Question of Op Amp application

[markcalaway]

understand ic ca3130 application

Hi. I am facing some problems in using an opamp. I hope someone can help me out here.

Basically, I am supposed to create a variable voltage regulator with these criteria.

1) Supply voltage of OpAmp needs to be a single supply voltage. ( Eg +Vcc = 10v, -Vcc = Gnd)

2) Input Voltage at the inverting input = 0 to 10v
3) Output Voltage = 0 to 5v

So basically whatever the input voltage is, the output voltage should be half of that. I thought the circuit below would be a perfect solution to this. But the problem is whenever I connected my circuit on the breadboard like so, the output voltage always give me erratic readings commonly around 0.6 - 0.8v. I have checked my circuit connection several times before and I am not really sure what is the problem. As for the Opamp, I am using ic LM 358N. I have also tried LM 258N. Both are dual low power opamps and they are supposedly powered by a single power supply. Could someone tell me where I could be possibly going wrong here or am I even using the right chip ??? Below is the link to the circuit diagram. Thanks to anyone who helps out.
**broken link removed**

 

[ptmsl]

This circuit cannot work with single supply because the output from the first opamp will invert the input with respect to the reference at the noninverting input. So any voltage between 0-10V will try to move the output to 0 to -5V. Remember that your output cannot go under 0V (it cannot even go to 0V for the opamps you tried because they are neither rail to rail input or output). Try moving the reference point of the opamp (the noninverting input) at a higher level (eg. 5V) so that the output will invert the input with respect to the reference of +5V. You can use a decoupled voltage devider to produce +5V. Use rail to rail opamps if your output needs to go close to 0V

 

[markcalaway]

Alright, just to make things clear. So as long as the Opamp is operated by a single supply, the output voltage can never go below 0 volts ? And sorry if this sounds noobish, but what does rail to rail really mean ? Thanks alot for taking the time to answers these questions. I am pretty new to these whole electronics stuff.

 

[sunking]

yes, the output can go to 0 even for rail to rail opa.
and for single supply, the output is around Vdd/2.
your spec is impossible.

 

[IanP]

Try to divide the input voltage (0-10V) using to identical resistors, say 1kΩ, and connect this divided voltage (0-5V) to non-inverting input of an opamp .. Also, connect inverting input to the opamps output (voltage follower configuration) .. And the last thing, the 358 opamp is not good for working close to 0V .. much better results you will have with the CA3130 (or similar) ..

Rail-to-Rail input/output voltage swing: see picture below ..

Regards,
IanP

 

[markcalaway]

Try to divide the input voltage (0-10V) using to identical resistors, say 1kΩ, and connect this divided voltage (0-5V) to non-inverting input of an opamp .. Also, connect inverting input to the opamps output (voltage follower configuration) .. And the last thing, the 358 opamp is not good for working close to 0V .. much better results you will have with the CA3130 (or similar) ..

Rail-to-Rail input/output voltage swing: see picture below ..

Regards,
IanP

I just realised this. If this is the configuration, what is the use of the opamp ? I mean the same result could be accomplished just from the voltage divider without the aid of the opamp. Am I right ??

 

[IanP]

In this configuration opamp works as buffer with very low output impedance (close to 0Ω) and can sink/source (in theory) infinite output currents ..
If you use just resistors without a buffer the impedance of a load will affect the voltage ..
Regards,
IanP

 

[ptmsl]

Sorry if my post was unclear. The problem with your circuit is that the first stage (and the second one)is an inverting opamp and the output is minus 0.5 times the input. So for a voltage of 5V at the input (not the input of the opamp but the input of the circuit) the output of the first opamp will try to go to -2.5 but since its negative supply is 0V the output of the opamp will approach the negative supply (0V) as much as its design will allow. This is why your reference voltage meaning the voltage at the noninverting input of the opamp should be at a midpoint of the supply. Another problem is that with most opamps the operating range of the inputs (both the inverting and noninveerting) does not include the supply rails even if the negative is "earth" (earth is just a convention- a reference voltage that we designate). Also the output of most opamps cannot reach the supply rails. A rail to rail input and output opamp is one that its outputs can reach the supply voltages at low output currents and its input operating range includes the supply voltages. The imput of some JFet opamps can operate even under the negative supplpy (but not above the positive).

 

[markcalaway]

Sorry if my post was unclear. The problem with your circuit is that the first stage (and the second one)is an inverting opamp and the output is minus 0.5 times the input. So for a voltage of 5V at the input (not the input of the opamp but the input of the circuit) the output of the first opamp will try to go to -2.5 but since its negative supply is 0V the output of the opamp will approach the negative supply (0V) as much as its design will allow. This is why your reference voltage meaning the voltage at the noninverting input of the opamp should be at a midpoint of the supply. Another problem is that with most opamps the operating range of the inputs (both the inverting and noninveerting) does not include the supply rails even if the negative is "earth" (earth is just a convention- a reference voltage that we designate). Also the output of most opamps cannot reach the supply rails. A rail to rail input and output opamp is one that its outputs can reach the supply voltages at low output currents and its input operating range includes the supply voltages. The imput of some JFet opamps can operate even under the negative supplpy (but not above the positive).

Thank you alot. Now I understand. Also thanks to the other guys like Ianp and sunking for clearing my doubt. I appreciate it.