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Here Operational amplifier (UA741CP) not working in buffer configuration I asked why my circuit is not working. The question was answered (it won't work with this voltages) but also it has been pointed out that it is all written in the datasheet. Can you please help me to read the datasheet in context of the given problem? Specifically I will go through the extra information given in the answer and point out what I do not understand.

The schematic is:

schematic

simulate this circuit – Schematic created using CircuitLab

The accepted answer is:

Datasheet excerpt

This opamp isn't specified for only 5 V supply.

These parameters shown above are for ±15 V supplies. The common mode input range only goes to within 3 V of either supply. Nothing is said how that scales to lower supply voltages, so assume it is at least that much. That means there is no common input range left with only a 5 V supply.

Likewise, the output can't drive to within 3 V of the supply rails with a 10 kΩ load, and not to within 5 V with a 2 kΩ load.

First I don't know if I understand what "Common-mode input voltage range is". Here it says:

Common-mode voltage range (CMVR) or Input Voltage Range (IVR): For signal processing devices with differential inputs, such as an op amp, CMVR is the range of common mode signal for which the amplifier's operation remains linear.

If we let the voltage present on the "-" input equal V1, and the voltage on the "+" input equal V2, then the common mode voltage is VCM = (V1+V2)/2.

So in my case, if I put 0.45 V to the + of OpAmp and - to the the output what is my common mode input range? (0.45 + 0.45)/2 = 0.45? But isn't it within the quoted \$\pm\$ 13 (knowing that is for 15 V supply voltage)? Can you clarify how the datasheet entry VICR being \$\pm\$12 or \$\pm\$13 leads to conclusion that this op amp will not work with 5 V supply?

Also I completely do not understand the part:

Likewise, the output can't drive to within 3 V of the supply rails with a 10 kΩ load, and not to within 5 V with a 2 kΩ load.

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  • \$\begingroup\$ You might also try this datasheet: uA741 which includes "Recommended Operating Conditions". Not sure why the Jameco version left that out. \$\endgroup\$ – Tut Feb 15 '17 at 21:40
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    \$\begingroup\$ Have you read Opamps for Everyone, see: focus.ti.com/lit/an/slod006b/slod006b.pdf ? If not you should ! It explains most of the properties you are unfamiliar with. Datasheets are for engineers which know how to use an opamp and what properties they need. You will know the basics of this after reading Opamps for Everyone. Best of all: it's free ! \$\endgroup\$ – Bimpelrekkie Feb 15 '17 at 21:48
  • \$\begingroup\$ Please note that to use an opamp the way you want it needs to have symmetrical power ie two rails one positive and one negative. Otherwise you won't have any accuracy close to ground. It's easy to achieve with two 9v batteries. \$\endgroup\$ – user1890202 Feb 15 '17 at 21:51
  • \$\begingroup\$ True, the 741 really needs a symmetrical supply. But that complicates matters. It can be done with a single supply and a "better"opamp like an MCP601 (that's a CMOS rail-to-rail input and output, designed for single 5 V supply, opamp). \$\endgroup\$ – Bimpelrekkie Feb 15 '17 at 21:53
  • \$\begingroup\$ The book that @FakeMoustache links to discusses dual and single supply in section 4.1. Single supply is more complicated but very common, however on the desk at home it isn't necessary. \$\endgroup\$ – user1890202 Feb 15 '17 at 21:55
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So in my case, if I put 0.45 V to the + of OpAmp and - to the the output what is my common mode input range? (0.45 + 0.45)/2 = 0.45?

Your common mode input is 0.45 V, assuming the op amp is operating as you expect (so that we can assume the negative feedback causes it to drive its inputs to the same voltage). The range of acceptable common mode input voltages is still dependent on the op amp and the power supplies you use.

The common mode voltage is VCM in the below schematic (taken from this answer):

schematic

simulate this circuit – Schematic created using CircuitLab

The two differential inputs \$V_1\$ and \$V_2\$ are split into a single common mode voltage \$V_{\text{CM}}\$ and two differential voltages of opposite polarity and amplitude \$V_{\text{D}}/2\$. VICR on the datasheet specifies the range of acceptable values for \$V_{\text{CM}}\$

But isn't it within the quoted \$\pm\$ 13 (knowing that is for 15 V supply voltage)?

No. The datasheet assumes supplies of +15 V and -15 V (30 V between the supply pins). You are only using +5 V and 0 V (GND) -- that's only 5 V between the supply pins. The datasheet is saying that the common mode voltage \$V_{\text{CM}}\$ must be no less than -12 V (i.e. 3 V above the -15 V supply) and no greater than +12 V (i.e. 3 V below the +15 V supply). With the negative supply pin at 0 V, your \$V_{\text{CM}} = 0.45\text{ V}\$ is not 3 V above the negative supply as required.

Can you clarify how the datasheet entry VICR being \$\pm\$12 or \$\pm\$13 leads to conclusion that this op amp will not work with 5 V supply?

Since \$V_{\text{CM}}\$ must be no less than 3 V above the negative supply, it must be at least 3 V above your 0 V negative supply. It also must be no greater than 3 V below your 5 V positive supply, which means that it must be less than 2 V. There is no \$V_{\text{CM}}\$ that is both greater than 3 V and less than 2 V, hence a single 5 V supply is insufficient for this op amp.

Also I completely do not understand the part:

Likewise, the output can't drive to within 3 V of the supply rails with a 10 kΩ load, and not to within 5 V with a 2 kΩ load.

A lower resistance load requires a higher output current in order to apply a given voltage. For example, to apply 1 V across a 10 kΩ load the op amp needs to supply 0.1 mA. To apply 1V across a 2 kΩ load, however, it must supply 0.5 mA. The op amp output can't swing as close to the supply voltages as this output current increases -- that's why it can swing to within 3 V of a 15 V supply with a 10 kΩ load but only to within 5 V of a 15 V supply with a 2 kΩ load.

Since the output can't swing to within 3 V of the supply voltages with a 10 kΩ load, you have the same problem as with the input common mode range: there is no output voltage that is 3 V away from both supply voltages when you are only using +5 V and 0 V.

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Here is a snippet of UA741 IC schematic. The inputs need to remain 4 diode drops above VDD-.

enter image description here

Count the arrows along the two paths. Each has 4 emitter-base diodes. Thus Vin needs to be 4 * 0.7v (2.8 volts), or 4 * 0.6v (2.4 volts), or 4 * 0.5v (2 full volts) above VDD-. Then the resistors, connected to VDD-, need some voltage. And, finally, to perform well as current sources, the transistors need some Vbase_collector, perhaps 0.5 volts; without original designer's (designers') device-curves (Ic & Vc plots, as Ibase steps over numerous linear increments), or without Vearly, we are just guessing, and 0.5 volt is a good guess. With 4 transistors, we have to add another 4*0.5 = 2.0 more volts.

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If you read your reference material's next paragraph:

Some op amps, for instance, will only allow the common mode voltage of a signal to come within a diode drop or so of the power supply rails.

In your case, its more than a diode drop, if either V1 or V2 gets to within a few volts of the power supply rails, the op-amp goes non-linear. Those few volts of no-man's land are there by design and do not scale with supply voltage as you have suggested.
At op-amp output, the same situation exists. If output voltage swings up towards supply voltage, it needs a few volts to remain linear. The output cannot supply current to pull the output voltage higher. Again, this required headroom doesn't scale with supply voltage.
Internally, op-amps use entirely different circuit topology to achieve rail-to-rail performance. For something like a hearing-aid, where supply voltage is a 1.5v battery, op-amp design is quite an art.

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  • \$\begingroup\$ if either V1 or V2 gets to within a few volts of the power supply rails, the op-amp goes non-linear. Hmm, it is actually the input and/or output saturating which means the opamp will have less gain. Opamps are very non-linear by themselves, we linearize them with feedback. But that feedback depends on enough gain. So if the gain drops (what happens when stages saturate) the feedback does not work anymore and the opamp circuit no longer works as expected. \$\endgroup\$ – Bimpelrekkie Feb 15 '17 at 21:58
  • \$\begingroup\$ @FakeMoustache Granted, an open-loop opamp is a squirrely non-linear beast, but monotonic (for DC) if you stay out of no-man's land. \$\endgroup\$ – glen_geek Feb 15 '17 at 22:55
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Interesting behavior shown by the UA741 output-circuit.

Look at the output circuitry. I drew little loops where the Vcb needs to remain above zero-volts so the transistor has enough Vce to act as a stiff current source [add at least 0.5v to the required voltage].

From VCC+ to Output, headroom is 3*Vbe + 0.5 = 3*0.6volts + 0.5 = 2.3 volts. [ I used Vbe=0.6volts because the entire opamp has limited Idd budget]. One can argue whether the short-circuit-protection transistor (NPN with emitter tied to Out) must be included in headroom.

From VCC- to Output, headroom is 4*Vbe * 0.5 = 4*0.6volts + 0.5 = 2.9 volts. And some unknown-function resistors, which I think are to boost the input impedance of their bipolars, to achieve maximum gain.

Now----for that interesting behavior. At high frequencies (above Unity Gain Bandwidth, where the OpAmp has in-out gain < one), what is the output resistance? As long as Iout << Ishortcircuit, Zout is Rem1 || Rem2.

Key: this is not true for a rail-rail MOSFET opamp, which MUST use FET drains to drive the output pin.

enter image description here

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