# 741 question

Discussion in 'General Electronics Chat' started by El_da, Dec 1, 2013.

1. ### El_da Thread Starter Member

Aug 17, 2009
10
0
Hello,

I have AC signal from current transformer :0,008 mV, 50 Hz. I want to amplifie it to 0,88 V with 741 op amp.
If I use NON-INVERTING AMPLIFIER I need 2 resistors:
GAIN (AV) = 1+(R2 / R1) = 1+(1000/100)= 1+ 10=11

My question, I have connected pin 4 to ground and pin 7 to +5V, do I have to connect pin 4 to -5V? Is it mandatory for 741 to have + and - power supply?

I have connected it on protoboard and it seems that on output I have continuos voltage even when I do not have connected 80 mV on input,

what am I doing wrong?

thanks

Jun 22, 2012
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3. ### #12 Expert

Nov 30, 2010
16,705
7,358
The 741 is a very old design. It will work with a single supply but it can not work within 1 volt of ground or 1.5 volts from the positive supply. You can either give it a dual supply or use a more modern chip that can understand a zero volt input.

Edit. The next guy is right. I was thinking about another circuit and got confused.

Last edited: Dec 1, 2013
4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,518
515
You have 80 mV signal. It is oscillating. I assume it is centered on zero. So the signal is from -40 mV to 40 mV. Your upper rail is +5 V. If you connect pin 4 to Ground, the lower rail will be 0 volts. The portion of the input signal, from -40 mV to 0 mV, will be blocked off. Do you want to amplify input that is from -40 to 0 mV? If answer is Yes, then you need -5 V lower rail so you need to connect -5 V to pin 4.

The reason you see output when nothing is connected to op-amp is because the op-amp is very sensitive and even miniscule voltages make it "do stuff". Basically you allowed the inputs to float. If you don't want this behavior, connect the inputs to ground.

5. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
You might barely get it to work with the single supply. But to do so you must bias the op-amp input at ~ 1/2 the supply voltage and feed the signal in through a coupling capacitor. But a newer rail to rail op-amp would be preferable.

You said 0,008 mv in and 0,88 volts out and a gain of 11.
It doesn't compute.
If you meant 8 mv in and 0,88 volts out the gain needs to be 110.

Last edited: Dec 1, 2013
6. ### El_da Thread Starter Member

Aug 17, 2009
10
0
This is my schema,

yes, my mistake, I meant 0,08 V not 0,008 V,

sorry,

Which "modern" amplifier would be better - TL071

thanks

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7. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
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You need just under a 2.5 volt p-p swing for a 0,88 volt AC output.
(I'm assuming the 0,88v was RMS because it was unspecified.)

With a single 5VDC supply the TLO71 looks like it could possibly reach that swing with a 10k or higher output load according to the datasheet. But at 2k or above load it can only swing about 1.5 volts p-p if I read the datasheet graph correctly.
(It appears the TLO71 can reach within ~ 1.75volts of the - and + supply rails with a >2k load, but higher with >10K)

Last edited: Dec 1, 2013
8. ### LvW Active Member

Jun 13, 2013
674
100
El_da, each opamp can be operated either with single or with double supply.
However, you have to watch two aspects:
* Some units can swing (nearly) until the rail limits and some cannot
* For single supply (Vcc) you have to bias the pos. input with Vcc/2 and two capacitors are necessary. One input coupling capacitor and one capacitor between the resistor R2 in your circuit and ground. Thus, the DC gain is unity and the bias voltage is transferred to the output.

9. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
El_da: Thanks for posting your schematic.

You are doing nothing wrong. Assuming R1 and R2 are the same value, the input is biased at 1/2 the supply, so the output will also be at 1/2 supply. This is a DC 'resting' level. The AC signal will modulate around this resting level..

We should add that a series coupling capacitor and load resistor are needed on the output for single-supply operation to work properly. Then with no signal you will see 0 VDC on the output load resistor with no signal present, but you will see the AC signal when it is applied.

Last edited: Dec 1, 2013
10. ### LvW Active Member

Jun 13, 2013
674
100
Tubeguy, don`t you think that for R1=R2 the DC output will be at Vcc/2*(1+R1/R2) ? That means: Vcc.
As I have mentioned, a capacitor is required between R2 and ground.
That is the classical bias scheme for non-inverting applications.
More than that, I do not see any necessity for a load resistor. Please, can you explain?

11. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
Sorry. my bad, I didn't look closely at the schematic and assumed the R1/R2 junction was going to the + input to act as a bias voltage divider. I missed a few things.

I agree that a cap at the junction of R1/R2 resistors is needed.
I mentioned the series coupling cap based on my incorrect assumption.

Last edited: Dec 1, 2013
12. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,229
The 50 Hz signal is 8 uV varying 4 uV above zero and 4 uV below zero?

If your signal is varying above and below zero, you will need dual power supplies. If you want your 0.88 V signal varying about a DC level, you can use a single power supply.

What does the OP really want to do? Is this some school assignment?

13. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
The OP corrected their spec. The OP has 80mv in and wants 880mv out (I think).

Last edited: Dec 1, 2013
14. ### tindel Active Member

Sep 16, 2012
576
196
Here is a way that might work to use the Op741 for what you want - you'll need two op amps though.

The first amplifier provides a pseudo split rail to center everything to the mid voltage point of the supply voltage. This voltage is used to make a reference. The second amplifier provides a gain of 11 of the input voltage. C1 provides AC coupling on the input, C2 provides AC coupling for the output, this may or may not be needed.

Note that the green trace is with a 5V source and saturates and the blue is with a 9V source. So as your power voltage gets larger, the better chance you have to get it to work. It does appear that the TL071 will work at either voltage (per the ti spice model, but sometimes the models don't simulate very well). Another spice model of the LM741 also shows it working... so it might work... only one way to find out. I wouldn't want to depend on it if you're making 1 Million of something though.

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15. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,403
1,229
Here's the way I would have done it using a 5 Volt source and one way of doing it with a dual +/- 5 V sources.

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16. ### LvW Active Member

Jun 13, 2013
674
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I think, I have to modify resp. to supplement my above quoted answer.
The mentioned capacitor in the feedback path (for producing a DC gain of unity) is not absolutely required.
Another alternative would be to retain the gain of two and to bias the non-inverting opamp input with 0.25*Vcc - using a resistive voltage division ratio of 1:3.
In this case, the ouput is biased (as desired) with 0.25*2*Vcc=0.5*Vcc.

17. ### tindel Active Member

Sep 16, 2012
576
196
I initially thought to do it that way also but ran into problems (in simulation) because I had set my bias point to divide the rail in half to be at 2.5V. The amplifier then wanted to output ~25Vdc... which won't work on a 5V rail. I will admit that I don't recall thinking to set the bias point lower. But I don't think setting the bias point lower will work though because of the common mode input voltage of the 741 does not extend to ground... in fact, worst case common mode input voltage is 3V from both the rails and this might not work at all on a 5V rail. You really want that input voltage to split the rail the best you can in this application, IMHO.

The problem with both of these approaches is that the PSRR sucks. I've been struggling with how to correct that issue. I think you'd probably want a commercially available rail splitter to improve that. Although, placing a filter cap across R3 in my schematic would filter out high frequency noise. You could also use a 2.5V precision reference in my circuit to filter out noise, but then it doesn't work as well as the supply voltage (the DC portion) changes.

18. ### tindel Active Member

Sep 16, 2012
576
196
Here's another way to do it that has a decent PSRR due to the large split-rail filter cap (C3)... and it only uses one op-amp.

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19. ### tindel Active Member

Sep 16, 2012
576
196
FYI, I did breadboard schematic from post #18 and I showed that the circuit appears to work, but the waveform is starting to saturate on the low side. No big deal really... I could add a trim pot on R6 to lower the resistance to boost that rail split voltage.

20. ### El_da Thread Starter Member

Aug 17, 2009
10
0
Hello to all and thanks for your help,

this is a home project. I use voltage from current transformer to detect when U < 0,08 V. I tried to do it with op amp and comparator function (PIC microcontroller). I have to amplify AC signal several times for PIC comparator.
Is ADC better way of sensing this low voltages? Is there PIC microcontroller which can work with 80 mV AC signal?

thanks,