
Technical Specification
Approved
Checked
Date
Rev
EN/LZT 146 3R9e4feRre5nCce August 2012
PKB 4000 series
SEC/D (Betty Wu)
ESECZHW
2009-2-9
B
© Ericsson AB
DC/DC converters, Input 36-75 V, Output 30 A/90 W
12 V/6 A Typical Characteristics
PKB 4713 PINB
Start-up
Shut-down
Start-up enabled by connecting VI at:
Top trace: output voltage (5 V/div.).
Bottom trace: input voltage (20 V/div.).
Time scale: (10 ms/div.).
Shut-down enabled by disconnecting VI at:
ref = +25°C, VI = 53 V,
IO = 6 A resistive load.
Top trace: output voltage (5 V/div.).
Bottom trace: input voltage (50 V/div.).
Time scale: (0.5 ms/div.).
Tref = +25°C, VI = 53 V,
T
I
O = 6 A resistive load.
Output Ripple & Noise
Output Load Transient Response
Output voltage ripple at:
Trace: output voltage (20 mV/div.).
Time scale: (2 µs/div.).
Output voltage response to load current step- Top trace: output voltage (200 mV/div.).
T
ref = +25°C, VI = 53 V,
O = 6 A resistive load.
change (1.5-4.5-1.5 A) at:
Bottom trace: load current (2 A/div.).
Time scale: (0.1 ms/div.).
I
Tref =+25°C, VI = 53 V.
Output Voltage Adjust (see operating information)
Passive adjust
Active adjust
The resistor value for an adjusted output voltage is calculated by using the
following equations:
The output voltage may be adjusted using a {current/voltage} applied to the
Vadj pin. This {current/voltage} is calculated by using the following equations:
Output Voltage Adjust Upwards, Increase:
⎛
Vdesired − 12 ⎞
V
adj = ⎜1.225 + 2.45 ×
⎟
⎟
⎜
⎛12
(
100 + ∆%
1.225 × ∆%
)
(100 + 2∆%) ⎞
12
⎝
⎠
Radj = 5.11⎜
−
⎟
⎟
V
⎜
∆%
⎝
⎠
k
Ω
Example: Upwards => 12.50 V
Example: Increase 2% =>Vout = 12.24 Vdc
⎛
12.50 − 12 ⎞
⎜1.225 + 2.45 ×
⎟
⎟
⎜
⎛12
(
100 + 2
1.225 × 2
)
(100 + 2 * 2) ⎞
12
⎝
⎠
Radj = 5.11⎜
−
⎟
⎟
V = 1.327 V
⎜
2
⎝
⎠
k
Ω
= 2287.2 k
Ω
Example: Downwards => 12.50 V
11.50 −12 V = 1.123 V
⎛
⎝
⎞
1.225 + 2.45×
⎜
⎟
⎠
Output Voltage Adjust Downwards, Decrease:
12
100
⎛
⎞
Radj = 5.11×
− 2 kΩ
⎜
⎝
⎟
∆%
⎠
Example: Decrease 4% =>Vout = 11.52Vdc
100
4
⎛
⎞
⎠
5.11×
− 2 kΩ = 117.5kΩ
⎜
⎝
⎟