MIC2171 datasheet(12/24 Pages) MICROCHIP

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MIC2171

DS20006355A-page 12



2022 Microchip Technology Inc. and its subsidiaries

5.6.2

OUTPUT DIODE

The output diode allows T1 to store energy in its

primary inductance (D2 blocked/reverse biased) and

release energy into C4 (D2 forward biased); the low

forward voltage drop of a Schottky diode minimizes

power loss in D2.

5.6.3

FREQUENCY COMPENSATION

A simple frequency compensation network consisting

of R3 and C2 prevents output oscillations.

High impedance output stages (transconductance

type) in the MIC2171 often permit simplified loop

stability solutions to be connected to circuit ground,

although a more conventional technique of connecting

the components from the error amplifier output to its

inverting input is also possible.

5.6.4

VOLTAGE CLIPPER

Extra care must be taken to minimize T1’s leakage

inductance, otherwise it may be necessary to add the

voltage clipper consisting of D1, R4, and C3 in order to

avoid second breakdown (failure) of the MIC2171’s

internal power switch.

5.6.5

DISCONTINUOUS CONDUCTION

MODE DESIGN

When designing a discontinuous conduction mode

flyback converter, first determine whether the device

can safely handle the peak primary current demand

drawn by the output power. Equation 5-10 finds the

maximum duty cycle required for a given input voltage

and output power. If the duty cycle is greater than 0.8,

discontinuous operation cannot be used.

EQUATION 5-10:

Then:

VIN(MIN) = VIN – (ICLIM × RSW)

VIN(MIN) = 4V – 0.78V

VIN(MIN) = 3.22V

δ ≥ 0.74 (76%), less than 0.8, so discontinuous is

permitted. A few iterations of Equation 5-10 may be

required if the duty cycle is found to be greater than

50% since the ICLIM is a function of duty cycle when δ

> 50%.

The next step is to calculate the maximum transformer

turns ratio

a, or NPRI/NSEC, that will guarantee a safe

operation of the MIC2171’s power switch.

EQUATION 5-11:

For the practical example:

VCE = 65V max. for the MIC2171

FCE = 0.8

VSEC = 5.6V

Then:

EQUATION 5-12:

Next, calculate the maximum primary inductance

required to store the needed output energy with the

power switch duty cycle of 76%.

EQUATION 5-13:

Then:



2 POUT



ICLIM VIN MIN



 V

–





---------------------------------------------------------------



Where:

POUT =

5.0V × 0.5A = 2.5W

VIN =

4.0V to 6.0V

ICLIM =

1.25A when δ < 50%

1.67 (2 – δ) when δ ≥ 50%

VCE FCE VIN MAX





–



VSEC

------------------------------------------------------------



Where:

a =

Maximum transformer turn ratio

VCE =

Power switch collector to emitter

maximum voltage

FCE =

Safety derating factor (0.8 for most

commercial and industrial applica-

tions)

VIN(MAX) =

Maximum input voltage

VSEC =

Transformer secondary voltage

(VOUT + VF)

a 65 0.8 6.0

–



5.6

-------------------------------



a 8.2



(NPRI/NSEC)

VIN MIN



 



ICLIM fSW



--------------------------------- LPRI

0.5 fSW VIN MIN





2 t



POUT

---------------------------------------------------------------------------



Where:

LPRI =

Maximum primary inductance

fSW =

Device switching frequency (100 kHz)

VIN(MIN) =

Minimum input voltage

tON =

Power switch on time

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