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I have a problem in a circuit

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7rots51

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who makes the 74lvt541

Hi
I have a problem in design a digital system.

I have a circuit that have 8 (TTL-74hc541) digital outpus(DO1~DO8).

I have to use two of these circuit in a system and connect the similar digital outputs together.(ex. DO1 of one
circuit is connected to DO1 of another one ,and so on).The digital outputs are inputs of another card.

at the same time only one of two circuit can be powered on ,the other one must be powered off.

Is this connection makes problem for the off circuit ?
(digital outputs are 74HC541)

or simply I say, can I connect outputs of two 74HC541 together and in state that on of them is powered off,is this make problem
for another 74hc541??
How can I make a better circuit with no risk of failure and design error??
Is open collector design a better choice?

Please give your comments.
 

klug

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Hi RTOS 51!


If your IC (74hc541) has open collector out stage maybe all will be ok.
But if it has push-pull output than it will be not correctly to connect two outputs together.

Try simple circuit - connect outputs through diodes - minuses of diodes connect together and in this point put resistor 10K to ground, take signal to your next stage from this point.

But you still will have problems with intermediate state when you will switch supply from one IC to another.

Best wishes! klug.
 

klug

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hmmm...

10K is too big to short next input to ground - it will be better 1K - 2K.
Use it with some risk - anyway this circuit is not correct enough...

It is better to use 7432 noninverting OR to unite these outputs.

Best Wishes! Klug.
 

7rots51

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If I connect two circuit (with 74541 TTL outputs,I think that the output is push pull) ,if only one circuit is on at a time and the other is off , is there problems or high risk of failure?
 

algilsan

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Hi,
if you connect two 74hc154 outputs at a time but only one is ON, there is no problem. Like you can say in datasheet of '541, it has a tristate outputs, so you only have to enable one chip and disable the other. There is no problem, maybe the world were so simply. It's a joke ;)
 

msmax

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You can use 2 diodes and a ref resistor (connect to ground or +5v) to connect both outputs together. If chip is not in "disable" state then still there is no problem.

Just take in account if both outputs are ON the input to the circuit is ON. Only if BOTH outputs are OFF the input will be OFF.
 

7rots51

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There are a switch that give power to one of them at a time .thus only one of them is on and the other is off.

I can say the problem in a simple way,I have two 74hc541 at two boards.
that the outputs are connected together ,at a time only one board has power and the other is off.

Is this have a risk of failure for the 74hc541?
it seems that algilsan idea is correct.

please give comment!
 

algilsan

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Well, Rtos61,

Like you said, you have two separate boards, with a hc541 inside of each one. And only one board is Powered at a time, so you can connect all outputs together at no risk OK!, because, fisrt one hc541 is at no power and second it has tristate outputs.

Only you will have a problem if you switch on two boards at a time, in this case you maybe will have a short circuit at hc541 outputs of them. If you want to be aware of this situation you only have to join outputs with a 100 ohms resistor. The circuit will function equal than if no resistors were inside but you don't damage hc541 outputs in case of switch on both boards by a mistake.

if you have any doubt, please ask me!
 

7rots51

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Thanks to algilsan and maziar

I get the result and I do what you said.
 

RegUser_2

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algilsan

Like you said, you have two separate boards, with a hc541 inside of each one. And only one board is Powered at a time, so you can connect all outputs together at no risk OK!, because, fisrt one hc541 is at no power and second it has tristate outputs.


Does it mean there are no protection diodes @ the outputs? If such exisits, there is a good possibility of shorts to gnd via dte unbiased protection diodes, even if the outputs are tristate.
 

badwang

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RegUser_2 is right, most 74HC chips i/o pins have ESD protection diodes. If two chip outputs are connected together, while one is power on and another is off, the active pin will be shorted to gnd via another pin's diode.
 

7rots51

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hi RegUser_2 ,badwang

please give me links on this topic if you have?

what about another ways? is there a safe way to implement this.

bye
 

klug

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Hi 7rots51!

I have given solutions to you.

You have 2 variants:

1. Use buffers with open collector output.

2. Use 7432 noninverting OR to unite outputs.

Variant #1 is better.

Best Wishes! klug.
 

wjhzhx

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Because 74HC series IC is not high impedenc when power is off, so your circuit will not work, change the 74HC541
to TI's correspond 74 series IC, such as 74LVCT541, 74LVT541, 74AHCT541 and 74ABT541.They are support hot swap(when it's power is off, it output pin and input pin are high impedenc)
 

alledauser

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Due to the protection diode from the pin to the vcc, when you switch off one of the boards, it may draw current from the outputs of the live chip through this diode and the VCC pin, eventually supplying other chips on the board which is turned off. This will probably not damage the chips but may disturb the output signals. You may put shotky diodes in series to the vcc pins of 541s to prevent this. And don't forget the decoupling caps.
regards
 

cdcll

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Hi all,

What about if I have 12 boards to be connected together, at a time only one board's output pins are active and the others should be tri-state, how should I design the circuit with no risk of failure and design error?

Regards,
cdcll
 

S

sick_man

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7rots51 said:
Hi
I have a problem in design a digital system.

I have a circuit that have 8 (TTL-74hc541) digital outpus(DO1~DO8).

I have to use two of these circuit in a system and connect the similar digital outputs together.(ex. DO1 of one
circuit is connected to DO1 of another one ,and so on).The digital outputs are inputs of another card.

at the same time only one of two circuit can be powered on ,the other one must be powered off.

Is this connection makes problem for the off circuit ?
(digital outputs are 74HC541)

or simply I say, can I connect outputs of two 74HC541 together and in state that on of them is powered off,is this make problem
for another 74hc541??
How can I make a better circuit with no risk of failure and design error??
Is open collector design a better choice?

Please give your comments.


use parallel shift registers and tie the enable output to the + vcc signal of the other unit

so it can be powered down

adding some switching diodes in series with each data port {ie d0 - what ever msb}

and also too the actual data {or just do this instead of adding the shift registers}



unit 1 D1 diodecathode>
>>>>>>>>>>>>>>>>>diode cathode---- data in
unit 2 D1 diodecathode>


ok???

but the problem then is data colusions if both units are on

if you are worried about the voltage leek across any diodes

also place a 220R resistor in series this will limit any harmfull current too a minimum on any digital i/o so gates wont get damaged so easy
 

7rots51

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hi !!!MONKEY!!!

The input of the target system is ttl,I think the circuit that you given will not be ok for the TTL inputs,I agree with you on using a current limiting resistor (220R) but the three diodes will not be suitable for TTL inputs.

BTW: only one of them is ON at a time(a switch toggles the power between the two circuits.
thanks
BYe
 

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