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Author Topic: Need ckt for 5 to 24 V iput output 5v or 0V  (Read 6088 times)
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senthils2k
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« on: October 11, 2010, 11:20:54 11:20 »

I Need Ckt for Convert the  Input Voltage Level From 5 -24 V to Output  0 and 5V using ULN2003A.

Any Help
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« Reply #1 on: October 11, 2010, 12:07:11 12:07 »

Do you want to build a LM7805 out of a ULN2003A Huh
if not reword request.
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« Reply #2 on: October 11, 2010, 01:32:29 13:32 »

Use a 5v zener between them.
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oldvan
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« Reply #3 on: October 11, 2010, 02:33:58 14:33 »

Use a 5v zener between them.

I'd suggest a 12V Zener.  Better noise/leakage immunity.
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« Reply #4 on: October 11, 2010, 02:57:21 14:57 »

UDN29xx serias source driver ic
http://www.allegromicro.com/en/Products/Part_Numbers/Archive/2580.pdf
5 to 24 volt or 24 to 5 volt output vs= output voltage
by
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« Reply #5 on: October 11, 2010, 08:18:48 20:18 »

UDN29xx serias source driver ic
http://www.allegromicro.com/en/Products/Part_Numbers/Archive/2580.pdf
5 to 24 volt or 24 to 5 volt output vs= output voltage by

Not sure how one could use that IC to easily answer the original request, but
"DISCONTINUED PRODUCT  For Reference Only"
is stamped in orange all over the data sheet.  

I Need Ckt for Convert the  Input Voltage Level From 5 -24 V to Output  0 and 5V using ULN2003A.

If I my understanding of the problem:

"I have an input that is 24V when H and 5V when L,
 I need an output that is 5V when H and 0V when L,
 I wish to accomplish this using a ULN2003"


is correct, then this is a correct answer:


An answer using an opto isolator instead of the ULN2003:
« Last Edit: October 11, 2010, 11:50:27 23:50 by oldvan » Logged

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senthils2k
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« Reply #6 on: October 12, 2010, 06:34:51 06:34 »

Not sure how one could use that IC to easily answer the original request, but
"DISCONTINUED PRODUCT  For Reference Only"
is stamped in orange all over the data sheet. 

If I my understanding of the problem:

"I have an input that is 24V when H and 5V when L,
 I need an output that is 5V when H and 0V when L,
 I wish to accomplish this using a ULN2003"


is correct, then this is a correct answer:


An answer using an opto isolator instead of the ULN2003:


ya almost correct

only difference is

Input H ---------->5 to 24V
input L------------->0V

Output  H----------->5V
Output L------------->0V

is it require the connection  between the 2 and 16 PIN?

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Amon
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« Reply #7 on: October 12, 2010, 08:06:40 08:06 »

Then you can't use the zener.
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« Reply #8 on: October 12, 2010, 08:35:18 08:35 »

Oldvan.
can i use the following ckt


this circuit work if R1 connected to  GND

if  R1  connected to 5V  it  not work

any suggestion ?
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« Reply #9 on: October 12, 2010, 02:35:42 14:35 »

can i use the following ckt

this circuit work if R1 connected to  GND
if  R1  connected to 5V  it  not work

any suggestion ?

Did you connect pin 8 of the ULN2003 to GND, or is it floating?



Each section of the ULN2003 functions as an inverter.  You can use just 1 section if
inverted output is acceptable, otherwise need to use a second section to flip it back
over.

If you are feeding a logic gate or other high impedance input, a pair of resistors will
get you out of the woods:


For a higher current need, a resistor and a Zener diode will do the trick:

The resistor limits current flow and the Zener holds H output voltage to 5V.
« Last Edit: October 12, 2010, 02:46:06 14:46 by oldvan » Logged

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« Reply #10 on: October 12, 2010, 07:46:52 19:46 »

Right solution is to subtract more than 5V (8.2V...) from input voltage. Vi according to datasheet can be as high as 30V in respect to substrate E pin 8 which is normally connected to GND. So ViH can be more than 5V while ViL must be lower than, let's say 1V.
Put the zener diode in series with the input pin and between input pin and GND connect a 10k resistor.
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« Reply #11 on: October 12, 2010, 10:30:11 22:30 »

Right solution is to subtract more than 5V (8.2V...) from input voltage. Vi according to datasheet can
be as high as 30V in respect to substrate E pin 8 which is normally connected to GND. So ViH can be
more than 5V while ViL must be lower than, let's say 1V.

 I beg to differ.  

Note that the "ABSOLUTE MAXIMUM RATINGS" section includes the footnote:
Quote
(1) Stresses beyond those listed under "absolute maximum ratings" may cause permanent damage to
    the device. These are stress ratings only, and functional operation of the device at these or
    any other conditions beyond those indicated under "recommended operating conditions" is not
    implied.
Exposure to absolute-maximum-rated conditions for extended periods may affect
    device reliability.

EDITED 10/13/2010 by Oldvan.  Reason:  I was incorrect.  See post below.

For reliable service, Input high should be 13V max per the datasheet:

The 24V max proposed by the initial problem minus the 12V Zener voltage I recommended gives
an input of 12 Volts to the ULN2003, safely within recommended range.  
  Cool

Subtracting only 8.2V as you suggested leaves 15.8V input on an input that is rated to operate
correctly long-term from up to 13V input.  PROBLEM!



...and between input pin and GND connect a 10k resistor.

WHY when a 12.9K load is already present inside the chip?  



It would be better to post only correct information, especially after
someone else has already posted the correct solution and schematic.  
EDIT:  It seems I was also talking to myself with that sentence.  
YIKES!
« Last Edit: October 13, 2010, 02:23:42 14:23 by oldvan » Logged

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« Reply #12 on: October 13, 2010, 07:50:26 07:50 »

Oldvan.
can i use the following ckt


this circuit work if R1 connected to  GND

if  R1  connected to 5V  it  not work

any suggestion ?

So what is the modification required in my circuit work for bot R1 connected GND and 5V

What is the required advice?

ULN2003A Com  connected to 5V
and 8th PIN Connected to GND
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« Reply #13 on: October 13, 2010, 08:52:37 08:52 »

Oldvan, you are not reading datasheet correctly. That 13V is valid for 2002 as a max guarantied Vi(ON) and not for max allowable Vi. Max allowable is 30V for all series.
Chips of this series has different circuits on input. 2002 has 7V zener in series with base of input transistor and that's why 13V is noted.
I do not mind about the additional zener diode voltage. It can be up to 18V if you wish but not less than 6V. It depends only on current capabilities of the driver.
Additional 10k resistor from input pin to GND I want just to minimize possible leakage currents.
And do not dramatize next time.
« Last Edit: October 13, 2010, 08:56:15 08:56 by borberk » Logged
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« Reply #14 on: October 13, 2010, 01:44:19 13:44 »

borberk:  Seems I should have done more homework, I'd have less a taste of crow and foot
in my mouth.

From the Toshiba version of the datasheet (since TI neglected to include the "Recommended
Operating Conditions" section in theirs):



Correctly understood, the chip will operate reliably with an input to 24 Volts.  Going to 30 Volts
is still not a good idea, but is not a concern for this circuit.

Here is the schematic modified so that:

If SW1 OR SW2 is H, output is 0V
If SW1 AND SW2 are L, output is 5V.
« Last Edit: October 13, 2010, 02:21:03 14:21 by oldvan » Logged

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« Reply #15 on: October 13, 2010, 03:55:50 15:55 »

guess it depends on the chip maker what the voltage really is and where you look in the DS  Smiley
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« Reply #16 on: October 14, 2010, 06:30:17 06:30 »

borberk:  Seems I should have done more homework, I'd have less a taste of crow and foot
in my mouth.

From the Toshiba version of the datasheet (since TI neglected to include the "Recommended
Operating Conditions" section in theirs):



Correctly understood, the chip will operate reliably with an input to 24 Volts.  Going to 30 Volts
is still not a good idea, but is not a concern for this circuit.

Here is the schematic modified so that:

If SW1 OR SW2 is H, output is 0V
If SW1 AND SW2 are L, output is 5V.


if this  ckt work under the Following Conditions Huh

if SW2 =0 V

Sw1:     0V           R3 ---------->5V

Sw1:   5-24V             R3------------>0V

and

if SW2 =5V

Sw1:     5-24V           R3 ---------->0V

Sw1:   0V             R3------------>5V

pls advice.
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« Reply #17 on: October 14, 2010, 02:11:38 14:11 »

To senthils2k, about the second half of your statement/question:

Look at the C1 voltage, if you put 5v on it with sw2, you have a High condition for the 2003.
It does not matter what you do with sw1 then.
If you set sw1 to 5-24v, you raise the C1 voltage further, and still have the High condition.
If you set sw1 to 0v, the voltage over C1 remains High (from sw2 via d1).
In all two cases the output on R3 will be zero or 0v.

If you want your statement to be true (sw2=5v, sw1=0v, R3=5v) then you should eliminate d2 and make R1 5k6.
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« Reply #18 on: October 14, 2010, 02:19:27 14:19 »

if this  ckt work under the Following Conditions Huh

if SW2 =0 V
(1) Sw1:     0V           R3 ---------->5V
(2) Sw1:   5-24V             R3------------>0V

and

if SW2 =5V
(3) Sw1:     5-24V           R3 ---------->0V
(4) Sw1:   0V             R3------------>5V
pls advice.

I'm guessing from your limited words that you want the circuit to work as described
above.

Therefore I made a logic table of your request in Excel to be sure I was looking at it
correctly:



Looking at the table, the state of SW2 is of no consequence to the output, so the
logic table can be simplified as follows:



Your circuit can be simplified to the following:



You leave an awful lot to guesswork with limited descriptions and an ever-growing
target.  It would have been better from the start to fully describe the problem.

If my description in this post is not what you desire, please supply a logic table of
what you DO want, so that I may get you a complete and correct answer.
« Last Edit: October 14, 2010, 02:41:24 14:41 by oldvan » Logged

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« Reply #19 on: October 15, 2010, 06:08:28 06:08 »

I'm guessing from your limited words that you want the circuit to work as described
above.

Therefore I made a logic table of your request in Excel to be sure I was looking at it
correctly:



Looking at the table, the state of SW2 is of no consequence to the output, so the
logic table can be simplified as follows:



Your circuit can be simplified to the following:



You leave an awful lot to guesswork with limited descriptions and an ever-growing
target.  It would have been better from the start to fully describe the problem.

If my description in this post is not what you desire, please supply a logic table of
what you DO want, so that I may get you a complete and correct answer.

my application is to Capture the Common Anode and Common Cathode signal of 7 Segment.

if Put the Input SW2  for changing the Input Signal ie CA or CC.

Ur Table is correct.

i Need correct Circuit(May be the Circuit with Didode help in one Configuration ie SW2 to GND its works but in other way ie Put in +24V or +5V  its not work).
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