L6562A

Application information

13/26

7

Application information

7.1

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

2.5V, also the voltage at pin INV will be 2.5V, then:

Equation 1

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:

Equation 2

The difference current

∆I

network and enter the error amplifier output (pin COMP). This current is monitored inside

the device and if it reaches about 24µA the output voltage of the multiplier is forced to

decrease, thus smoothly reducing the energy delivered to the output. As the current

exceeds 27µA, 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 maintained

until the current falls below approximately 7µA, 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:

Equation 3

∆V

- 6

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 13%, i.e. 13% tolerance on

∆Vo. Since ∆Vo << Vo, the tolerance on the absolute

value will be proportionally reduced.

Example: Vo = 400V,

∆Vo = 40V. Then: R1 = 40V/27µA

R2 = 1.5 M

Ω ·2.5/(400-2.5) = 9.43kΩ. The tolerance on the OVP level due to the L6562A will

be 40·0.13 = 5.3V, that is ± 1.2%.

I

R2

I

R1

2.5

R2

--------

V

O

2.5

–

R1

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

==

=

I'

R1

V

O

2.5

–

V

O

∆

+

R1

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

=