LTC1624CS8 datasheet(11/28 Pages) LINER

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LTC1624

what is limiting the efficiency and which change would

produce the most improvement. Percent efficiency can be

expressed as:

%Efficiency = 100% – (L1 + L2 + L3 + ...)

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. LTC1624 VIN current

2. I2R losses

3. Topside MOSFET transition losses

4. Voltage drop of the Schottky diode

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

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 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.

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.

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

loss is thus reduced by the duty cycle.) For example, at

50% DC, if RDS(ON) = 0.05Ω, RL = 0.15Ω and RSENSE =

0.05

Ω, 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.

3. Transition losses apply only to the topside MOSFET(s),

and only when operating at high input voltages (typically

20V or greater). Transition losses can be estimated

from:

Transition Loss = 2.5(VIN)

1.85 (IMAX)(CRSS)(f)

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%.

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

losses generally account for less than 2% total additional

loss.

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, VOUT immediately shifts

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

a stability problem. The ITH external components shown in

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

µF) supply bypass capacitors. The

discharged bypass capacitors are effectively put in parallel

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