L6562

8/16

Tj (°C)

(V)

-50

0

50

100

150

10

11

12

13

14

15

Vcc = 20 V

Tj (°C)

-50

0

50

100

150

0.5

0.6

0.7

0.8

0.9

1

1.1

Vcc = 0 V

Vpin7

(V)

4

Application Information

4.1 Overvoltage protection

Under steady-state conditions, the voltage control loop keeps the output voltage Vo of a PFC pre-regulator

close to its nominal value, set by the resistors R1 and R2 of the output divider. Neglecting ripple compo-

the error amplifier is internally referenced at 2.5V, also the voltage at pin INV will be 2.5V, then:

.

If the output voltage experiences an abrupt change

∆Vo > 0 due to a load drop, the voltage at pin INV will

be kept at 2.5V by the local feedback of the error amplifier, a network connected between pins INV and

COMP that introduces a long time constant to achieve high PF (this is why

∆Vo can be large). As a result,

the current through R2 will remain equal to 2.5/R2 but that through R1 will become:

.

The difference current

ter the error amplifier output (pin COMP). This current is monitored inside the L6562 and if it reaches about

37 µA the output voltage of the multiplier is forced to decrease, thus smoothly reducing the energy deliv-

ered to the output. As the current exceeds 40 µA, the OVP is triggered (Dynamic OVP): the gate-drive is

forced low to switch off the external power transistor and the IC put in an idle state. This condition is main-

tained until the current falls below approximately 10 µA, which re-enables the internal starter and allows

switching to restart. The output

∆Vo that is able to trigger the Dynamic OVP function is then:

.

An important advantage of this technique is that the OV level can be set independently of the regulated

output voltage: the latter depends on the ratio of R1 to R2, the former on the individual value of R1. Another

advantage is the precision: the tolerance of the detection current is 12%, that is 12% tolerance on

∆Vo.

Since

∆Vo << Vo, the tolerance on the absolute value will be proportionally reduced.

Example: Vo = 400 V,

∆Vo = 40 V. Then: R1=40V/40µA=1MΩ; R2=1MΩ·2.5/(400-2.5)=6.289kΩ. The tol-

erance on the OVP level due to the L6562 will be 40·0.12=4.8V, that is 1.2% of the regulated value.

I

R2

2.5

R2

--------

I

R1

Vo

2.5

–

R1

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

==

=

I'

R1

Vo

2.5

–

Vo

∆

+

R1

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

=

Vo

∆

R 1 40 10

6

–

⋅⋅

=