LTC1703 View Datasheet(PDF) - Linear Technology

Part Name

Description

Manufacturer

LTC1703

Dual 550kHz Synchronous 2-Phase Switching Regulator Controller with 5-Bit VID

Linear Technology 

LTC1703

APPLICATIO S I FOR ATIO

from CIN (time point A). 50% of the way through, TG2

turns on and the total current is 13A (time point B).

Shortly thereafter, TG1 turns off and the current drops to

10A (time point C). Finally, TG2 turns off and the current

spends a short time at 0 before TG1 turns on again (time

point D).

( ) ( ) IAVG = 3A • 0.5 + 13A • 0.16 +

( ) ( ) 10A • 0.16 + 0A • 0.18 = 5.18A

Now we can calculate the RMS current. Using the same

waveform we used to calculate the average DC current,

subtract the average current from each of the DC values.

Square each current term and multiply the squares by the

same period percentages we used to calculate the aver-

age DC current. Sum the results and take the square root.

The result is the approximate RMS current as seen by the

input capacitor with both sides of the LTC1703 at full load.

Actual RMS current will differ due to inductor ripple

50%

16% 16% 18%

7.8

4.8

0

–2.2

– 5.2

0

A

B CD

TIME

1703 SB2

Figure SB2. AC Current Calculation

current and resistive losses, but this approximate value is

adequate for input capacitor calculation purposes.

( ) ( ) –2.182 • 0.5 + 7.822 • 0.16 +

( ) ( ) IRMS = 4.822 • 0.16 + –5.182 • 0.18

= 4.55ARMS

If the circuit is likely to spend time with one side operating

and the other side shut down, the RMS current will need

to be calculated for each possible case (side 1 on, side 2

off; side 1 off, side 2 on; both sides on). The capacitor

must be sized to withstand the largest RMS current of the

three—sometimes this occurs with one side shut down!

Side1only:

( ) ( ) IAVE1 = 3A • 0.67 + 0A • 0.33 = 2.01A

( ) ( ) IRMS1 = 12 • 0.67 + –22 • 0.33 = 1.42ARMS

Side 2 only:

( ) ( ) IAVE2 = 10A • 0.32 + 0A • 0.68 = 3.2A

( ) ( ) IRMS2 = 6.82 • 0.32 + –3.22 • 0.68

= 4.66ARMS > 4.55ARMS

Consider the case where both sides are operating at the

same load, with a 50% duty cycle at each side. The RMS

current with both sides running is near zero, while the

RMS current with one side active is 1/2 the total load

current of that side.

In our hypothetical 1.6V, 10A example, we'd set the ripple

current to 40% of 10A or 4A, and the inductor value would

be:

L = ( tON(QB) VOUT ) = (1.2µs)(1.6V) = 0.64µH

IRIPPLE

3A

with tON(QB) = ⎛⎝⎜1− 15.6VV⎞⎠⎟ / 550kHz = 1.2µs

The inductor must not saturate at the expected peak

current. In this case, if the current limit was set to 15A, the

inductor should be rated to withstand 15A + 1/2 IRIPPLE,

or 17A without saturating.

1703fa

20

Share Link: