LM358 op-amp

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

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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

Reply to
Allen Bong
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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

Reply to
Bob Monsen

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

John

Reply to
John Larkin

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"
Reply to
Jamie

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

Reply to
Allen Bong

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.

Reply to
PhattyMo

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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

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Reply to
Allen Bong

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

Reply to
Bob Monsen

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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

Reply to
Allen Bong

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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

Reply to
Allen Bong

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

Reply to
Bob Monsen

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