You can, but you do not need to connect OA output to any pin.
You should not do it if you do not know what to do with it.

--- In m...@yahoogroups.com, "abufadel" wrote:
>
> Greetings,
> If I want to connect the OA0 output to ADC10 input, I can connect them
> to external pin A1 for example. What happens to the noise on the pin
> in this case? Do I need to connect anything externally on A1 to
> prevent outside noise sources from degrading the signal?
>
> Thanks,
>
> A
>

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



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--- In m...@yahoogroups.com, "old_cow_yellow"
wrote:
>
> You can, but you do not need to connect OA output to any pin.
> You should not do it if you do not know what to do with it.
>

I guess I am having trouble routing the OA output signals. I am
trying to implement a 2 OA Differential Amp as in the 2xx user manual
with its output connected to the ADC10 input w/o connecting to any
external pins.

My question is about OAADCx. When OAFCx > 0, OAADCx = 0 is used for
internal routing. To connect to channel A1 for ADC10, OAADCx = 3, but
it is also connected to external pins. How can I avoid this?

For sanity check, here are a few lines of code of what I am doing:

//Setup OpAmps as Differential Amp
OA0CTL0 = OAPM_1 + OAADC0; // +ve input OA0I0
(default) and internal routing for output and slow slew rate
OA0CTL1 = OAFC_7; // Differential Amp
OA1CTL0 = OAN_3 + OAP_2 + OAPM_1 + OAADC3;//-ve input as OA1IB +ve
input as OA1I1, slow slew rate and ADC10 input A1
OA1CTL1 = OAFBR_7 + OAFC_6; // inverting with PGA = 15

//Setup ADC10 with internal reference 1.5V and interrupts
ADC10CTL0 = SREF_1 + ADC10SHT_2 + REFON + ADC10ON + ADC10IE;
ADC10CTL1 = INCH_1;

Does it make sense?

Thank you,

-- A
------------------------------------



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Near the end of the Data-sheet, there is an entire section describing
how internal signals (such as the output of the OA) are routed to the
pins. The title of this section is "APPLICATION INFORMATION". It has
schematics and tables; both try to tell you what control-bits/signals
determine what internal signals are connected to I/O pins.

--- In m...@yahoogroups.com, "abufadel" wrote:
>
> --- In m...@yahoogroups.com, "old_cow_yellow"
> wrote:
> >
> > You can, but you do not need to connect OA output to any pin.
> > You should not do it if you do not know what to do with it.
> > I guess I am having trouble routing the OA output signals. I am
> trying to implement a 2 OA Differential Amp as in the 2xx user manual
> with its output connected to the ADC10 input w/o connecting to any
> external pins.
>
> My question is about OAADCx. When OAFCx > 0, OAADCx = 0 is used for
> internal routing. To connect to channel A1 for ADC10, OAADCx = 3, but
> it is also connected to external pins. How can I avoid this?
>
> For sanity check, here are a few lines of code of what I am doing:
>
> //Setup OpAmps as Differential Amp
> OA0CTL0 = OAPM_1 + OAADC0; // +ve input OA0I0
> (default) and internal routing for output and slow slew rate
> OA0CTL1 = OAFC_7; // Differential Amp
> OA1CTL0 = OAN_3 + OAP_2 + OAPM_1 + OAADC3;//-ve input as OA1IB +ve
> input as OA1I1, slow slew rate and ADC10 input A1
> OA1CTL1 = OAFBR_7 + OAFC_6; // inverting with PGA = 15
>
> //Setup ADC10 with internal reference 1.5V and interrupts
> ADC10CTL0 = SREF_1 + ADC10SHT_2 + REFON + ADC10ON + ADC10IE;
> ADC10CTL1 = INCH_1;
>
> Does it make sense?
>
> Thank you,
>
> -- A
>

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



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