April 2005

11

M9999-062205

MIC2168A

Micrel, Inc.

100

60

37.5

15

7.5

30

30

60

1000000

100

f

Figure 4. The Gain Curve for G(s)

100

10

10

10

10

10

10

10

10

150

100

50

0

0

180

1000000

100

f

Figure 5. Phase Curve for G(s)

It can be seen from the transfer function G(s) and the gain

curve that the output inductor and capacitor create a two pole

system with a break frequency at:

1

=

f

C

ff

2 × π √ L × C

OUT

By looking at the phase curve, it can be seen that the output

system by introducing a zero at:

1

=

f

ZERO

ff

2 × π × ESR × C

OUT

From the point of view of compensating the voltage loop, it is

recommendedtousehigherESRoutputcapacitorssincethey

provide a 90° phase gain in the power path. For comparison

purposes, Figure 6, shows the same phase curve with an

ESR value of 0.002Ω.

100

150

100

50

0

0

180

1000000

100

f

Figure 6. The Phase Curve with ESR = 0.002Ω

It can be seen from Figure 5 that at 50kHz, the phase is

approximately –90° versus Figure 6 where the number is

–150°. This means that the transconductance error ampli-

fier has to provide a phase boost of about 45° to achieve a

closed loop phase margin of 45° at a crossover frequency

of 50kHz for Figure 4, versus 105° for Figure 6. The simple

RC and C2 compensation scheme allows a maximum error

amplifier phase boost of about 90°. Therefore, it is easier to

stabilize the MIC2168A voltage control loop by using high

ESR value output capacitors.

It is undesirable to have high error amplifier gain at high

frequencies because high frequency noise spikes would be

picked up and transmitted at large amplitude to the output,

thus, gain should be permitted to fall off at high frequencies.

At low frequency, it is desired to have high open-loop gain to

attenuate the power line ripple. Thus, the error amplifier gain

should be allowed to increase rapidly at low frequencies.

The transfer function with R1, C1, and C2 for the internal

equation:

1+ R1 × S × C1

×

g

m

s × (C1 + C2)

Error Amplifier (z) -









1 + R1 × C1

× C2 × S

C1 + C2

The above equation can be simplified by assuming

C2<<C1,

1+ R1 × S × C1

×

g

m

s × (C1)(1+ R1 × C2 × S)

Error Amplifier (z) -









From the above transfer function, one can see that R1 and

C1 introduce a zero and R1 and C2 a pole at the following

frequencies:

Figures 7 and 8 show the gain and phase curves for the above

transfer function with R1 = 9.3k, C1 = 1000pF, C2 = 100pF,

amplifier exhibits approximately 45° of phase margin.