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Could you Review This Schematic Please?

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ilovecplusplus

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I'm sorry for the crudeness of the drawing...

There are some HM 6116 Static-RAMs that I have, and I wanted to see if they worked.

I'm only going to use two address lines and one I/O bus line to simplify things.


Is there anything wrong that I'm doing? I'm especially worried about that I/O bus line and the LED.



thank you
 

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My brain isn't working very well right now, vbut I thinks you should incorporate a tri-state inverter and/or buffer in your design such as this one: https://www.futurlec.com/74LS/74LS368.shtml

The chip is not likely able to operate the LED either. An Open collector inverter is a typical way to drive a LED, but a ULN2003 is one of my favorite. The former's inputs would float high,this for an open condition, you would get a lit LED. In the later a floating inpput will result in an unlit LED.

You can look at the values with the ULN2003 or the inverter.

But for input, you need to use the tri-state part. Only when the system is write enabled, would you enable the tri-state buffer.

My brain is a litle foggy right now. Hopefully I gave you some ideas.
 
I hope I got this right

I hope I'll get this right the first time--is it?


**broken link removed**

When I want to read the value, I "open" the tristate buffer.


Oh wait, and you also said that the chip's not going to operate the LED... well, ignoring that for a moment, did I get it right?
 
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NA!!

Connect a standard inverter to the I/O pin to a LED like you did. i.e. remove the 10K, the CNTRL and the switch and replace the IC with an open collector inverter.

~WE of the 6116 controls the CTRL signal. When ~WE is enabled, the tri-state buffer should be enabled so you may need an extra inverter for that. (or an OC inverter with a pull-up) So a connection nneds to be made from ~WE of the 6116 with or without an inverter to enable the tri-state thingy when ~CE is low.

The input of the buffer should be +5 resistor, and a switch to ground. This will get you a logic level depending on the position of the switch.

The OUTPUT of the tri-state buffer/inverter, would then connect to the I/O pin.

Aside: using OC. inverters might be easier. Remember a resistor turns an OC inverter into an inverter and a diode converts a normal inverter to an OC inverter.
 
"remove the 10K, the CNTRL and the switch and replace the IC with an open collector inverter."

After that, it's this:

|||-----LED-----resistor1.2K------oc inverter-------\\\\\I-O\\pin\\\\\

But then, you started to talk about the tri-state buffer, which you told me to get rid of.



Please, be really patient with me, and don't abandon me! I'm trying to do my best with my feeble mind. I know what you're going through... I develop software (hence my name) and in my programming forum, it's pretty annoying if somebody doesn't understand something when I had explained it as simple as it would get.
I think it'd rather be quicker if you could draw something up quickly in Paint or something. If I have it in my face, it'd be easier to understand.

Thank you so much for all the help so far.
 
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yea, like double-bubble sorts, semephores tokens and queue's. then there is bit banging and words, nibbles and bits. And of course, big and little endian.

All of the drawing packages I have so far really suck. I really loved Vectorworks which is too expensive for my budget. It was so intuitive. This was done in DIPTRACE which is a little dippy. Ascii text instead of continuing, you have to place it on the next line.

OK, so I hacked something together. I didn't follow the logic completely through because I
didn't select a Tri-state device.

OOPS. I forget the resistor in series with the LED.
 

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**broken link removed**

This is what I got. I think I understand it much better now. Thank you very much for such a kind answer.

Although... one thing that irks me still is why the 5V is going through the LED, through the ULN2003, and to the IO pin. Isn't it supposed to be ground--since the IO pin is going to "output" the signal? Or is the IO pin completing a circuit when its value is 1 and thus, the LED is lighting up that way?
 
I said I missed a resistor. There should be (+5) (resistor) (LED) (ULN2003). The ULN2003 has a ground pin, but no power pin.
Ressistor = (5 - 0.6 -1.2)/15e-3 or so ohms.

When the input to the ULN2003 is high, the led will light, because the OUTPUT of the ULN2003 sinks to ground.

When the input to the ULN2003 is not connected, the LED will be off.

When the input is above the TTL threshold, the LED will be on.

Just think of the ULN2003 as a bunch of transistors. The base is the input of the ULN2003 with a series resistor. The output is the Open collector. All of the emitters are connected to ground. The COM pin has to do with relay suppression.
 
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Ok. I understand much better now.

**broken link removed**

So... this is the final, I guess. I can start building this, right?

Also, the 10KOhm resistor will be enough, I hope.



thank you for all the help you have given me.
 
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