AN3168

Examples of negative power supplies

Doc ID 17193 Rev 1

16/20

operates in discontinuous mode. Discontinuous mode has the advantage of reducing the

size and the cost of the inductance with respect to continuous mode.

Equation 8 gives a simplified relation between the average output DC current and the

inductance value for discontinuous mode (VIPer power losses not considered).

Equation 8

VIPer16 device, the typical peak current limitation is 400 mA. The minimum inductance in

discontinuous mode is given by:

Equation 9

The ratio between output and input voltages is used to define the VIPer duty cycle.

Equation 10

For a 230 V mains voltage, a 12 V output voltage and a 60 kHz switching frequency

(VIPer16 typical frequency), the average output current versus different inductances and the

typical standby power consumption are:

The advantage of a buck-boost converter compared to a buck converter is that there is no

need for an added output load resistance or output Zener (see Reference 3). For both

topologies, the feedback capacitor is still discharged with the IC feedback pin current,

whereas the output capacitor is not discharged if the output current is zero. The feedback

capacitor then indicates a lower output level than reality. Furthermore, this drawback is

amplified by the buck topology as the output capacitor is charged during each MOSFET on

time, whereas the feedback capacitor is not. So output voltage can increase to an excessive

value and has to be clamped.

The added resistance or clamping diode is then required at the buck output to avoid an

excessive output voltage in case of no-load or very light load.

With a buck-boost converter, the output capacitor is not charged during MOSFET on time.

Furthermore, the output voltage capacitor (C6) can not be charged if the feedback capacitor

(C2) voltage is lower than C6, as diode D2 is blocked. So there is no risk of an excessive

output voltage and the clamping device is not required.

Table 5.

Calculated average output DC current (

losses

L2 inductance

900 µH

800 µH

700 µH

600 µH (min. value)

100 mA

115 mA

130 mA

155 mA

OUT

2

)

peak

(

mains

2

OUT

V

f

2

L

2

V

I

×

×

×

×

α

=

f

I

V

2

L

lim

D

)

peak

(

mains

(min)

×

α

×

=

α

−

α

=

1

V

V

e

s

e

s

s

V

V

V

+

=

α

so