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Sci.Electronics.Basics -> LM358 op-amp

There are 11 messages in this thread.
You are currently looking at messages 1 to 11.


Author: Allen Bong
Date: 06:40 15-04-08


I have constructed a circuit as below to convert the input of 12 to
24V as 0 to 12V and show the voltage on a DC volt-meter. The
schematics is shown below:


VCC +24V
+
|
.-.
| | +24V
2k2| | VCC
'-' +
| |
| 10K |\|
| +12V-|___|-+--|-\LM358
| ___ | | >---+-----+
IN +-|___|---+-----|+/ | |
| 10K | | |/| | |
| | | === | |
.-. | | GND | |
| | | | ___ | |
2k2| | .-. +--|___|--+ /+\
'-' | | 10K ( )
| 10K| | VOLT \-/
| '-' |
=== | |
GND | ===
=== GND
GND

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


I have the +12V reference taken from a TL431.

When the input voltage at "IN" is +12V I was not able to get a 0V on
the voltmeter. Instead I am getting 0.58V at the output. Does this
has something to do with the CMRR of the chip? How would I solve
this problem? Would changing the op-amp to LM748 and connecting a
pot between the balance inputs of the LM748 help to reduce the
voltage to O?

Thanks In Advance!

Allen
Author: Bob Monsen
Date: 12:00 15-04-08

"Allen Bong" <allenbsf6502@gmail.com> wrote in message
news:156490eb-d94f-49fa-a6ed-aca020d20b0f@a5g2000prg.googlegroups.com...
>I have constructed a circuit as below to convert the input of 12 to
> 24V as 0 to 12V and show the voltage on a DC volt-meter. The
> schematics is shown below:
>
>
> VCC +24V
> +
> |
> .-.
> | | +24V
> 2k2| | VCC
> '-' +
> | |
> | 10K |\|
> | +12V-|___|-+--|-\LM358
> | ___ | | >---+-----+
> IN +-|___|---+-----|+/ | |
> | 10K | | |/| | |
> | | | === | |
> .-. | | GND | |
> | | | | ___ | |
> 2k2| | .-. +--|___|--+ /+\
> '-' | | 10K ( )
> | 10K| | VOLT \-/
> | '-' |
> === | |
> GND | ===
> === GND
> GND
>
> (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
>
>
> I have the +12V reference taken from a TL431.
>
> When the input voltage at "IN" is +12V I was not able to get a 0V on
> the voltmeter. Instead I am getting 0.58V at the output. Does this
> has something to do with the CMRR of the chip? How would I solve
> this problem? Would changing the op-amp to LM748 and connecting a
> pot between the balance inputs of the LM748 help to reduce the
> voltage to O?
>
> Thanks In Advance!
>
> Allen


The LM358 is a single supply opamp, but is is not going to pull all the way
down to 0V without a load. The low side output driver is a PNP, so it should
be able to get within the saturation voltage (i.e., Vce(sat)) of the ground
rail. However, if you put a load to ground on it, it'll be able to pull all
the way to ground, I believe.

Regards,
Bob Monsen
Author: John Larkin
Date: 12:11 15-04-08

On Tue, 15 Apr 2008 03:40:51 -0700 (PDT), Allen Bong
<allenbsf6502@gmail.com> wrote:

>I have constructed a circuit as below to convert the input of 12 to
>24V as 0 to 12V and show the voltage on a DC volt-meter. The
>schematics is shown below:
>
>
> VCC +24V
> +
> |
> .-.
> | | +24V
> 2k2| | VCC
> '-' +
> | |
> | 10K |\|
> | +12V-|___|-+--|-\LM358
> | ___ | | >---+-----+
> IN +-|___|---+-----|+/ | |
> | 10K | | |/| | |
> | | | === | |
> .-. | | GND | |
> | | | | ___ | |
> 2k2| | .-. +--|___|--+ /+\
> '-' | | 10K ( )
> | 10K| | VOLT \-/
> | '-' |
> === | |
> GND | ===
> === GND
> GND
>
>(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
>
>
>I have the +12V reference taken from a TL431.
>
>When the input voltage at "IN" is +12V I was not able to get a 0V on
>the voltmeter. Instead I am getting 0.58V at the output. Does this
>has something to do with the CMRR of the chip? How would I solve
>this problem? Would changing the op-amp to LM748 and connecting a
>pot between the balance inputs of the LM748 help to reduce the
>voltage to O?
>
>Thanks In Advance!
>
>Allen

This opamp needs a negative power supply here, if you expect it to
swing all the way to ground.

John
Author: Jamie
Date: 17:17 15-04-08

Allen Bong wrote:
> I have constructed a circuit as below to convert the input of 12 to
> 24V as 0 to 12V and show the voltage on a DC volt-meter. The
> schematics is shown below:
>
>
> VCC +24V
> +
> |
> .-.
> | | +24V
> 2k2| | VCC
> '-' +
> | |
> | 10K |\|
> | +12V-|___|-+--|-\LM358
> | ___ | | >---+-----+
> IN +-|___|---+-----|+/ | |
> | 10K | | |/| | |
> | | | === | |
> .-. | | GND | |
> | | | | ___ | |
> 2k2| | .-. +--|___|--+ /+\
> '-' | | 10K ( )
> | 10K| | VOLT \-/
> | '-' |
> === | |
> GND | ===
> === GND
> GND
>
> (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
>
>
> I have the +12V reference taken from a TL431.
>
> When the input voltage at "IN" is +12V I was not able to get a 0V on
> the voltmeter. Instead I am getting 0.58V at the output. Does this
> has something to do with the CMRR of the chip? How would I solve
> this problem? Would changing the op-amp to LM748 and connecting a
> pot between the balance inputs of the LM748 help to reduce the
> voltage to O?
>
> Thanks In Advance!
>
> Allen
You need a little (-) offset at the common rail of the chip..

use a (-) supply rail to get it or create a little (-) rail
voltage using a 555 or something..

--
http://webpages.charter.net/jamie_5";
Author: Allen Bong
Date: 18:13 15-04-08

On Apr 16, 5:17=A0am, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
> Allen Bong wrote:
> > I have constructed a circuit as below to convert the input of 12 to
> > 24V as 0 to 12V and show the voltage on a DC volt-meter. =A0The
> > schematics is shown below:
>
> > =A0 =A0 VCC +24V
> > =A0 =A0 =A0+
> > =A0 =A0 =A0|
> > =A0 =A0 .-.
> > =A0 =A0 | | =A0 =A0 =A0 =A0 =A0 =A0 =A0 +24V
> > =A02k2| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 VCC
> > =A0 =A0 '-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0+
> > =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =A0| =A0 =A0 =A0 10K =A0 =A0 |\|
> > =A0 =A0 =A0| +12V-|___|-+--|-\LM358
> > =A0 =A0 =A0| =A0___ =A0 =A0 =A0 | =A0| =A0>---+-----+
> > =A0 IN +-|___|---+-----|+/ =A0 =A0| =A0 =A0 |
> > =A0 =A0 =A0| =A010K =A0 =A0| =A0| =A0|/| =A0 =A0| =A0 =A0 |
> > =A0 =A0 =A0| =A0 =A0 =A0 =A0 | =A0| =A0 =3D=3D=3D =A0 | =A0 =A0 |
> > =A0 =A0 .-. =A0 =A0 =A0 =A0| =A0| =A0 GND =A0 | =A0 =A0 |
> > =A0 =A0 | | =A0 =A0 =A0 =A0| =A0| =A0 ___ =A0 | =A0 =A0 |
> > =A02k2| | =A0 =A0 =A0 .-. +--|___|--+ =A0 =A0/+\
> > =A0 =A0 '-' =A0 =A0 =A0 | | =A0 =A0 10K =A0 =A0 =A0 ( =A0 )
> > =A0 =A0 =A0| =A0 =A0 10K| | =A0 =A0 =A0 =A0 =A0 VOLT \-/
> > =A0 =A0 =A0| =A0 =A0 =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 =3D=3D=3D =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0 GND =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =3D=3D=3D
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 =3D=3D=3D =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0GND=

> > =A0 =A0 =A0 =A0 =A0 =A0 =A0 GND
>
> > (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
>
> > I have the +12V reference taken from a TL431.
>
> > When the input voltage at "IN" is +12V I was not able to get a 0V on
> > the voltmeter. =A0Instead I am getting 0.58V at the output. =A0Does this=

> > has something to do with the CMRR of the chip? =A0How would I solve
> > this problem? =A0Would changing the op-amp to LM748 and connecting a
> > pot between the balance inputs of the LM748 help to reduce the
> > voltage to O?
>
> > Thanks In Advance!
>
> > Allen
>
> You need a little =A0(-) offset at the common rail of the chip..
>
> =A0 =A0use a (-) supply rail to get it or create a little (-) rail
> voltage using a 555 or something..
>
> --http://webpages.charter.net/jamie_5";- Hide quoted text -
>
> - Show quoted text -

Thanks to all who advised me on the problem. I'll try first to get a
little (-) on the negative supply rail. How much is a little -- is 1
volt good enough to offset the 0.58V that I got?

Next I'll google to some schematics on how to set up the 555 to get a
negative voltage.

Thanks too all again!

Allen
Author: PhattyMo
Date: 01:44 16-04-08

Allen Bong wrote:
> On Apr 16, 5:17 am, Jamie
> <jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
>> Allen Bong wrote:
>>> I have constructed a circuit as below to convert the input of 12 to
>>> 24V as 0 to 12V and show the voltage on a DC volt-meter. The
>>> schematics is shown below:
>>> VCC +24V
>>> +
>>> |
>>> .-.
>>> | | +24V
>>> 2k2| | VCC
>>> '-' +
>>> | |
>>> | 10K |\|
>>> | +12V-|___|-+--|-\LM358
>>> | ___ | | >---+-----+
>>> IN +-|___|---+-----|+/ | |
>>> | 10K | | |/| | |
>>> | | | === | |
>>> .-. | | GND | |
>>> | | | | ___ | |
>>> 2k2| | .-. +--|___|--+ /+\
>>> '-' | | 10K ( )
>>> | 10K| | VOLT \-/
>>> | '-' |
>>> === | |
>>> GND | ===
>>> === GND
>>> GND
>>> (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
>>> I have the +12V reference taken from a TL431.
>>> When the input voltage at "IN" is +12V I was not able to get a 0V
on
>>> the voltmeter. Instead I am getting 0.58V at the output. Does this
>>> has something to do with the CMRR of the chip? How would I solve
>>> this problem? Would changing the op-amp to LM748 and connecting a
>>> pot between the balance inputs of the LM748 help to reduce the
>>> voltage to O?
>>> Thanks In Advance!
>>> Allen
>> You need a little (-) offset at the common rail of the chip..
>>
>> use a (-) supply rail to get it or create a little (-) rail
>> voltage using a 555 or something..
>>
>> --http://webpages.charter.net/jamie_5";- Hide quoted text -
>>
>> - Show quoted text -
>
> Thanks to all who advised me on the problem. I'll try first to get a
> little (-) on the negative supply rail. How much is a little -- is 1
> volt good enough to offset the 0.58V that I got?

If so,you might be able to get away with stacking a couple diodes off of
the ground rail- anode to 0V/ground,cathode to the new '-V' rail.
If you connect the supply voltage between -V and +V,your -V rail will be
approx 0.6V (per diode) below the 0V rail and/or '0V' will be ~0.6V
above the -V rail.

If ya get my drift.. kinda hard to explain,need a drawing.

>
> Next I'll google to some schematics on how to set up the 555 to get a
> negative voltage.
>
> Thanks too all again!
>
> Allen
Author: Allen Bong
Date: 03:33 16-04-08

On Apr 16, 1:44=A0pm, PhattyMo <phatt...@not.net> wrote:
> Allen Bong wrote:
> > On Apr 16, 5:17 am, Jamie
> > <jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
> >> Allen Bong wrote:
> >>> I have constructed a circuit as below to convert the input of 12 to
> >>> 24V as 0 to 12V and show the voltage on a DC volt-meter. =A0The
> >>> schematics is shown below:
> >>> =A0 =A0 VCC +24V
> >>> =A0 =A0 =A0+
> >>> =A0 =A0 =A0|
> >>> =A0 =A0 .-.
> >>> =A0 =A0 | | =A0 =A0 =A0 =A0 =A0 =A0 =A0 +24V
> >>> =A02k2| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 VCC
> >>> =A0 =A0 '-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0+
> >>> =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> >>> =A0 =A0 =A0| =A0 =A0 =A0 10K =A0 =A0 |\|
> >>> =A0 =A0 =A0| +12V-|___|-+--|-\LM358
> >>> =A0 =A0 =A0| =A0___ =A0 =A0 =A0 | =A0| =A0>---+-----+
> >>> =A0 IN +-|___|---+-----|+/ =A0 =A0| =A0 =A0 |
> >>> =A0 =A0 =A0| =A010K =A0 =A0| =A0| =A0|/| =A0 =A0| =A0 =A0 |
> >>> =A0 =A0 =A0| =A0 =A0 =A0 =A0 | =A0| =A0 =3D=3D=3D =A0 | =A0 =A0 |
> >>> =A0 =A0 .-. =A0 =A0 =A0 =A0| =A0| =A0 GND =A0 | =A0 =A0 |
> >>> =A0 =A0 | | =A0 =A0 =A0 =A0| =A0| =A0 ___ =A0 | =A0 =A0 |
> >>> =A02k2| | =A0 =A0 =A0 .-. +--|___|--+ =A0 =A0/+\
> >>> =A0 =A0 '-' =A0 =A0 =A0 | | =A0 =A0 10K =A0 =A0 =A0 ( =A0 )
> >>> =A0 =A0 =A0| =A0 =A0 10K| | =A0 =A0 =A0 =A0 =A0 VOLT \-/
> >>> =A0 =A0 =A0| =A0 =A0 =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> >>> =A0 =A0 =3D=3D=3D =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0
=A0=
|
> >>> =A0 =A0 GND =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0
=3D=3D=3D=

> >>> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =3D=3D=3D =A0 =A0 =A0 =A0 =A0 =A0 =A0
=A0G=
ND
> >>> =A0 =A0 =A0 =A0 =A0 =A0 =A0 GND
> >>> (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
> >>> I have the +12V reference taken from a TL431.
> >>> When the input voltage at "IN" is +12V I was not able to get
a 0V on
> >>> the voltmeter. =A0Instead I am getting 0.58V at the output. =A0Does
th=
is
> >>> has something to do with the CMRR of the chip? =A0How would I solve
> >>> this problem? =A0Would changing the op-amp to LM748 and connecting a
> >>> pot between the balance inputs of the LM748 help to reduce the
> >>> voltage to O?
> >>> Thanks In Advance!
> >>> Allen
> >> You need a little =A0(-) offset at the common rail of the chip..
>
> >> =A0 =A0use a (-) supply rail to get it or create a little (-) rail
> >> voltage using a 555 or something..
>
> >> --http://webpages.charter.net/jamie_5";- Hide quoted text -
>
> >> - Show quoted text -
>
> > Thanks to all who advised me on the problem. =A0I'll try first to get a
> > little (-) on the negative supply rail. =A0How much is a little -- is 1
> > volt good enough to offset the 0.58V that I got?
>
> If so,you might be able to get away with stacking a couple diodes off of
> the ground rail- anode to 0V/ground,cathode to the new '-V' rail.
> If you connect the supply voltage between -V and +V,your -V rail will be
> approx 0.6V (per diode) below the 0V rail and/or '0V' will be ~0.6V
> above the -V rail.
>
> If ya get my drift.. kinda hard to explain,need a drawing.
>
>
>
>
>
> > Next I'll google to some schematics on how to set up the 555 to get a
> > negative voltage.
>
> > Thanks too all again!
>
> > Allen- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

Do you mean adding a diode or two this way will offset the V- which is
connected to 0V as if it were connected to -1.2V ?


VCC
+
|
|\|
+12V ref -|-\
| >--------+ V2
+12V input -|+/ |
|/| V1 /+\
+---+ ( )
V | \-/
- | |
| /+\ |
V ( ) |
- \-/ |
| | |
+---+-----+
=3D=3D=3D
GND

V1=3D1.2V

V2=3D0.58 offset by 1.2V

(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
Author: Bob Monsen
Date: 12:42 16-04-08

"Allen Bong" <allenbsf6502@gmail.com> wrote in message
news:156490eb-d94f-49fa-a6ed-aca020d20b0f@a5g2000prg.googlegroups.com...
>I have constructed a circuit as below to convert the input of 12 to
> 24V as 0 to 12V and show the voltage on a DC volt-meter. The
> schematics is shown below:
>
>
> VCC +24V
> +
> |
> .-.
> | | +24V
> 2k2| | VCC
> '-' +
> | |
> | 10K |\|
> | +12V-|___|-+--|-\LM358
> | ___ | | >---+-----+
> IN +-|___|---+-----|+/ | |
> | 10K | | |/| | |
> | | | === | |
> .-. | | GND | |
> | | | | ___ | |
> 2k2| | .-. +--|___|--+ /+\
> '-' | | 10K ( )
> | 10K| | VOLT \-/
> | '-' |
> === | |
> GND | ===
> === GND
> GND
>
> (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
>
>
> I have the +12V reference taken from a TL431.
>
> When the input voltage at "IN" is +12V I was not able to get a 0V on
> the voltmeter. Instead I am getting 0.58V at the output. Does this
> has something to do with the CMRR of the chip? How would I solve
> this problem? Would changing the op-amp to LM748 and connecting a
> pot between the balance inputs of the LM748 help to reduce the
> voltage to O?
>
> Thanks In Advance!
>
> Allen


The LM358 includes a 50uA pull-down on the output, to get you to 0V.
However, you are pulling up on the output (with the reference through the
two 10k resistors) using 600uA. So, increase all the 10k resistors to 1MEG.
That way, your pull-up will only be 6uA, and it'll be able to get all the
way down to 0V. You can also increase the divider (which uses 2.2k
resistors) to using 1MEG resistors.

When you run into this kind of problem, go to the manufacturer, and see if
they have a datasheet for the device. They often include a schematic in the
datasheet, and using it, you can sometimes figure out what's up.

Regards,
Bob Monsen
Author: Allen Bong
Date: 19:45 16-04-08

On Apr 17, 12:42=A0am, "Bob Monsen" <rcmon...@gmail.com> wrote:
> "Allen Bong" <allenbsf6...@gmail.com> wrote in message
>
> news:156490eb-d94f-49fa-a6ed-aca020d20b0f@a5g2000prg.googlegroups.com...
>
>
>
>
>
> >I have constructed a circuit as below to convert the input of 12 to
> > 24V as 0 to 12V and show the voltage on a DC volt-meter. =A0The
> > schematics is shown below:
>
> > =A0 =A0VCC +24V
> > =A0 =A0 +
> > =A0 =A0 |
> > =A0 =A0.-.
> > =A0 =A0| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 +24V
> > 2k2| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 VCC
> > =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0+
> > =A0 =A0 | =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0 | =A0 =A0 =A0 10K =A0 =A0 |\|
> > =A0 =A0 | +12V-|___|-+--|-\LM358
> > =A0 =A0 | =A0___ =A0 =A0 =A0 | =A0| =A0>---+-----+
> > =A0IN +-|___|---+-----|+/ =A0 =A0| =A0 =A0 |
> > =A0 =A0 | =A010K =A0 =A0| =A0| =A0|/| =A0 =A0| =A0 =A0 |
> > =A0 =A0 | =A0 =A0 =A0 =A0 | =A0| =A0 =3D=3D=3D =A0 | =A0 =A0 |
> > =A0 =A0.-. =A0 =A0 =A0 =A0| =A0| =A0 GND =A0 | =A0 =A0 |
> > =A0 =A0| | =A0 =A0 =A0 =A0| =A0| =A0 ___ =A0 | =A0 =A0 |
> > 2k2| | =A0 =A0 =A0 .-. +--|___|--+ =A0 =A0/+\
> > =A0 =A0'-' =A0 =A0 =A0 | | =A0 =A0 10K =A0 =A0 =A0 ( =A0 )
> > =A0 =A0 | =A0 =A0 10K| | =A0 =A0 =A0 =A0 =A0 VOLT \-/
> > =A0 =A0 | =A0 =A0 =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > =A0 =A0=3D=3D=3D =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|
> > =A0 =A0GND =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =3D=3D=3D
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D=3D=3D =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0GND
> > =A0 =A0 =A0 =A0 =A0 =A0 =A0GND
>
> > (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
>
> > I have the +12V reference taken from a TL431.
>
> > When the input voltage at "IN" is +12V I was not able to get a 0V on
> > the voltmeter. =A0Instead I am getting 0.58V at the output. =A0Does this=

> > has something to do with the CMRR of the chip? =A0How would I solve
> > this problem? =A0Would changing the op-amp to LM748 and connecting a
> > pot between the balance inputs of the LM748 help to reduce the
> > voltage to O?
>
> > Thanks In Advance!
>
> > Allen
>
> The LM358 includes a 50uA pull-down on the output, to get you to 0V.
> However, you are pulling up on the output (with the reference through the
> two 10k resistors) using 600uA. So, increase all the 10k resistors to 1MEG=
.
> That way, your pull-up will only be 6uA, and it'll be able to get all the
> way down to 0V. You can also increase the divider (which uses 2.2k
> resistors) to using 1MEG resistors.
>
> When you run into this kind of problem, go to the manufacturer, and see if=

> they have a datasheet for the device. They often include a schematic in th=
e
> datasheet, and using it, you can sometimes figure out what's up.
>
> Regards,
> =A0Bob Monsen- Hide quoted text -
>
> - Show quoted text -

Hi Bob,

Thanks for your good advice. I actually read through the datasheet and
the application notes, but there are a lot of things in the datasheet
I dont understand. Now reading the datasheet again, I see the 50uA at
the 'Output Current Sink' at Vo =3D 200mV and V+ =3D 15V.

I'll try it out tonight and report the result here.

Thank you again for helping!

Allen
Author: Allen Bong
Date: 06:20 17-04-08

On Apr 17, 7:45=A0am, Allen Bong <allenbsf6...@gmail.com> wrote:
> On Apr 17, 12:42=A0am, "Bob Monsen" <rcmon...@gmail.com> wrote:
>
>
>
>
>
> > "Allen Bong" <allenbsf6...@gmail.com> wrote in message
>
> >news:156490eb-d94f-49fa-a6ed-aca020d20b0f@a5g2000prg.googlegroups.com...
>
> > >I have constructed a circuit as below to convert the input of 12 to
> > > 24V as 0 to 12V and show the voltage on a DC volt-meter. =A0The
> > > schematics is shown below:
>
> > > =A0 =A0VCC +24V
> > > =A0 =A0 +
> > > =A0 =A0 |
> > > =A0 =A0.-.
> > > =A0 =A0| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 +24V
> > > 2k2| | =A0 =A0 =A0 =A0 =A0 =A0 =A0 VCC
> > > =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0+
> > > =A0 =A0 | =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > > =A0 =A0 | =A0 =A0 =A0 10K =A0 =A0 |\|
> > > =A0 =A0 | +12V-|___|-+--|-\LM358
> > > =A0 =A0 | =A0___ =A0 =A0 =A0 | =A0| =A0>---+-----+
> > > =A0IN +-|___|---+-----|+/ =A0 =A0| =A0 =A0 |
> > > =A0 =A0 | =A010K =A0 =A0| =A0| =A0|/| =A0 =A0| =A0 =A0 |
> > > =A0 =A0 | =A0 =A0 =A0 =A0 | =A0| =A0 =3D=3D=3D =A0 | =A0 =A0 |
> > > =A0 =A0.-. =A0 =A0 =A0 =A0| =A0| =A0 GND =A0 | =A0 =A0 |
> > > =A0 =A0| | =A0 =A0 =A0 =A0| =A0| =A0 ___ =A0 | =A0 =A0 |
> > > 2k2| | =A0 =A0 =A0 .-. +--|___|--+ =A0 =A0/+\
> > > =A0 =A0'-' =A0 =A0 =A0 | | =A0 =A0 10K =A0 =A0 =A0 ( =A0 )
> > > =A0 =A0 | =A0 =A0 10K| | =A0 =A0 =A0 =A0 =A0 VOLT \-/
> > > =A0 =A0 | =A0 =A0 =A0 =A0'-' =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 |
> > > =A0 =A0=3D=3D=3D =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|=

> > > =A0 =A0GND =A0 =A0 =A0 =A0| =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =3D=3D=3D
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D=3D=3D =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0GN=
D
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0GND
>
> > > (created by AACircuit v1.28.6 beta 04/19/05www.tech-chat.de)
>
> > > I have the +12V reference taken from a TL431.
>
> > > When the input voltage at "IN" is +12V I was not able to get a
0V on
> > > the voltmeter. =A0Instead I am getting 0.58V at the output. =A0Does th=
is
> > > has something to do with the CMRR of the chip? =A0How would I solve
> > > this problem? =A0Would changing the op-amp to LM748 and connecting a
> > > pot between the balance inputs of the LM748 help to reduce the
> > > voltage to O?
>
> > > Thanks In Advance!
>
> > > Allen
>
> > The LM358 includes a 50uA pull-down on the output, to get you to 0V.
> > However, you are pulling up on the output (with the reference through th=
e
> > two 10k resistors) using 600uA. So, increase all the 10k resistors to 1M=
EG.
> > That way, your pull-up will only be 6uA, and it'll be able to get all th=
e
> > way down to 0V. You can also increase the divider (which uses 2.2k
> > resistors) to using 1MEG resistors.
>
> > When you run into this kind of problem, go to the manufacturer, and see =
if
> > they have a datasheet for the device. They often include a schematic in =
the
> > datasheet, and using it, you can sometimes figure out what's up.
>
> > Regards,
> > =A0Bob Monsen- Hide quoted text -
>
> > - Show quoted text -
>
> Hi Bob,
>
> Thanks for your good advice. I actually read through the datasheet and
> the application notes, but there are a lot of things in the datasheet
> I dont understand. =A0Now reading the datasheet again, I see the 50uA at
> the 'Output Current Sink' at Vo =3D 200mV and V+ =3D 15V.
>
> I'll try it out tonight and report the result here.
>
> Thank you again for helping!
>
> Allen- Hide quoted text -
>
> - Show quoted text -

I have changed all the 10K resistors to 1 Mohm and the circuit worked
as expected even though I kept the voltage divider at 2.2K (there is a
reason I can't change the value of these 2 resistors). I will solder
it up onto a PCB and test it for a few days to observe its
stabilities.

In order to learn more about this op-amp, I am also looking into the
"supplying negative voltage into the V- using a 555" method. I will
ask more questions if I bump into more problems.

Thanks very, very much !

Allen
Author: Bob Monsen
Date: 18:53 17-04-08

"Allen Bong" <allenbsf6502@gmail.com> wrote in message
news:cb54d3d6-0e79-468f-bdc9-5b0fc90602ed@f63g2000hsf.googlegroups.com...

How are you powering it? Also, what it is the output going to be doing
(besides making your multimeter register 0)?

If you are trying to shift from the range [12,24] to the range [0,12], then
you might be able to simply reference your output ground up to 12V, and use
the actual input rather than messing around with this level shifter

If the input you are driving is very high resistance, you can just use a
voltage divider to generate the 'virtual' ground.

Regards,
Bob Monsen
1


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