Converting a signals voltage range

I must Convertia signals voltage range from 0-4.8 mV to 0-5 V.
I have a LM741 OpAmp.
Can i use it?
how?

Thanks
N

Yup, you can use that op amp.

(you can ignore Rz)

The gain is determined by A=1+Rf/Rin. So to amplify the signal by 10 times, you would need Rf/Rin = 9.

So could use a Rf = 10k, and Rin = 1.2k, and that would get you a gain of 9.333, which is pretty close.

May I ask what you are using it for? If it is to feed a signal into an Arduino's ADC, you could just set the ADC reference to external, and then connect a 270k resistor from the AREF pin to 5v.

I need 1250 times from 4.8 mV to 5V
I have a LM741.
i have 120kohm resistor and 95.3 ohm

i’ve just use all how in tou schema, but i have 1.9 V when sensor (loading cell) is 0.
what is the problem?
Mi cell is:
http://www.emmeshop.it/product.php?id_product=239
http://www.phidgets.com/documentation/Phidgets/3132_0_Datasheet.pdf

What else can i do?
thanks
N

First, you are asking for an amplification of about 1000 (0 - 0.0048 => 0 - 4.8 V) which really requires at least two stages of amplification to be stable (IMO). Second the 741 can not approach its power rails, so in order to achieve a range of 0-4.8V the power rails have to be both below 0V (negative) and above 5V... If you don't already have such power rails available, it would be better to choose another opamp with rail to rail capability.

Insert my schema.

I use 0 and 5V.

contecavour: I need 1250 times from 4.8 mV to 5V

Actually 4.8mV to 5V requires an amplification of 1041.6666667

contecavour:
I use 0 and 5V.

Then as I said the 741 WILL NOT produce the output range you want. Select another opamp

Can you suggest an opAmp? There is a two stages in a single chip?

I have 0-5 V from arduino e 0-4.8mv from cell.

Thanks N

There are what are known as dual or quad op-amps which have either two or four opamps per IC.

One rail to rail that should work is the TI LMV358 (dual) or the TI LMV324 (quad). Personally I would go with the quad and have three stages each with an amplification of 10... (using the values mentioned above).

In general the higher the amplification one attempts in a single stage, the more likely problems will be encountered--about a 100 amplification per stage is a good max for rule of thumb

Here are just a small sampling of other possible OP AMPS you could use...

MCP6002-I/P MCP602-I/P MCP607-I/P TLV272IP TLV2372IP TLC2272CP TLV2462CP OPA2340PA OPA2743PA OPA2350PA OPA2336P OPA2251PA OPA2241PA OPA2704PA

Thanks wanderson . I'd like to have simpliest circuit. If i use TI LMV358 (dual) with 35 of amplification for each stage -> 35*36 = 1260 that is what i need. what do you think?

N

contecavour:
Thanks wanderson .
I’d like to have simpliest circuit.
If i use TI LMV358 (dual) with 35 of amplification for each stage → 35*36 = 1260 that is what i need.
what do you think?

N

Not quite; first consider that available components (resistors) and their tolerances will affect the actual gain versus the theoretical gain. And again, your math is off, you do not want a gain of 1260… which would take a signal of 4.8mV (0.0048) and produce an output of 6.048V… Actually what would happen is that since your positive supply is only 5V, any signal over (5V/1260) or about 3.97mV would produce a 5V output… This is called clipping… You need a gain of approximately 1000 for your sensor…

So for a two stage amp you want a gain of approximately 32 for each stage.

you could just set the ADC reference to external, and then connect a 270k resistor from the AREF pin to 5v.

Can you (anyone) give a bit more background info, please? Is there a defined and precise internal resistance between AREF and GND, such that this high resistor builds a reliable voltage divider ?

What about setting analogReference(INTERNAL);   ( = 1.1 V ? ), and use a 2 stage dual opamp ?

contecavour: I need 1250 times from 4.8 mV to 5V

Sorry about that, I missread what you wrote as 0.48V

Also i have missread.

http://www.phidgets.com/documentation/Phidgets/3132_0_Datasheet.pdf Rated Output 0.8±0.1 mv/V Excitation Voltage 5 VDC

0.8 x 5 = 4.0 mV

Arduino needs 0 to 5V

5V / 4mV = 1250

Nic

Tom, not sure I agree. IIRC, the minimum AREF voltage that can be externally supplied is 1V for the Atmel 328. Op-amps hence have their place, but amplifying by a factor of ~1000 will be interesting in terms of keeping the noise down.

Tom, not sure I agree. IIRC, the minimum AREF voltage that can be externally supplied is 1V for the Atmel 328. Op-amps hence have their place, but amplifying by a factor of ~1000 will be interesting in terms of keeping the noise down.

What can i do. I have a signal by 4mV and i need a signal 5V

N

contecavour: Also i have missread.

http://www.phidgets.com/documentation/Phidgets/3132_0_Datasheet.pdf Rated Output 0.8±0.1 mv/V Excitation Voltage 5 VDC

0.8 x 5 = 4.0 mV

Arduino needs 0 to 5V

5V / 4mV = 1250

Nic

Okay, you might also want to consider looking at all of those error terms, such as the ±0.1 mv/V and the usual 5% for 5V supplies (4.75V - 5.25V) when doing your calculations.

For instance

0.7mV * 5 = 3.5mV (5/0.0035 ~ 1428) 0.9mV * 5 = 4.5mV (5/0.0045 ~ 1111)

and then including the 5V tolerance;

0.7 * 4.75 = 3.325 (4.75/0.003325 ~ 1428) 0.7 * 5.25 = 3.675 (5.25/0.003675 ~ 1360) ...

It is good to understand the boundary conditions ...

I have Arduino Mega analogReference(INTERNAL); became analogReference(INTERNAL1V1);

Is it correct?

N