Offset voltage on voltage follower

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Turbo46
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Offset voltage on voltage follower

Hi<p>I need a bit of help on a circuit im trying to build. I will try to explain my problem below.<p>Circuit requirements:<p>*input to op-amp = 0 to 5v
*output from op-amp = 0.5v to 3.5v<p>
Im am currently using a CA3140E opamp with a 0.5v offset in it, which is quoted in the data sheet.<p>Pin wirring on chip:<p>Pin 1: to left leg of 10k pot
pin 2: connected to pin 6 with no series resistor
pin 3: varible input voltage, 0 to 5v
pin 4: to middle leg of above 10k pot and connected to GND
pin 5: to right leg of above 10k pot
pin 6: connected to pin 2 with no series resistor, and then this voltage is dropped across 2 resistors in series to produce the output voltage ratio<p>
Im at the stage in the circuit, that when you increase the input voltage from o to 5v, the output is alway 0.5v higher than the input which is correct. At this point i drop the output voltage over a potential divider and i get the correct output voltage ratio over this. The problem starts when a 1k load resistor is taken from the potential divider. At this point the voltage drops. When the load resistor is removed the circuit produces the correct output voltage again.<p>I think the problem is the circuit cant provide enough corrent, but i cant see how to change this.<p>Any ideas?

dyarker
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Re: Offset voltage on voltage follower

self deleted - see others and my new post below<p>[ August 19, 2004: Message edited by: Dale Y ]</p>
Dale Y

Ron H
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Re: Offset voltage on voltage follower

If you're using the CA3140's offset null to inject 0.5v offset, you're barking up the wrong tree. The offset null is meant to compensate for a few millivolts of offset error inside the op amp. The op amp may not even work correctly if you try to get 0.5v, and it certainly won't be temperature stable.
What you need is a voltage divider, as you surmised, but it will be a 3-resistor divider which includes a resistor to the positive supply rail. This will need to be on the input side of the op amp, so that you will be able to drive your 1k load on the output side. If your source can't drive a voltage divider, you may first have to buffer the source with another op amp.
What is your supply voltage? Be advised that the output of the CA3140 can only reach 3 volts on the high side if your supply is only 5v. If it is higher, then this shouldn't be a problem.
BTW, why do you need to get 0.5-3.5v from a 0-5v input?

cato
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Re: Offset voltage on voltage follower

You are expecting the voltage at the node of the voltage divider to remain the same even if you add load to it. It wont. The op amp is controlling the voltage at the input to the divider. The first resistor limits the available output current. When you add the load resistor, you are splitting the available current between the second part of the divider and the load resistor. In effect you are simply lowering the resistance of the second voltage divider resistor by placing your load resistor in parallel with it.

dyarker
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Re: Offset voltage on voltage follower

Put a feedback resistor from pin 6 to pin 2.
Remove the output voltage divider resistors.
Put a resistor in series with your 0V - 5V source. It's value needs to have the same ratio to the feedback resistor as the two divider resistors have now.<p>ie instead of dividing the output (and making the impedance higher), set the gain to less than one for the same output voltage (and keep the impedance low).<p>Cheers,<p>[ August 19, 2004: Message edited by: Dale Y ]</p>
Dale Y

Turbo46
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Re: Offset voltage on voltage follower

The reason for having the input voltage different than the output voltage:<p>Currently i have a 0 - 5V transducer and it keeps breaking. I have now found a 0.5 to 3.5v tranducer which is more robust and dosnt break. Therefore im trying to fall the reading system, so it thinks its reading the 0 - 5v, instead of the 0.5 to 3.5v.<p>ere i think thats proper english. lol<p>Thanks to all for your help. This site always comes back with lots of good answers!!

Turbo46
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Re: Offset voltage on voltage follower

I have tried the mods, but i cant get the circuit to work properly. Its the 0.5v voltage increase on the output is the part i dont understand. No op-amp cookbook has this weird application. All the null offsets are used for error correction and not output voltage increases.<p>could the replies go back to the begining and start again, and also give resistor values if possible.<p>cheers for your help.

dyarker
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Re: Offset voltage on voltage follower

From transducer to op amp non-inverting input a resistor 20 to 50 times the transducer impedance.<p>The feedback resistor (op amp output to inverting input) a resistor 5/3 the value the transducer input resistor. This sets gain so 3V swing of new transducer becomes a 5V swing out.
0.5V * 5 / 3 = 0.833V
3.5 * 5 / 3 = 5.833V
5.833V - 0.833V = 5V<p>From an accurate positive voltage source a resistor also to inverting input. The value of this resistor must be such that it's current is exactly the same as 0.833V causes in the feedback resistor. This subtracts 0.833V from the output.<p>Before connecting that resistor to an accurate voltage, and before connecting the transducer resistor to the transducer, ground them both; and adjust the offset pot for zero volts out of the op amp.<p>Now when you put a 1K Ohm load on the output, the voltage will try to go down, but this decreases the current thru the feedback resistor to the INVERTING input which raises the voltage back up.
In actual fact the output voltage does decrease but the error is the output voltage divided by the open loop gain of the op amp. That's 100,000 to 10,000,000! Unmeasurable by mere mortals.<p>I might try to figure real values for you, but I'll need the impedance of the transducer.<p>C U L -
Dale Y

Turbo46
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Re: Offset voltage on voltage follower

Cheers for your help. Its a good learning curve for me.<p>The input impedance at the 5v end is 10k, and at the output, 3.5v end the impedance is 100k. Also the op-amp will hopefully be powered by 0 and 12v, and try and steer away from -12v. This will hopefully make things easier for me.<p>Hope all this seems logical.<p>cheers

rosborne
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Re: Offset voltage on voltage follower

Have you considered:
A 3/5 voltage divider for your 0-5 volt signal. Hoping for a 0-3 volt output.
Divide your DC voltage down to 0.5 VDC.
Then taking those two inputs into an OPAMP configured as an adder to get 0-3.5VDC out?
7 resistors and an OPAMP.
Would that work?
-Rick

MrAl
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Re: Offset voltage on voltage follower

Hello there,<p>
When you stated:<p>QUOTE
-----------------------------------------------------------------
Currently i have a 0 - 5V transducer and it keeps breaking.
I have now found a 0.5 to 3.5v tranducer which is more robust
and dosnt break. Therefore im trying to fall (fool) the reading system,
so it thinks its reading the 0 - 5v, instead of the 0.5 to 3.5v.
-----------------------------------------------------------------
END QUOTE<p>what it sounds like you mean is that you need an amplifier that
takes a new input 0.5 to 3.5v and makes it 0.0 to 5.0v because
you have a new transducer that now puts out a lower voltage and
also has an offset.<p>This can be done using a pretty simple circuit that uses an op amp.
Only problem is, we dont know what kind of supply voltages you
have to use with your soon to be new op amp circuit. We also
dont know what kind of time response you need or what you had.<p>Assuming you have a plus 10 volt supply and at least a 5 volt negative
supply you can create a low impedance voltage divider from ground to
minus 5 volts that gives you minus 0.5 volts. You then connect the
you've already simply and effectively eliminated the offset of 0.5 volts.
The output of the transducer will now look like 0 to 3 volts to the
remainder of the circuit, which can simply be a non inverting amplifier
with a gain of 5/3. With a resistive divider made of two resistors
900 ohms and 100 ohms with the 900 ohms run to minus 5 volts and
the 100 ohm connected to ground you have at the center of the two
resistors minus 0.5 volts which is where the ground connection from
the new transducer connects.
Because you say your new transducer has an internal impedance of 10k
we'll adjust the parallel combination to be equal to 10k.
The gain ratio will be 5/3-1 or 2/3, and because we want the parallel
combo to be equal to 10k this gives us two equations:<p>10000=R1*R2/(R1+R2)
and
R2/R1=2/3<p>Solving the second equ for R1 gives us:
R1=R2*3/2<p>Substituting this for R1 in the first equ leads to:
R2=50000/3=16667 ohms
Substituting this back into the second equ gives us
R1=25000 ohms<p>Going over the connections...
Connect the op amp V+ supply to +10v and the V- supply to -5v.
Connect R2 as feedback, from op amp output to op amp inverting input.
Connect R1 from op amp inverting input to ground.
Connect transducer positive output to op amp non-inverting input.
Connect 900 ohms in series with 100 ohms which forms a junction.
Connect loose end of 900 ohms to minus 5 volt supply.
Connect loose end of 100 ohms to ground.
Connect transducer minus (ground) lead to junction of 100 and 900 ohms.<p>Turn power on <p>
Take care,
Al
LEDs vs Bulbs, LEDs are winning.

dyarker
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Re: Offset voltage on voltage follower

I'm confused! Are we reading the transducer, or driving the transducer???<p>Read the posts. It does say "sort of" output TO new transducer 0.5 to 3.5. But if this is the case, what does the 1K load have to do with anything??<p>It might help if I knew what that transducer was.<p>Cheers,<p>[started working on this post before MrAl's excellent post (except I thought you said transducer is 100K]<p>[ August 20, 2004: Message edited by: Dale Y ]</p>
Dale Y

Turbo46
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Re: Offset voltage on voltage follower

Hi again<p>The 100k is the transducer, and 10k is the instrument. When i first built the circuit i put 1k in to test it. Then i was going to put a 100k in, but i didnt have one available.<p>Sorry to muck you around.<p>Thanks again.

Turbo46
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Re: Offset voltage on voltage follower

Back again<p>Circuit requirements:<p>*input to op-amp = 0 to 5v, from instrument
*output from op-amp = 0.5v to 3.5v, to transducer<p>This is in relation to my above post.<p>cheers.

dyarker
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Re: Offset voltage on voltage follower

I think I've got a mental block over this and maybe can't be of assistance.<p>"Currently i have a 0 - 5V transducer and it keeps breaking. I have now found a 0.5 to 3.5v tranducer which is more robust and dosnt break. Therefore im trying to fall (fool) the reading system, so it thinks its reading the 0 - 5v, instead of the 0.5 to 3.5v."<p>Which leads me to assume the transducer is some kind of sensor that outputs 0.5V to 3.5V depending on some kind of physical phenomenon (temperature, humidity, pressure, acceleration, etc). The transducer could have two leads (output and common), or three leads (power from from the instrument, output to the instrument, and common).<p>A transducer can be an output. An example is piezio(sp?) electric crystal that outputs ultrasonic waves when it gets a signal from the instrument. If this is the case, then change the ratio to 3/5 (vs. 5/3) in the formulas provided by MrAl. In fact your first circuit would work if the total of the divider resistors is around 2K. (Even though you're mis-using the offset null to get the 0.5V as RonH said.) The problem was the 1K load resistor drawing 100 times more current through the upper divider resistor than the 100K transducer would.<p>On the other hand, if the transducer is a sensor then:
"*input to op-amp = 0 to 5v, from instrument
*output from op-amp = 0.5v to 3.5v, to transducer"