LTC1624
U
W U U
APPLICATIONS INFORMATION
what is limiting the efficiency and which change would
producethemostimprovement. Percentefficiencycanbe
expressed as:
loss is thus reduced by the duty cycle.) For example, at
50% DC, if RDS(ON) = 0.05Ω, RL = 0.15Ω and RSENSE
=
0.05Ω, then the effective total resistance is 0.2Ω. This
results in losses ranging from 2% to 8% for VOUT = 5V
as the output current increases from 0.5A to 2A. I2R
losses cause the efficiency to drop at high output
currents.
%Efficiency = 100% – (L1 + L2 + L3 + ...)
whereL1, L2, etc. aretheindividuallossesasapercentage
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:
3. Transition losses apply only to the topside MOSFET(s),
andonlywhenoperatingathighinputvoltages(typically
20V or greater). Transition losses can be estimated
from:
1. LTC1624 VIN current
2. I2R losses
Transition Loss = 2.5(VIN)1.85 (IMAX)(CRSS)(f)
3. Topside MOSFET transition losses
4. Voltage drop of the Schottky diode
4. The Schottky diode is a major source of power loss at
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%.
1. The VIN current is the sum of the DC supply current IQ,
given in the Electrical Characteristics table, and the
MOSFET driver and control currents. The MOSFET
driver current results from switching the gate
capacitanceofthepowerMOSFET. EachtimeaMOSFET
gate is switched from low to high to low again, a packet
of charge dQ moves from INTVCC to ground. The
resulting dQ/dt is a current out of VIN which is typically
much larger than the control circuit current. In
continuous mode, IGATECHG = f (QT + QB), where QT and
QB are the gate charges of the topside and internal
bottom side MOSFETs.
As expected, the I2R losses and Schottky diode loss
dominate at high load currents. Other losses including
CIN and COUT ESR dissipative losses and inductor core
lossesgenerallyaccountforlessthan2%totaladditional
loss.
Checking Transient Response
By powering BOOST from an output-derived source
(Figure 10 application), the additional VIN current
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 INTVCC current results in
approximately 1.5mA of VIN current. This reduces the
midcurrent loss from 5% or more (if the driver was
powered directly from VIN) to only a few percent.
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. Whenaloadstepoccurs, VOUT immediatelyshifts
by an amount equal to (∆ILOAD • ESR), where ESR is the
effective series resistance of COUT. ∆ILOAD also begins to
charge or discharge COUT which generates a feedback
error signal. The regulator loop then acts to return VOUT to
its steady-state value. During this recovery time VOUT can
be monitored for overshoot or ringing that would indicate
astabilityproblem. TheITH externalcomponentsshownin
theFigure1circuitwillprovideadequatecompensationfor
most applications.
2. I2R losses are predicted from the DC resistances of the
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 RSENSE multiplied by the duty
cycle can simply be summed with the resistance of L to
obtain I2R losses. (Power is dissipated in the sense
resistor only when the topside MOSFET is on. The I2R
Asecond, moreseveretransient, iscausedbyswitchingin
loads with large (>1µF) supply bypass capacitors. The
dischargedbypasscapacitorsareeffectivelyputinparallel
11
Linear [ Linear ]
