L6258EX

PWM current control loop

21/32

In this case the Bx block has a DC gain equal to the open loop and equal to zero at a

frequency given by the following formula:

In order to cancel the pole of the load, the zero of the Bx block must be located at the same

frequency of 163Hz; so now we have to find a compromise between the resistor and the

capacitor of the compensation network.

Considering that the resistor value defines the gain of the Bx block at the zero frequency, it

is clear that this parameter will influence the total bandwidth of the system because,

annulling the load pole with the error amplifier zero, the slope of the total transfer function is

-20dB/decade.

So the resistor value must be chosen in order to have an error amplifier gain enough to

guarantee a desired total bandwidth.

In our example we fix at 35dB the gain of the Bx block at zero frequency, so from the

formula:

where: Rb = 20k

Ω

we have: Rc = 1.1M

Ω

Therefore we have the zero with a 163Hz the capacitor value:

Now we have to analyse how the new Aloop transfer function with a compensation network

on the error amplifier is.

The following bode diagram shows:

–

the Ax function showing the position of the load pole

–

the open loop transfer function of the Bx block

–

the transfer function of the Bx with the RC compensation network on the error

amplifier

–

the total Aloop transfer function that is the sum of the Ax function plus the transfer

function of the compensated Bx block.

Bx

Zc

Rb

--------

–

Rc

j

1

2

π fCc

⋅⋅

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

–

Rb

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

–

==

Fzero

1

2

π Rc Cc

⋅⋅

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

=

Bx_gain

@ zero freq.

20

Rc

Rb

--------

log

⋅

=

Cc

1

2

π Fzero Rc

⋅

⋅

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

1

6,28

163

1,1

10

6

–

⋅⋅

⋅

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

880pF

==

=