non-inverting OPAMP

[floflo]

Hello everybody, I will have a work about op-amps. I must design a non-inverting amplifier with a voltage gain of 11. When the input voltage is larger than 10mV a LED must illuminate and when the output reach 11V a LED must illuminate also.
Actually we didn't do op-amps as comparator in class yet, so I started to check by myself on the net and in my electronic book.
But I am sure at all if I do understand. I tried to draw a design using 2 op-amps (lm741) as comparator for the input and another one op-amps (lm741) using the inverting as feedback to reduce the voltage gain at 11 and amplify.
Is my drawing definitely wrong?
thank you very much.

 

[JoeJester]

At 11V is the same LED lit or is it a second LED being energized?

Can you type exactly what your professor had in the problem statement?

your drawing is an attempt at a window comparator feeding an amplifier with a gain of 11. you have your reference at ground and you have supply voltages of only 11V. You will need rail to rail opamps to get close to 11V output using an 11 V supply.

 

[crutschow]

As Joe noted you need a higher supply voltage with a 741.
The old 741 output can go no closer than about 2-3V of the positive and negative supply voltages before the output saturates.

 

[floflo]

thanks for your answer. Oh yes I read that this op-amps is old and not efficient, only school still using it, It means that I have to supply my op-amps at around 14V even 15V?
Only my LED at the right far must lit at 11V, and my first LED at 10mV.
design a non-amplifier with a voltage of 11 with the following features, when the input voltage is larger than 11mV p-p a red LED must lit to indicate signal presents, and when the output reaches 11V p-p a green LED must lit to indicate that the maximum voltage is reached. That's my problem statement...
I thought to use a window comparator because the voltage at the input is very small to compare (10mV) with one op-amps, is it a bad idea?
thank you.

 

[JoeJester]

You have one amplifier with a voltage gain of 11. This has NOTHING to do with the supply voltage being 15V. In fact, with the 741, crutchow explained the output will be lower than the supply voltage by 2 to 3 volts.

You can setup two comparators ... one for the > 10 mV and one for the > 11V. The signal feeding both comparators is the output of your x11 (voltage gain of 11) amplifier. Are you restricted to using the 741 as your comparator as well?

Are there any constraints on the resistor series? (see attached file)

 

[floflo]

ha okay sorry I do understand what you mean now with the output voltage.
yes we only have to use the 741.
we only use E12 series resistors.
I tried to redraw my circuit, does it make more sens?
Thank you very much

 

[JoeJester]

That is one way. Your reference could be a single string of resistors, using three resistors. However, being restricted to the E12 series, it might be best using it the way you have it drawn.

With my comments on the output voltage ... it was the output of the x11 amplifier. With 15V supplies, you will get 2-3 volts lower than 15 volts at the output of that amplifier.

Go ahead and fill in the remaining values of the resistors. Repost your schematic when your done.

Are you using a single or dual (+ and -) 15 volt supplies?

 

[floflo]

I choose this method because my electronic book only shows up 3 methods : voltage divider reference, battery reference, zener diode set reference. I didn't know about a single string resistors thank you .
I am using a + and - 15 volt supplies, I suppose it's a dual supply?... ( sorry english is not my first language).
I repost the schematic with my resistors value, and also the list of the resistors that I am allowed to use.
Thank you very much.

 

[atferrari]

You attached two lists of resistors.

 

[JoeJester]

Are we to assume you are using the 1/4 W 5% (E24 series) resistors ?

You calculated the 10 mV input improperly.

Correlate your diagram to the problem statement.

 

[floflo]

Do you see the same picture? maybe I did a mistake. One picture in the list of the resistors that I am allowed to use.
And the second picture is my schematic with my chosen resistors.
thank you

 

[floflo]

Yes I use these resistors.
ha really... I only see this formula in my book, I will search more if I can find out. thank you

 

[JoeJester]

Look at the words on your problem statement. You should be able to point each of those, plus more, on your schematic.

 

[floflo]

That's true... With this drawing I have 2 output comparator and none one output/one input. I think I see my mistake thank you. But can I keep these resistors value for my input voltage divider comparator?
We were supposed to do this lesson last week, but because some student are striking the school is close, sorry to ask a lot of questions.

 

[JoeJester]

If you have a 9 mV DC signal at the input to the amplifier, what is the output of the amplifier, and is the low LED energized?

 

[floflo]

My output will be low, and my LED will be off. is that right?

 

[JoeJester]

No. That is not how you designed it.

With 9mV at the input to the op amp, what is the input voltage to the comparator?

Re-read your requiements.

 

[floflo]

I redraw my schematic. My book says that : when Vin exceeds the reference voltage, the output goes to his maximum voltage.

To my understanding if Vin for my output comparator exceed my Vref (11V) my LED goes ON.

And for my input comparator when Vin exceed my Vref (10mV) my LED goes ON.

 

[JoeJester]

There was nothing wrong with the previous topology.

What you had incorrect was Using 10 mV as the threshold.

I wanted you to realize the 10 mV into the x11 amplifier produced 110 mV at the output and 110mV was the threshold you needed and not 10 mV.

What input to the amplifier causes the amplifiers output to be the maximum voltage ... Assume the maximum voltage to be two volts lower than your selected Vdd.

 

[floflo]

Ha yes of course it makes a lot of sense... sorry to didn't see this mistake.

(for the 11V output comparator) 1v to my input causes to be the maximum voltage to the output because 1v * 11(AV) = 11V and the threshold is 11V.

With 9 mV input, the output of the amplifier * 11 = 99mV, with my wrong threshold is 10mV<99 mV my LED is ON.

Thank you very much .