# 741 Question

Discussion in 'The Projects Forum' started by Piggins, May 5, 2014.

1. ### Piggins Thread Starter New Member

May 5, 2014
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Welcome to AAC!

A thread belongs to the OP (original poster). Trying to take over someone elses thread is called hijacking, which is not allowed at All About Circuits. I have therefore given you a thread of your very own.

Can someone clarify to me what happens with this circuit when ,for example, a 1mV signal is brought to the input and Vcc is 5Volts so 2.5Volts at the noninverting input?

If I understood the guide correctly that op amp is trying to hold both inputs at 2.5Volts so the inverting input would be at 2.5Volts and the output at gain*2.5?

But how does the 1mV signal get amplified then? How does it effect the operation?

Last edited by a moderator: May 6, 2014
2. ### AnalogKid Distinguished Member

Aug 1, 2013
4,686
1,297
The circuit is doing two things at once. It amplifies the *difference* between the + and - inputs, and it offsets the output a fixed DC amount. The + input is fixed at 1/2 Vcc. The - input DC value can float up to 1/2 Vcc because of the 0.1uF input coupling capacitor. So now the DC values of the two inputs are the same and the reference input is 1/2 Vcc so the output DC value is 1/2 Vcc.

Another way to think of this is that at DC the input capacitor looks like an open circuit so the 1K resistor at the - input no longer does anything and the circuit reduces to a voltage follower. At DC the circuit has a gain of 1, even though at audio frequencies it has a gain of 100. The input 2.5V becomes a 2.5V output, and an input 10 mV sinewave becomes an output 1 V sinewave.

Something to consider is that in order for the circuit to have a gain of 1 at DC and 100 at audio frequencies, the circuit is acting as a highpass filter. The corner frequency is formed by the input capacitor and resistor.

ak

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3. ### Piggins Thread Starter New Member

May 5, 2014
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So if I was to connect the output to an oscilloscope it would give me 2 curves, one for the DC gain and one for AC?

4. ### Piggins Thread Starter New Member

May 5, 2014
26
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I could not find the edit button so Ill make a new reply.

I could not find the open loop gain for 741cn in the data sheet but if we say its 100 000 does that mean that the DC signal difference between the inverting input and the noninverting input is 2.5/100000 = 25uV?

Is this 25uV the difference where we add or take the instantaneus AC signal voltages from and amplify the result with the amplification made by the resistors? And after the amplification add them to the 2.4999Volts at output?

Does the amplifier take the AC signal, amplify it by around 100 and then add it to the voltage if its a positive or decrease it from the voltage if its a negative one?

So at if a wave is a sinewave with high peak of 1mV and low peak of -1mV it would be:

At high peak: 1mV+25uV times 100 = 0.1001 Volts high at output

Low peak: -1mv+25uV times 100 = -0.0975 Volts at low

So we would get a voltage swinging in the range of 2.5Volts(at point 6 in the picture)+0.1001Volts and 2.5-0.0975volts?

So 2.6001high 2.4025low? And after the decoupling cap its a wave swinging in both positive and negative side providing that something can supply that negative side?

And if the gain was set to 1000 it vould swing in about +1volt -1volt? (after cap)

http://html.alldatasheet.com/html-pdf/53593/FAIRCHILD/741CN/813/2/741CN.html

5. ### atferrari AAC Fanatic!

Jan 6, 2004
2,663
784
Hola Piggins,

My friendly suggestion: after you got the answer from the Kid, could you try to simulate the circuit (should be easy) or even implement it?

I would do that and THEN come with more questions if any.

6. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,436
1,625
Another friendly suggestion: throw the 741 into the trash bin of obsolete parts and get something designed in the current century. It was designed at a time when "analog" meant you have + and - 15 volt rails... well, + and - 12 volts minimum.

While there is no way that device will do any useful work for you when only given 5 volts, there are many many current devices designed to work off a single supply at low voltages.

Mar 24, 2008
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8. ### Veracohr Well-Known Member

Jan 3, 2011
559
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In general, a 1mV signal at the input would give you a 100mV signal at the output centered on 2.5V (before the capacitor), or in other words, 2.4-2.6V. After the capacitor you'd have a -100mV-100mV signal. In reality the post-capacitor signal may still be offset a bit, and the imbalance of resistor values may cause some offsets.