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LTC1775

losses ranging from 2% to 8% as the output current

increases from 0.5A to 2A for a 5V output. I2R losses

cause the efficiency to drop at high output currents.

3. Transition losses apply only to the topside MOSFET,

and only when operating at high input voltages (typi-

cally 20V or greater). Transition losses can be esti-

mated from:

Transition Loss = (1.7)(VIN2)(IO(MAX))(CRSS)(f)

4. LTC1775 VIN supply current. The VIN current is the DC

supply current to the controller excluding MOSFET gate

drive current. Total supply current is typically about

850

µA. If EXTVCC is connected to 5V, the LTC1775 will

draw only 330

µA from VINandtheremaining520µAwill

come from EXTVCC. VIN current results in a small

(< 1%) loss which increases with VIN.

Other losses including CIN and COUT ESR dissipative

losses, Schottky conduction losses during dead time 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 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, and COUT begins to charge or dis-

charge. The regulator loop acts on the resulting feedback

error signal to return VOUT to its steady-state value. During

this recovery time VOUT can be monitored for overshoot or

ringing which would indicate a stability problem. The ITH

pin external components shown in Figure 1 will provide

adequate compensation for most applications.

A second, more severe transient is caused by connecting

loads with large (> 1

µF) supply bypass capacitors. The

discharged bypass capacitors are effectively put in parallel

with COUT, causing a rapid drop in VOUT. No regulator can

deliver enough current to prevent this problem if the load

switch resistance is low and it is driven quickly. The only

solution is to limit the rise time of the switch drive in order

to limit the inrush current to the load.

cycle skipping can occur with correspondingly larger

current and voltage ripple.

Efficiency Considerations

The efficiency of a switching regulator is equal to the

output power divided by the input power (

×100%). Per-

cent efficiency can be expressed as:

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

where L1, L2, etc. are the individual losses as a percentage

of input power. It is often useful to analyze individual

losses to determine what is limiting the efficiency and

which change would produce the most improvement.

Although all dissipative elements in the circuit produce

losses, four main sources usually account for most of the

losses in LTC1775 circuits:

1. INTVCC current. This is the sum of the MOSFET driver

and control currents. The driver current results from

switching the gate capacitance of the power MOSFETs.

Each time a MOSFET gate is switched on and then off,

a packet of gate charge Qg moves from INTVCC to

ground. The resulting current out of INTVCC is typically

much larger than the control circuit current. In continu-

ous mode, IGATECHG = f(Qg(TOP) + Qg(BOT)).

By powering EXTVCC from an output-derived source,

the additional VIN current resulting from the driver and

control currents will be scaled by a factor of Duty Cycle/

Efficiency. For example, in a 20V to 5V application at

400mA load, 10mA of INTVCC current results in ap-

proximately 3mA of VIN current. This reduces the loss

from 10% (if the driver was powered directly from VIN)

to about 3%.

2. DC I2R Losses. Since there is no separate sense resis-

tor, DC I2R losses arise only from the resistances of the

MOSFETs and inductor. In continuous mode the aver-

age output current flows through L, but is “chopped”

between the top MOSFET and the bottom MOSFET. If

the two MOSFETs have approximately the same RDS(ON),

then the resistance of one MOSFET can simply be

summed with the resistance of L to obtain the DC I2R

loss. For example, if each RDS(ON) = 0.05Ω and RL =

0.15

Ω, then the total resistance is 0.2Ω. This results in

APPLICATIO S I FOR ATIO

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