TI-83

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Hi i'm trying to connect a TI-83 to an avr so i can communicate with different TI's via rfm12 from hoperf.

My doubts are now relative to the TI Link Port hardware because in here (http://zastava.student.utwente.nl/linkguide/hardware.html) and other sources it states Vcc -> 10k resistor -> diode -> A

I'm assuming that the transistor is off given i get 5.3V in point A and makes sense according to protocol. To test if this circuit was like this or not i tried using different values of resistance to the ground
Vp -> 10k resistor -> diode -> A -> My resistor -> ground (common to calc ofc)

Bu the readings I get are for instance:
MyR - V point A
9.89K - 3.489V
995 - .858V
4.94K - 2.587V

And from this values I am not able to verify that circuit, because it seems more like:
Vp -> 5219Ohm -> point A -> ...

i = (Vp - Vd - Vm) / Rin

For i=352uA (MyR=9.89k) with Rin=10K the drop on the diode would be -1.7V...

So maybe the 10K value was wrong so i used 2 measure i values for different MyR values and made Rin=Rin

Rin = (Vp-Vd-Vm)/i

i1=i2 <-> (Vp-Vd-Vm1)/i1 = (Vp-Vd-Vm2)/i2 wich gives me Vd=0.0194V => Rin=5175ohm (ouch)

Could it there be such a low drop diode?
Is anything wrong with my thinking here?
Appreciate any help, thanks! :)

As for the connection to the AVR for instance in the fastAvrlink http://zastava.student.utwente.nl/linkguide/cable.html the pins from the AVR go directly to point A. But the VA when high is higher than 5V, isn't this bad to the port? I know the pins can take for instance Vcc+0.5 but is it wise to take this to the limit?
In other pic of the schematic of that same fastAvrLink a pull up to Vcc can be seen in connected to the avr pins connected to the TI but again if Vp > Vcc current would flow to Vcc...

So i thought of something like
point A -> ?ohm resistor -> uC pin -> Zener4v7 -> ground

Again, any help here is much appreciated!

Tiago

I'm having trouble trying to understand what you're asking.

Does the TI pull the signals above 5V? If not, then you don't have a problem. Adding a 4V7 diode would add some protection.

If you think a link between two TI calculators, one with fresh batteries (4x1.6V=6.4V) and one with near empty batteries (I don't know how low it goes but lets assume 4x1.0V=4V). They do not go up in smoke despite another has more than VCC on it's inputs.

The thing is the 10k resistor. Or whatever the value is. CMOS inputs have input clamping diodes, so they clamp to VCC+0.5V approximately as you said, and can clamp to that voltage when current flows. The 10k resistor just limits the current.

6V VCC -> diode -> 10k resistor -> CMOS input -> diode -> 4V VCC. Assuming diodes were ideal, there would be 2V drop over 10k resistor. That is 200uA, which the input diode will clamp happily. A 2k resistor instead of 10k would limit the current to 1mA, that should be within safe limits too.

A zener would help yes, but they have a very soft knee, so 4V7 is too low, try 5V1 or 5V6 zener since currents are so low.

You are right, it is not a good thing to design things to depend on input clamping diodes. But, there is no other protection between two TIs either when they have different voltages.

Edit: you do know that voltage drop over diode depends on current trough it, so the voltage drop is different when you have a different test resistor there. Best thing to measure the internal circuitry is to just measure battery voltage, voltage on pin when unloaded (multimeters have around 10-20 Mohms input resistance) and measure current when you short-circuit to ground, and approximate from there. You can also think the diode is ideal, since we don't know if it is a schottky or normal diode inside, unless you measure voltage drop over it.

Kartman wrote:
I'm having trouble trying to understand what you're asking.

Does the TI pull the signals above 5V? If not, then you don't have a problem. Adding a 4V7 diode would add some protection.

Yes an no. Depends on the battery state!

Jepael

So are you saying i can connect the point A directly to avr pins because of the avr internal claping diode pin -> diode -> vcc and thus looking at the vp -> resistor -> diode -> point A -> pin the voltage in the pin should get increase to bigger than Vcc+0.5? Another doubt here is, the current wouldn't be running to my 7805 this way?

yes i get what you say about two TI calculators and they not going to smoke because of the diode.

What do you mean soft knee?

I get 1.02mA when connecting point A directly to ground.

which is weird because Vp - R*i - Vd = 0
and Vp - R*i would give -4.9V :/

Thanks a lot :)

If I were making an AVR link cable right now, I would just connect the pins directly to AVR, without any components, it would work. But for completeness and now when the TI internal circuit is known, I'd put the 470 ohm series resistor there and maybe 10 kohm pull-ups to 5V, maybe via diodes too.

In fact, back then when I built the LPT port cable with just two diodes, I now would build it without any diodes, just straight wire, but it would require different LPT port pins so existing software would not be compatible with it.

Surely the AVR consumes about 10mA, so even 2mA of clamping current would still leave 8mA to be sourced from the 7805. Current won't be running towards 7805.

Yes the internal diodes in AVR would clamp the voltage to about VCC+0.5V.

Soft knee :
A zener is not an ideal device, with 0 current before treshold or infinite current after treshold.
It has the given voltage drop like 5.1V over at some specified current, like 10mA. Less current, and voltage is lower, more current, and voltage is higher. It means that the zener could have about 4.5V over it when 1mA current flows, i.e. it starts to conduct early.

1.02mA means there could be about 4.7kohm pull-up resistor inside, as voltage was 5.3V with no load. We don't know your battery voltage, and even then we don't know the diode type so calculations are correct enough. The actual resistance value does not matter, it high enough and just means there is a bit more current available that needs to get clamped inside avr.

But if avr was in low power state consuming less than 1mA then that current would make a difference?

I get what you say about the zenner i think, since the current will be low a higher drop voltage zener will be require so that the real voltage across the zener would be considered a high value by the avr?

Why would you put those pull-up with diodes?

thanks a lot for the help :)

Eviltelm wrote:
But if avr was in low power state consuming less than 1mA then that current would make a difference?

Yes, it would. As an example, I once had to work with improperly designed hardware, where AVR was powered with 3.3V, but a 5V device was connected to AVR input via about 1k resistance. Well, at least AVR had more supply voltage, and some leds got extremely bright.

On the other hand, there is little point putting the AVR to low current sleep, as the 7805 will draw 4 to 10 mA by itself, even if nothing is connected after it. Not very low power compared to the AVR.

Eviltelm wrote:

I get what you say about the zenner i think, since the current will be low a higher drop voltage zener will be require so that the real voltage across the zener would be considered a high value by the avr?

Eviltelm wrote:

Why would you put those pull-up with diodes?

thanks a lot for the help :)

Well, are there diodes in the TI too? That's why. They prevent current flowing to supply voltage via pull-up resistors if input voltage is higher than supply. Great in the case the AVR is powered off while cable is connected, as batteries won't drain through resistors.

Then maybe i can power the avr from a more efficient way? The final purpose is to put a pcb with the avr and the rf inside the ti case and powered by it's batteries. So current consumption would be an issue... unless i use a switch to turn the circuit on or off whenever i want hmm

Yes i get what the diode job is, but why to put the pull-ups there?

thanks :)

You could get an AVR that can work with the voltage range the batteries give. Most work up to 5.5V, but you could just put a diode there to drop the voltage, so you could use it without a regulator. Other options are better linear regulators or switch mode regulators.

Why pull-ups? Since you seem to connect the AVR to the link port permanently, indeed, you might not need them at all.

But I just want to remind, that the link port is meant to operate with another identical device, with all the pull-ups and diodes etc.

Yes you are right, but maybe i can do it without the zener as you stated previously mostly because when not pulling the line low it will be sourcing current through the zener wasting the batteries :/

So maybe i use the Vcc->10k->diode->pointB and another resistor before the pin, but should the resistor be between pointB and the pin or after pointB?

Thanks for the help :)

So i've been thinking a way to reduce consumption when in idle and how about this way?