ADP3811AR datasheet(15/16 Pages) AD

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ADP3810/ADP3811

–15–

REV. 0

Step 9. Iterate CC1:

Because fZ1 is very close to fCV, it will increase the error ampli-

fier gain in a nonnegligible amount at the 0 dB point. The in-

crease in gain is calculated as:

× log 1+

f Z1

f CV









= 7.1dB

Now, the total error amplifier gain loss required is:

GLOSS = 37.6 dB + 7.1 dB = 44.7 dB

With this, the new fP1 can be calculated from the equation in

Step 5.

fP1 = 0.58 Hz

Finally, CC1 is recalculated using the equation in Step 6.

CC1 =

π× 400 kΩ× 0.58 Hz

≈ 0.7 µF

Following these steps gives a cookbook method for calculating

the compensation components for the voltage loop. As men-

tioned above, these components can be optimized in the actual

circuit. The results of a PSpice

1 analysis of the loop is shown in

Figure 32. The open loop gain of the loop is 108 dB as calcu-

lated. The crossover frequency is 100 Hz with a phase margin

of 52

°. The graph shows the phase leveling off at 90°. In reality

the phase will continue to fall as higher frequency parasitic poles

take effect.

1PSpice is a trademark of MicroSim Corporation.

FREQUENCY – Hz

–100

0.01

100k

0.1

100

200

100

180

0dB CROSSOVER

PHASE MARGIN = 52

Figure 32. Voltage Loop Gain/Phase Plots

Voltage Loop Compensation, Battery Present

When the battery has finished charging and is still connected to

the charging circuitry, the system is said to be “floating” the bat-

tery. The loop is maintaining a constant output voltage equal to

the battery voltage, and the output current has dropped to

nearly zero. This case is actually the easiest to compensate be-

cause the battery’s capacitance creates a very low frequency

dominant pole, giving a single pole response. For example, if the

battery is modeled as a 10 Farad capacitor, the dominant pole

will be 1/(2

π × 1.2 kΩ × 10 F) = 0.013 MHz. This very low fre-

quency pole causes the system to cross over 0 dB at less than

10 Hz, giving a stable single pole system. The compensation

components have little effect on this response, so no further

calculations are needed for this case.

Current Loop Compensation

Now that the voltage loop compensation is complete, it is time

to add the compensation for the current loop. The definitions

for modulator gain and error amplifier gain are the same as be-

fore; but now, the controlling error amplifier is GM1 in Figure

31, as opposed to GM2, for the voltage loop. Otherwise, the

calculations are very similar.

Step 10. Calculate the dc loop gain (GLOOP), fPM, and fZM:

GMOD = 20 × log GM 3 × ITXOC × RF × AV 2 ×GM 4 × RCS

[]

GMOD = 20 × log

6 mA /V

× 0.36 × 3.3 kΩ×

0.333

×1.0 A /V × 0.25 Ω









=−4.5 dB

GEA = 20 × log GM1× R5

[]=20log

8.3 mA /V

× 400 kΩ

[]=70.4dB

GLOOP = –6.1 dB + 70.4 dB = 64.3 dB

f PM =

π× R

CS + RF1

()×C

π×0.35 Ω×1mF

= 450 Hz

f ZM =

π× R

F1 × CF1

π×0.1Ω×1mF

=1.6 kHz

Step 11. Pick the current loop crossover frequency, fCI:

From Step 2 in the voltage loop calculations, fCI ~ 1.9 kHz.

Step 12. Calculate GMOD at fCI:

The modulator gain of –4.5 dB is the dc gain. The modulator

pole reduces this gain above fPM.

GMOD = 1.9 kHz

()=G

MOD dc

()−20×log 1+ fCI

f PM









+ 20 × log 1+

f CI

f ZM









GMOD = 1.9 kHz

()=−4.5dB−12.7dB+3.8dB = –13.4dB

If the 1 mF capacitor has a much higher ESR, e.g., 1

Ω, the

modulator zero, fZM, will be lower in frequency than the modu-

lator pole, fPM. This causes the loop gain and bandwidth to in-

crease and could cause instability. One possible solution to this

scenario is to use a much higher value (47 nF) for the CF = 1 nF

capacitor. The pole of this capacitor would then be in the 1 kHz

range and would reduce the loop gain. If the ESR is much less

than 0.1

Ω, the bandwidth of the loop will decrease slightly.

Step 13. Calculate gain loss of GEA at fCI:

The gain loss of GEA in the current loop is a combination of the

loss due to CC1, RC1 and the additional loss from CC2, RC2. To

calculate the contribution of gain roll-off needed from CC2, RC2,

the effective gain of GEA must first be calculated. Since the gain

is calculated at 1.9 kHz, the impedance of CC1 is 120

Ω. Thus,

the gain becomes:

GEA 1.9 kHz

()=20×log GM1× R6+R

C1 + 120 Ω

()

[]

GEA 1.9 kHz

()=20×log 8.3mA/V × 10320 Ω

()

[]= 38.9dB

GLOSS = GEA (1.9 kHz) – GMOD (1.9 kHz)

= 38.9 dB – 13.4 dB = 25.5 dB

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