MIC5190YMM-TR Datasheet MIC5190YMM, MIC5190YML-TR, MIC5190YMM-TR | P7-P9

MIC5190 Micrel

Applications Information

Designing with the MIC5190

Anatomy of a transient response

The measure of a regulator is how accurately and effectively

it can maintain a set output voltage, regardless of the load's

power demands. One measure of regulator response is the

load step. The load step gauges how the regulator responds

to a change in load current. Figure 2 is a look at the transient

response to a load step.

Figure 2. Typical Transient Response

At the start of a circuit's power demand, the output voltage is

regulated to its set point, while the load current runs at a

constant rate. For many different reasons, a load may ask for

more current without warning. When this happens, the regu-

lator needs some time to determine the output voltage drop.

This is determined by the speed of the control loop. So, until

enough time has elapsed, the control loop is oblivious to the

voltage change. The output capacitor must bear the burden

of maintaining the output voltage.

Since this is a sudden change in voltage, the capacitor will try

to maintain voltage by discharging current to the output. The

first voltage drop is due to the output capacitor's ESL (equiva-

lent series inductance). The ESL will resist a sudden change

in current from the capacitor and drop the voltage quickly. The

amount of voltage drop during this time will be proportional to

the output capacitor's ESL and the speed at which the load

steps. Slower load current transients will reduce this effect.

Placing multiple small capacitors with low ESL in parallel can

help reduce the total ESL and reduce voltage droop during

high speed transients. For high speed transients, the greatest

voltage deviation will generally be caused by output capacitor

ESL and parasitic inductance.

After the current has overcome the effects of the ESL, the

output voltage will begin to drop proportionally to time and

inversely proportional to output capacitance.

Output voltage variation will depend on two factors: loop

bandwidth and output capacitance. The output capacitance

will determine how far the voltage will fall over a given time.

With more capacitance, the drop in voltage will fall at a

decreased rate. This is the reason that more capacitance

provides a better transient response for the same given

bandwidth.

The time it takes for the regulator to respond is directly

proportional to its bandwidth gain. Higher bandwidth control

loops respond quicker causing a reduced drop on the supply

for the same amount of capacitance.

Final recovery back to the regulated voltage is the final phase

of transient response and the most important factors are gain

and time. Higher gain at higher frequency will get the output

voltage closer to its regulation point quicker. The final settling

point will be determined by the load regulation, which is

proportional to DC (0Hz) gain and the associated loss terms.

There are other factors that contribute to large signal tran-

sient response, such as source impedance, phase margin,

and PSRR. For example, if the input voltage drops due to

source impedance during a load transient, this will contribute

to the output voltage deviation by filtering through to the

output reduced by the loops PSRR at the frequency of the

voltage transient. It is straightforward: good input capaci-

tance reduces the source impedance at high frequencies.

Having between 35° and 45° of phase margin will help speed

up the recovery time. This is caused by the initial overshoot

in response to the loop sensing a low voltage.

Compensation

The MIC5190 has the ability to externally control gain and

bandwidth. This allows the MIC5190 design to be individually

tailored for different applications.

In designing the MIC5190, it is important to maintain ad-

equate phase margin. This is generally achieved by having

the gain cross the 0dB point with a single pole 20dB/decade

roll-off. The compensation pin is configured as Figure 3

demonstrates.

Error Amplifier Driver

3.42MΩ

20pF

Internal

External Comp

Figure 3. Internal Compensation

∆V L

∆V

idt=∫

∆V

idt↓=

↑

∫

∆V

idt↓= ∫ ↓

∆V L

↓=

↑

∆V L

↓= ↓

Time

∫

idt

V =

Load Current

Output Voltage

AC-Coupled

Output voltage vs. time

during recovery is

directly proportional to

gain vs. frequency.

∆ V = L

December 2005 7 M9999-120105

MIC5190 Micrel

This places a pole at 2.3 kHz at 80dB and calculates as

follows.

MpF

F kHz

××

2 3 42 20

2 32

Ω

-20

100

0.01 0.1 1 10 100 1000 10000 100000

Frequency (KHz)

Gain (dB)

-45

135

180

225

Phase (Deg)

Figure 4. Internal Compensation

Frequency Response

There is single pole roll off. For most applications, an output

capacitor is required. The output capacitor and load resis-

tance create another pole. This causes a two-pole system

and can potentially cause design instability with inadequate

phase margin. External compensation is required. By provid-

ing a dominant pole and zero–allowing the output capacitor

and load to provide the final pole–a net single pole roll off is

created, with the zero canceling the dominant pole. Figure 5

demonstrates placing an external capacitor (C

COMP

) and

resistor (R

COMP

) for the external pole-zero combination.

Where the dominant pole can be calculated as follows:

Error Amplifier Driver

3.42MΩ

20pF

Internal

External Comp

COMP

Figure 5. External Compensation

M C

COMP

××

2 3 42

.Ω

And the zero can be calculated as follows:

COMPCOMP

××

This allows for high DC gain, and high bandwidth with the

output capacitor and the load providing the final pole.

Figure 6. External Compensation

Frequency Response

It is recommended that the gain bandwidth should be de-

signed to be less than 1 MHz. This is because most capaci-

tors lose capacitance at high frequency and becoming resis-

tive or inductive. This can be difficult to compensate for and

can create high frequency ringing or worse, oscillations.

By increasing the amount of output capacitance, transient

response can be improved in multiple ways. First, the rate of

voltage drop vs. time is decreased. Also, by increasing the

output capacitor, the pole formed by the load and the output

capacitor decreases in frequency. This allows for the increas-

ing of the compensation resistor, creating a higher mid-band

gain.

Figure 7. Increasing Output Capacitance

This will have the effect of both decreasing the voltage drop

as well as returning closer and faster to the regulated voltage

during the recovery time.

MOSFET Selection

The typical pass element for the MIC5190 is an N-Channel

MOSFET. There are multiple considerations when choosing

a MOSFET. These include:

• V

to V

OUT

differential

• Output current

• Case size/thermal characteristics

• Gate capacitance (C

ISS

<10nF)

• Gate to source threshold

-20

100

0.01 0.1 1 10 100 1000 10000 100000

Frequency (KHz)

Gain (dB)

-45

135

180

225

Phase (Deg)

The Dominant Pole

External Zero

LOAD

OUT

Pole

comp

× ×

42.32

comp

-20

100

0.01 0.1 1 10 100 1000 10000 100000

Frequency (KHz)

Gain (dB)

-45

135

180

225

Phase (Deg)

Increasing C

OUT

reduces

the load resistance and

output capacitor pole

allowing for an increase

in mid-band gain.

December 2005 8 M9999-120105

MIC5190 Micrel

The V

(min) to V

OUT

ratio and current will determine the

maximum R

DSON

required. For example, for a 1.8V (±5%) to

1.5V conversion at 5A of load current, dropout voltage can be

calculated as follows (using V

(min)):

1 71V 1 5V

R m

DSON

IN OUT

OUT

DSON

−

()

−

()

42 Ω

Running the N-Channel in dropout will seriously affect tran-

sient response and PSRR (power supply ripple rejection). For

this reason, we want to select a MOSFET that has lower than

42mΩ for our example application.

Size is another important consideration. Most importantly,

the design must be able to handle the amount of power being

dissipated.

The amount of power dissipated can be calculated as follows

(using V

(max)):

= (V

– V

OUT

) × I

OUT

= (1.89V – 1.5V) × 5A

= 1.95W

Now that we know the amount of power we will be dissipating,

we will need to know the maximum ambient air temperature.

For our case we’re going to assume a maximum of 65°C

ambient temperature. Different MOSFETs have different

maximum operating junction temperatures. Most MOSFETs

are rated to 150°C, while others are rated as high as 175°C.

In this case, we’re going to limit our maximum junction

temperature to 125°C. The MIC5190 has no internal thermal

protection for the MOSFET so it is important that the design

provides margin for the maximum junction temperature. Our

design will maintain better than 125°C junction temperature

with 1.95W of power dissipation at an ambient temperature of

65°C. Our thermal resistance calculates as follows:

So our package must have a thermal resistance less than

31°C /W. Table 1. shows a good approximation of power

dissipation and package recommendation.

Package Power Dissipation

TSOP-6 <850mW

TSSOP-8 <950mW

TSSOP-8 <1W

PowerPAK™1212-8 <1.1W

SO-8 <1.125W

PowerPAK™ SO-8 D-Pack <1.4W

TO-220/TO-263 (D

Pack) >1.4W

Table 1. Power Dissipation and

Package Recommendation

In our example, our power dissipation is greater than

1.4W, so we’ll choose a TO-263 (D

Pack) N-Channel

MOSFET. θ

is calculated as follows.

= θ

+ θ

Where θ

is the junction-to-case resistance, θ

is the

case-to-sink resistance and the θ

is the sink-to-ambient

air resistance.

In the D

package we’ve selected, the θ

is 2°C/W. The

, assuming we are using the PCB as the heat sink, can

be approximated to 0.2°C/W. This allows us to calculate

the minimum θ

= θ

– θ

= 31°C/W – 0.2°C/W – 2°C/W

= 28.8°C/W

Referring to

Application Hint 17, Designing PCB Heat

Sinks

, the minimum amount of copper area for a D

Pack

at 28.8°C/W is 2750mm

(or 0.426in

). The solid line

denotes convection heating only (2 oz. copper) and the

dotted line shows thermal resistance with 250LFM air-

flow. The copper area can be significantly reduced by

increasing airflow or by adding external heat sinks.

Figure 8. PC Board Heat Sink

Another important characteristic is the amount of gate

capacitance. Large gate capacitance can reduce tran-

sient performance by reducing the ability of the MIC5190

to slew the gate. It is recommended that the MOSFET

used has an input capacitance <10nF (C

ISS

T max T ambient

125 C65 C

1.95W

C W

()

−

()

°− °

=°31 /

PC Board Heat Sink

Thermal Resistance vs. Area

December 2005 9 M9999-120105

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MIC5190YMM-TR

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