How does an Error amplifier works?

[dann11]

I am using an ISL6840 and I want to understand its internal operation. And internally it has an error amplifier, Can you please explain to me how an error amplifier work?

 

[KlausST]

Hi,

An error amplifier is a simple "mathematical difference"

On one input there is the set point (here: 2.5V)
On the other input is the output voltage (named feedback)
--> on the output of the error amplifier you see the deviation of output voltage to set point. Or in other words: the error voltage of the output signal .

In the real circuit the output may be amplified and it may contain offset..

The use of wikipedia is always a good hint: https://en.m.wikipedia.org/wiki/Error_amplifier_(electronics)

Klaus

 

[dann11]

thanks.. but I cant still image how the output waveform will be like..

 

[KlausST]

Hi,

mathematically it is: V_out = V_offs + (V_FB - V_Ref) * Gain

To simplify the example let´s say offset = 0V and gain = 1, V_Ref = 2.5V

V_FB --> V_out
---------------------
2.5V --> 0 V (no error)
3.5V --> 1.0 V (1 V error)
1.5V --> -1.0 V (-1 V error)
2.501V --> 0.001V ( 1mV error)
and so on.

Klaus

 

[dick_freebird]

It's just an op amp, which may have one input pinned
to a reference (so no longer a classical, anything-goes
input range; likewise the output load should be well
known and constrained; the error amp may be a
somewhat degenerate form as a result, since its
application is quite fixed).

A transconductance error amp is easy to externally
compensate. But hardly anyone likes that type as
a piece-part for board use, since its output drive is
weak and its frequency response so dependent on
board level externalities.

 

[dann11]

Hi,

mathematically it is: V_out = V_offs + (V_FB - V_Ref) * Gain

To simplify the example let´s say offset = 0V and gain = 1, V_Ref = 2.5V

V_FB --> V_out
---------------------
2.5V --> 0 V (no error)
3.5V --> 1.0 V (1 V error)
1.5V --> -1.0 V (-1 V error)
2.501V --> 0.001V ( 1mV error)
and so on.

Klaus

Does this means that my output voltage will be the difference between my reference voltage and my input voltage?

 

[KlausST]

Hi,

Does this means that my output voltage will be the difference between my reference voltage and my input voltage?

-->
Yes, like said in post#2.

Klaus

 

[dick_freebird]

Does this means that my output voltage will be the difference between my reference voltage and my input voltage?

Input voltage difference times AVOL, or times gm*Rout,
centered on output common-mode point, depending on
the amplifier form (voltage op amp, or transconductance
amp).

 

[arya.jagadeesh]

Input voltage difference times AVOL, or times gm*Rout,
centered on output common-mode point, depending on
the amplifier form (voltage op amp, or transconductance
amp).

when both terminal of err amp are same what is your output is common mode output voltage..now on this common mode if you apply some differential signal..change in your output is AVOL times input signal difference