LTC1624CS8 Linear Technology, LTC1624CS8 Datasheet - Page 11

• Manufacturer Part Number: LTC1624CS8
• Description: IC SW REG CONTROLLER N-CH 8-SOIC
• Manufacturer: Linear Technology
• Type: Step-Down (Buck), Step-Up (Boost), Inverting, Sepicr

Specifications of LTC1624CS8

• Internal Switch(s): No
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.19 ~ 30 V
• Current - Output: 2A
• Frequency - Switching: 200kHz
• Voltage - Input: 3.5 ~ 36 V
• Operating Temperature: 0°C ~ 70°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Power - Output: -

APPLICATIONS
what is limiting the efficiency and which change would
produce the most improvement. Percent efficiency can be
expressed as:
where L1, L2, etc. are the individual losses as a percentage
of input power.
Although all dissipative elements in the circuit produce
losses, four main sources usually account for most of the
losses in LTC1624 circuits:
1. The V
2. I
1. LTC1624 V
2. I
3. Topside MOSFET transition losses
4. Voltage drop of the Schottky diode
given in the Electrical Characteristics table, and the
MOSFET driver and control currents. The MOSFET
driver current results from switching the gate
capacitance of the power MOSFET. Each time a MOSFET
gate is switched from low to high to low again, a packet
of charge dQ moves from INTV
resulting dQ/dt is a current out of V
much larger than the control circuit current. In
continuous mode, I
Q
bottom side MOSFETs.
By powering BOOST from an output-derived source
(Figure 10 application), the additional V
resulting from the topside driver will be scaled by a
factor of (Duty Cycle)/(Efficiency). For example, in a
20V to 5V application, 5mA of INTV
approximately 1.5mA of V
midcurrent loss from 5% or more (if the driver was
powered directly from V
MOSFET, inductor and current shunt. In continuous
mode the average output current flows through L but is
“chopped” between the topside main MOSFET/current
shunt and the Schottky diode. The resistances of the
topside MOSFET and R
cycle can simply be summed with the resistance of L to
obtain I
resistor only when the topside MOSFET is on. The I
%Efficiency = 100% – (L1 + L2 + L3 + ...)
2
B
R losses are predicted from the DC resistances of the
2
are the gate charges of the topside and internal
R losses
IN
2
current is the sum of the DC supply current I
R losses. (Power is dissipated in the sense
IN
current
U
GATECHG
INFORMATION
U
IN
SENSE
) to only a few percent.
IN
= f (Q
current. This reduces the
multiplied by the duty
W
T
+ Q
CC
CC
IN
which is typically
current results in
B
to ground. The
), where Q
IN
U
current
T
and
2
Q
R
,
3. Transition losses apply only to the topside MOSFET(s),
4. The Schottky diode is a major source of power loss at
Checking Transient Response
The regulator loop response can be checked by looking at
the load transient response. Switching regulators take
several cycles to respond to a step in DC (resistive) load
current. When a load step occurs, V
by an amount equal to ( I
effective series resistance of C
charge or discharge C
error signal. The regulator loop then acts to return V
its steady-state value. During this recovery time V
be monitored for overshoot or ringing that would indicate
a stability problem. The I
the Figure 1 circuit will provide adequate compensation for
most applications.
A second, more severe transient, is caused by switching in
loads with large (>1 F) supply bypass capacitors. The
discharged bypass capacitors are effectively put in parallel
loss is thus reduced by the duty cycle.) For example, at
50% DC, if R
0.05 , then the effective total resistance is 0.2 . This
results in losses ranging from 2% to 8% for V
as the output current increases from 0.5A to 2A. I
losses cause the efficiency to drop at high output
currents.
and only when operating at high input voltages (typically
20V or greater). Transition losses can be estimated
from:
high currents and gets worse at high input voltages.
The diode loss is calculated by multiplying the forward
voltage drop times the diode duty cycle multiplied by
the load current. For example, assuming a duty cycle of
50% with a Schottky diode forward voltage drop of
0.5V, the loss is a relatively constant 5%.
As expected, the I
dominate at high load currents. Other losses including
C
losses generally account for less than 2% total additional
loss.
IN
Transition Loss = 2.5(V
and C
OUT
DS(ON)
ESR dissipative losses and inductor core
2
= 0.05 , R
R losses and Schottky diode loss
OUT
TH
LOAD
external components shown in
which generates a feedback
IN
OUT
)
• ESR), where ESR is the
1.85
L
. I
OUT
= 0.15 and R
(I
MAX
LOAD
immediately shifts
LTC1624
)(C
also begins to
RSS
)(f)
OUT
OUT
11
SENSE
OUT
= 5V
can
2
to
R
=