Re: Parallel op amps current

On Apr 10, 3:57 am, "Winfield" <winfieldh...@xxxxxxxxx> wrote:
Richard Crane wrote:
Winfield wrote...
At low frequencies this can be nicely done using the old
ballast-resistor current-sharing trick,

. -- feedback network ---,
. normal opamp |
. ------| >--+--/\/--+-+----
. _______| |
. | add-on opamps |
. | as followers |
. +--| >-----/\/--+
. | |
. +--| >-----/\/--+
. : :

Will this work if the first stage (only) has gain and a
filter cap in the feedback? IOW the second stage is unity.

Or, do both stages need to be identical?

Well, the 2nd opamp is inside the feedback loop along
with the first. We've added an output-current-sense
resistor to the first opamp and made the second one
deliver the same current. For half of an LM358 with
RC feedback it'd be something like this.

. ,-----||------, feedback network
. +----/\/\/----+-----,
. | __ |
. ---+--|- \ 4.7 |
. | >--+-/\/\--+-+----
. ---|+_/ | |
. _________| |
. | __ |
. '--|+ \ 4.7 |
. | >--+-/\/\--'
. ,--|-_/ |
. |_________| LM358

A 4.7-ohm resistor will drop 100mV at 20mA, and for an
LM358 with 7mV input-offset voltage max, the standing
opamp-opamp output current would be Vos / 2Ro = 700uA.


Doing this with faster op-amps will sometimes lead to unstable
operation. Changing the location of the feddback capacitor so that
the output of the first op-amp is used for it instead of the over all
output will often solve this. The LM358 is slow enough that I doubt
it will be an issue.

.

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