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convert a negative and positive square wave to be only positive.

Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?
 
D

Don Bowey

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?

How near to 15V must the resulting signal be?

Seems like a diode from the bipolar signal would do what you want.

Don
 
L

Lord Garth

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?

Can you get away with a diode and a resistor? Zero will actually be -.6
if you will connect the cathode to your input signal and the anode to the
series resistor and then ground the other end of the resistor. When the
signal swings negative, the diode conducts and clamps the input.
 
J

John Fields

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?
 
P

petrus bitbyter

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?

There are more ways to skin a cat, but I'd prefer a real comparator for the
job. Nevertheles, see below.

Ri___
in---|___|--+------- out
|
-
^ D
|
Gnd---------+-------

Clamping the negative part of the square. During that negative part the
output wil be -Vd. Use a Schottky diode if -0.6V of the general purpose
silicium is to much.


R___
+-|___|-+
| |
Ri___ R___ | |\| |
----|___|--+----|___|-+-|-\ |
| | >--+----
V +-|+/
- D | |/|
| |
-----------+----------+------------

Clamping the positive part of the square and inverting the resulting
negative part.


R ___
+--|___|-+
| +--------- out
| |
| -
R ___ | |\| ^ D
in ---|___|-++--|-\ |
| | >--+
| +-|+/ |
| | |/| -
| | ^ D
| | |
+-)-------+
Gnd -----------+----------------

An active half wave rectifier does the same.


R___ R___
in ----|___|---+-----|___|---+
| |
R___ | |\| |
-15V----|___|---+-----|-\ |
| >----+------out
+-|+/
| |/|
|
Gnd---------------+----------------

An analog solution.

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

No rocket science. Standard textbooks on opamp circuits will provide this
and other circuits.


petrus sbitbyter
 
Hi, thanks for all your fast and good answer, the output that i have to
drive is an optocoupler the square wave is the signal control for an
igbt finally i have done this,
http://img65.imageshack.us/img65/9913/dibujovt1.jpg, from the
comparator i got a square wave with two values 0, and -15V, and after
i use a transistor to proportionate enough current for the optocoupler,
after the transistor there is a wire that goes to one optocoupler, and
the other wire goes to another transistor to invert the signal and
finally to the other optocoupler.

The circuit in orcad works, but i cant understand a couple of things,
when i connect the comparator to the transistor i got a square with two
values (0, -15), but if a connect the comparator to a resistor and the
resistor to ground i get a square wave but the values are (15V,-15V),
can anyone explain me this? and the other thing, is why i get an invert
output from the transistor. And finally i want to use a fast transistor
and a fast comparator, can anyone recommend to me one?, agian thanks :).
 
L

Lord Garth

Jan 1, 1970
0
Hi, thanks for all your fast and good answer, the output that i have to
drive is an optocoupler the square wave is the signal control for an
igbt finally i have done this,
http://img65.imageshack.us/img65/9913/dibujovt1.jpg, from the
comparator i got a square wave with two values 0, and -15V, and after
i use a transistor to proportionate enough current for the optocoupler,
after the transistor there is a wire that goes to one optocoupler, and
the other wire goes to another transistor to invert the signal and
finally to the other optocoupler.

The circuit in orcad works, but i cant understand a couple of things,
when i connect the comparator to the transistor i got a square with two
values (0, -15), but if a connect the comparator to a resistor and the
resistor to ground i get a square wave but the values are (15V,-15V),
can anyone explain me this? and the other thing, is why i get an invert
output from the transistor. And finally i want to use a fast transistor
and a fast comparator, can anyone recommend to me one?, agian thanks :).


When Q1's base was at +15v, the transistor was on and the collector was
pulled
to ground less Vce (sat). When the base was at -15v, the transistor was off
so the
collector is pulled to +15v by R29.
 
P

petrus bitbyter

Jan 1, 1970
0
Hi, thanks for all your fast and good answer, the output that i have to
drive is an optocoupler the square wave is the signal control for an
igbt finally i have done this,
http://img65.imageshack.us/img65/9913/dibujovt1.jpg, from the
comparator i got a square wave with two values 0, and -15V, and after
i use a transistor to proportionate enough current for the optocoupler,
after the transistor there is a wire that goes to one optocoupler, and
the other wire goes to another transistor to invert the signal and
finally to the other optocoupler.

The circuit in orcad works, but i cant understand a couple of things,
when i connect the comparator to the transistor i got a square with two
values (0, -15), but if a connect the comparator to a resistor and the
resistor to ground i get a square wave but the values are (15V,-15V),
can anyone explain me this? and the other thing, is why i get an invert
output from the transistor. And finally i want to use a fast transistor
and a fast comparator, can anyone recommend to me one?, agian thanks :).


It's still not clear to me what signal source you actualy want to use. But
supposing you have that +15/-15V square *and* you want to drive only
optocouplers there is no use for comparators and extra electronics
whatsoever. If you are not sure the signalsource can provide the 10mA the
optocouplers need, you can use one 1x amplifier. The TLE2024 can provide all
the current you need the drive the optocouplers.


+---------+
| |
| |\| |
+----|-\ | 1k___
| >-+--|___|---+---------+
in----|+/ | |
|/| TLE2024 V opto - opto
- LED1 ^ LED2
| |
Gnd----------------+---------+
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Whenever you build this circuit make sure both opto LEDs are connected,
otherwise you will blow the connected one. If you need only one optocoupler,
replace the others LED by a normal one or a general purpose Si diode.

petrus bitbyter
 
Answer for Lord Garth, now is clear thanks for your answer :).

It's still not clear to me what signal source you actualy want to use. But
supposing you have that +15/-15V square *and* you want to drive only
optocouplers there is no use for comparators and extra electronics
whatsoever. If you are not sure the signalsource can provide the 10mA the
optocouplers need, you can use one 1x amplifier. The TLE2024 can provide all
the current you need the drive the optocouplers.


+---------+
| |
| |\| |
+----|-\ | 1k___
| >-+--|___|---+---------+
in----|+/ | |
|/| TLE2024 V opto - opto
- LED1 ^ LED2
| |
Gnd----------------+---------+
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Whenever you build this circuit make sure both opto LEDs are connected,
otherwise you will blow the connected one. If you need only one optocoupler,
replace the others LED by a normal one or a general purpose Si diode.

petrus bitbyter

Hi, thanks for your answer, so you mean a schematic like this?,
http://img214.imageshack.us/img214/7971/circuitoxk0.png, the comparator
is needed because with the comparator i get the square wave, with the
comparator i'm doing a spwm, i'm comparing a sinusoidal wave with a
triangular wave, to obtain the square wave to control the igbt's.

i have done a simulation in orcad with this circuit, in the output of
the comparator i get a square wave with two values (15V, -15V), after
the 1x amplifier i get the same square wave but the values are (-4v,
4V), and with that low voltage i can't turn off and turn on the
optocouplers, i hope you understand me. Can you explain me why i get
differents values in the output if i change the output of the AO, i
mean if in the output of the comparator i put another A0(input
impedance = infinity) the output is a square wave with (-15v,15V), but
if in the output of the comparator i put a resistor of 1k connected to
ground i get the same square wave,but with lower values (-4v,4v).
 
B

Bob Masta

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V, to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong, or another
solution to get the signal that i want, in addition i want to invert
the signal, is there any ic with i can change the voltage and also
invert the signal?

The following may not be too useful for your particular task at
the moment, but one common approach is to simply use a
voltage divider on the comparator output, but with a twist:
The high end of the (equal value) divider goes to +15 and the
bottom end goes to the op-amp output. When the amp is
putting out +15, you of course get +15 from the "divider", which
is then just 2 resistors in parallel from +15. When the amp is
putting out -15, the divider has 30V across it and the output is
in the center at 0V. Obviously, this approach is not for where
you need a "hard" zero, but its a good trick to remember when
interfacing bipolar stuff to CMOS, for example.

Best regards,


Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Home of DaqGen, the FREEWARE signal generator
 
P

petrus bitbyter

Jan 1, 1970
0
Answer for Lord Garth, now is clear thanks for your answer :).



Hi, thanks for your answer, so you mean a schematic like this?,
http://img214.imageshack.us/img214/7971/circuitoxk0.png, the comparator
is needed because with the comparator i get the square wave, with the
comparator i'm doing a spwm, i'm comparing a sinusoidal wave with a
triangular wave, to obtain the square wave to control the igbt's.

i have done a simulation in orcad with this circuit, in the output of
the comparator i get a square wave with two values (15V, -15V), after
the 1x amplifier i get the same square wave but the values are (-4v,
4V), and with that low voltage i can't turn off and turn on the
optocouplers, i hope you understand me. Can you explain me why i get
differents values in the output if i change the output of the AO, i
mean if in the output of the comparator i put another A0(input
impedance = infinity) the output is a square wave with (-15v,15V), but
if in the output of the comparator i put a resistor of 1k connected to
ground i get the same square wave,but with lower values (-4v,4v).

Looks like a simulator error. According to the Texas datasheet, the TLE2024
should be able to sink and source up to 40mA. Unless I do not understand
Ohms law anymore, a load of 1k at 15V should take 15/1=15mA which is well
below that maximum of 40mA. So you don't even need that extra 1x amplifier.
Maybe you can try another model like TLE2024/301/TI or TLE2024/302/TI.

BTW the LED of that optocoupler (like all LEDs I'm aware of) are current
driven. According to the datasheet 10mA should be enough although it can
handle up to 80mA. The voltage across it at 15mA will become somewhere
between 1V and 1.5V.

I strongly advise you to get the datasheets of the components you want to
use and learn to interprete them.

petrus bitbyter
 
J

jasen

Jan 1, 1970
0
Hi, i have a square signal, with positive at 15V, and negative with
-15V, and i will like to convert this signal so that 15V will be 15V,
and the -15V will be 0V,

how about even voltage divider between your signal and +15V
to do this i have uses an op482 as omparator,
the configuration is this, +IN = -5V, -IN = Square wave, V+ = 15V, V- =
0V, but in the output i get a signal with a high value of 1.3, and a
low value of 0.3, some suggestion about what doing wrong,

your input voltage is outside the allowable range (it's more negative than
-in)
or another solution to get the signal that i want,
in addition i want to invert the signal,

15V
try this: |
|\|
in --10K--+-----|-\
1K | >----
+5V-------+-----|+/
|/|
|
0V

you won't get exactly 15V or 0V. probably more like 14.7 an 0,3 is that
close enough?

Bye.
Jasen
 
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