Datasheet

LTC4366-1/LTC4366-2
436612fd
For more information www.linear.com/LTC4366
17
applicaTions inForMaTion
After the charge pump is active the V
SS
current increases
to 160µA (worst-case 230µA, see Table 1) current while
the final value OUT voltage is equal to the minimum supply
voltage. The C1 voltage is clamped at 5.7V (worst-case
6.0V):
R
SS(MAX)
=
V
IN(MIN)
– V
Z(OUT)
I
VSS(CP)
R
SS(MAX)
=
18V – 6V
230µA
= 52.3k
Step 2: Determine R
IN
The value for resistor R
IN
is calculated using the calculated
R
SS
value. R
IN
is chosen to provide enough headroom to
sufficiently charge C1 to 4.9V the maximum undervoltage
lockout 2 threshold (V
UVLO2
) which starts the charge pump.
The parameters that determine R
IN
include: minimum
supply voltage, the final C1 voltage, MOSFET threshold
voltage, R
SS
, 72µA maximum V
SS
pin current (regulation
amplifier on, I
VSS(AMP)
), and finally the 13µA maximum
start-up current in the V
DD
pin (I
VDD(STHI)
):
R
IN(MAX)
=
V
IN(MIN)
– V
UVLO2
− V
D
− V
TH
− I
SS(AMP)
•R
SS
( )
I
VDD(STHI)
R
IN(MAX)
=
18V − 4.9V − 0.58V − 5V − 72µA • 52.3k
( )
13µA
R
IN(MAX)
= 287k
Table 1. Electrical Parameters Used in Design Example
SYMBOL PARAMETER CONDITIONS TYP MAX
V
Z(OUT)
OUT Shunt Reg. Voltage I = 1mA, BASE = 0V 5.7V 6.0V
V
UVLO2
OUT Undervoltage Lockout 2 Rising 4.75V 4.9V
I
VSS(CP)
V
SS
Pin Current – Charge Pump On –160µA –230µA
I
VSS(AMP)
V
SS
Pin Current – Regulation Amplifier On –45µA –72µA
I
VDD(STHI)
V
DD
Pin Current – Start-Up, Gate High GATE Open, V
DD
= 7V, OUT = 0V 9µA 13µA
I
GATE(ST)
GATE Pin Current – Start-Up GATE = OUT = 0V –7.5µA –11µA
V
UVLO1
OUT Undervoltage Lockout 1 Rising 2.55V 2.75V
Step 3: Find R
SS(MAX)
In some cases this value for R
SS
is too large to charge C1
and power the overvoltage amplifier before the maximum
input voltage passes to the output. The voltage at the V
SS
pin when I
RIN
= I
RSS
is called the match point (V
SS(MATCH)
).
Choosing the match point (with supply at the maximum)
sufficiently below V
REG
(by at least 7V), allows C1 to charge
up in time to protect the load from overvoltage:
R
SS(MAX)
=
R
IN
• V
REG
– 7V
( )
V
IN(MAX)
−5V − V
REG
R
SS(MAX)
=
287k • 43V – 7V
( )
250V – 5V – 43V
= 51.1k
In this case the R
SS
value of 52.3k calculated in Step 1
is too large.
Step 4: Iterate Smaller R
SS
Using 51.1k (R
SS(MAX)
) as the next guess for R
SS
, we can
now calculate R
IN
and R
SS(MAX)
:
R
IN
=
18V – 4.9V − 0.58V − 5V − 72µA • 51.1k
( )
13µA
R
IN
= 294k
R
SS(MAX)
294k • 43V – 7V
( )
250V – 5V − 43V
= 52.3k
In this case the R
SS
value of 51.1k is less than R
SS(MAX)
and the solution is acceptable.