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Summary: 2 Replies, Latest post by mmoron21 on 08 Oct 2010 01:24 PM PDT
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user_48311473's picture
71 posts

In the doccuments AN58827 AN58304 talks about the internal paths the Users Modules track to reach
the GPIO and the equivalent path resistance.
Concretely into the Pin Selection paper (AN58304) states in page 4:

Several direct routes bypass the analog muxbus, analog
globals, and the analog local bus to connect analog blocks
directly to GPIOs....
The op-amps also have direct connections between their
terminals and dedicated GPIOs.

OPAMP  GPIO Pin (Non-inverting)    GPIO Pin(Inverting)    GPIO Pin (Output)
Opamp0 P0[2] P0[3] P0[1]
Opamp1 P3[5] P3[4] P3[6]
Opamp2 P0[4] P0[5] P0[0]
Opamp3 P3[3] P3[2] P3[7]

These direct routes are limited to Port0 and Port3.

But before in page 3 sais:
Analog blocks may be interconnected to each other or to GPIOs. The small circles
indicate switches that connect the paths that intersect. The On resistance
of the switches colored red is about 200 ohms. The On resistance of the white
switches is about 870 ohms. Note that the connection between the pins and
analog globals or AMUXBUS is with the lower resistance switches.

What is confirmed in the Routing Considerations paper (AN58827):

[b]Small Switch (colored white) ~ 870Ω
Large Switch (colored red) ~ 200Ω[/b]
Analog Global AGL[7:0], AGR[7:0] ~ 200Ω
Analog Mux Bus AMUXBUSR, AMUXBUSL ~ 100Ω
Analog Local Bus AbusR[3:0], AbusL[3:0] ~ 100Ω

My question is:
Following the internal routing schemes (Page 2 or 5 previous related documents), the OpAms
are suggested to use with ports P0 or P3 because are shorter and "bypass the analog muxbus, analog
globals, and the analog local bus to connect analog blocks
directly to GPIOs....". But after that the conexions are tied to GPIO with white
switch, which are 870Ω. You can see in the 4 opamps (upper left and right).

Why is possible they try to shorter the path and after that they add the larger
resistance switch 870Ohms (white), instead the red one (200 Ohms).
That has some meaning or I'm wrong?

I attach the doccuments

Thanks a lot.

user_48311473's picture
71 posts

I got this answer in PsOC developer (by abitkin):

Well, so here's the thing, the white switches are high impedance, where as the red switches are low impedance.

If you check the inputs of the OpAmp, both negative and positive sides all the connections are using high impedance switches. So, if you use a directly connected pin, you'll go through one high impedance switch. If you use a pin that's not directly connected you'll get again a high impedance switch from the analog global wire/analog mux bus as well as a low impedance switch to get on that wire.

So not only is your total resistance more (high impedance + low impedance) you also use more power to drive the wire. As for the output of the OpAmp, that of course is fixed.

There isn't a path to the OpAmp input that doesn't include at least one high impedance switch, including the follower mode switch.

user_48311473's picture
71 posts

I have another question.
I have calculate the Rint due to the white switch before the OpAmp1 output reach the port P0[1]. It is stated
in the papers it mus to be arround 870 Ohms and I got 10,8Ohms.
That measure match with my measurements of the OpAmp output  saturation voltage drop.
Instead of 3.3V I have 3.252V and myoutput OpAmp current is 3mA.

Its true they don't say you in the AN58827 paper the tolerance margins, but we are talking
about a diference arround 1000% percent.
That's possible?

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