did you try to change the C for another R? 

Hi Javier, think a little more :-)

See pg 24 in

https://www.researchgate.net/publication/281480600_Theory_of_electronic_circuits?ev=prf_pub

Book Theory of electronic circuits

Ok, but what are you trying to do? the image says that is an noninverting amplifier, that circuit works without any C

The circuit will not work!

ok, lets start over, what are you trying to do? i read the file, is about AO basics and characteristics, 

This is the basic trick question. This circuit will not work - and you should be able to say why? In my "apprenticeship years" (45 years ago) above said problem nicely bothered me :-)

In you circuit any polarization datail is missing.

If you circuit is an AC equivalent, the AC equivalent of the circuit that bias V+ terminal should be present.

If you do not plan to use any bias for V+, I don't know why it should work, since V+ DC current is blocked by the input capacitance.

sim

P.S. After writing my answer I have read pag. 24 of your book, and I found the same answer written by you. So where is the trick ?

Dear Mr. Simone Orcioni,

Maybe I did not express full exactly. In that case, I apologize. Of course, the issue is that there must be "DC Way" also to the non-inverting input. The mentioned situation is quite clear. The situation is less clear if the capacitor is "hidden" in the signal source - and it's not obvious at first sight.

To solve the problem, we have to add the proper resistance from the non-inverting input to ground - this will substantially change the input resistance of the structure.

With modern OPAMP I have succesfully used biasing resistances up to 10 mega ohm. It is enough for the most of the AC application that I know.

Yes, and if necessary, we can use the "bootstrap".

Nice exercise. A pity I missed it due to being offline some time :(

Nice basic op-amp use question :)

Another problem with op-amps is using the gain product bandwidth deliberately as a first order low-pass filter. Doesn't always give the expected result.

Hi Susan

Yes, I like the bootstrap. On the 'high' frequencies behaves like an infinite resistance (ideally) and at low frequencies as inductor in series with resistance. See for example

https://www.researchgate.net/publication/281495071_Narrow-band-reject_filter_with_a_real_operational_amplifier

To using the gain product bandwidth deliberately as a first order low-pass filter see:

https://www.researchgate.net/publication/283329017_The_fifth_journey_-_a_short_look_in_unreliable_country_of_R_filters

Josef

Conference Paper Narrow-band-reject filter with a real operational amplifier

Research The fifth journey – a short look in unreliable country of “R filters”

Susan and Josef ,

interesting examples!

We even can exploit the pole(s) of an opamp for realizing "R-oscillators".

And there is another story about ideal vs. real opamps:  There is a well-known oscillator structure that will oscillate for real opamps ONLY. It will not oscillate if "everything" is ideal. Funny.

If the amplifier is ideal, does not "produce" a frequency dependent phase changes, thus it is "not able to oscillate". Then we must use the traditional good feedback circuits :-) Their properties determine required - not by adverse variance parameters of amplifiers. In practice, it is advisable to work where the amplifier is somehow perfect.

If you allow me, I will go back to the initial essential point and will supplement it with a few more questions.

IMO it is interesting to see why these "10 mega ohm biasing resistances" (Simone) are not internally implemented. Are they really "biasing"? If so, how they bias the op-amp?

MOhms are difficult to implement in silicon. Though today some microcontrollers have the resistor (or an equivalent) 'on board'.

Regarding the resistor it cannot bias, but taking into account the parasitic input capacitance (single-digit pF) you get an RC constant in the multi-µs range - effectively biasing the input at around VCC/2.

U. Dreher,

"MOhms" can be implemented as constant-current elements (transistors)... but nevertheless it is not accepted to integrate them internally. Why?

Strictly speaking, the alone resistor cannot bias; the voltage source connected in series with this resistor is that which can bias. But where is this voltage source here?

Hi Cyril

Depending on the type of input transistors - serves as a source of supply for this resistor "negative or positive supply voltage".

Of course, everything "works well" only when the DC strong negative feedback exists.

Josef

Hi Josef,

Your question is excellent but there is a need of a reasonable explanation of the role of the input "biasing" resistors connected between the op-ap inputs and ground (or other appropriate point). My opinion is that they are not biasing resistors; if they were such, they would determine the bias current... but they don't...

Hi Cyril,

my point of view - see annex

Josef

A picture is worth a thousand words... and a series of five pictures is even more:)

It now remains only to summarize all this wisdom in one conclusion...

Hi Cyril, there is five pictures there.

And conclusion everyone is able do yourself :-)

Josef

OK, I have corrected it... and I am eager to see the conclusion:)

Conclusion: we must pay great attention to quiescent conditions of OPAs (all circuits generally) - to function properly.

Dear Josef,

Frankly, I expected more... Well, obviously I myself have to write what I want to read:)

As you have shown in your nice pictures, the transistors of the op-amp input differential stage are biased from the side of the emitters by inserting there a bias current source. It injects the bias current directly into the common emitters of the two transistors forming the long-tailed pair. Thus the emitter current source forces the transistors to adjust their base-emitter voltages, accordingly their base currents, so that to pass the half emitter currents through their collector-emitter junctions (a cascode configuration). In this way, both bases remain vacant for the input signals and the "soft" emitter current source allows the emitters to follow the (common-mode) base variations. I have considered this mechanism in detail below:

https://www.researchgate.net/post/What_is_the_truth_about_the_exotic_current_feedback_amplifier_Is_it_something_new_or_just_a_well_known_old_Is_it_really_a_current_feedback_device/1

https://en.wikipedia.org/w/index.php?title=Differential_amplifier&oldid=441555401#Biasing

Now at crux of the matter... Note that this biasing from the side of the emitters is prepared in advance but it comes into play only when we somehow close the paths of the base bias currents from the side of the bases. For this purpose, we have to connect some "jumpers" between the bases and ground (or other "stiff" point)... and in no case we should leave the inputs unconnected because the op-amp will saturate.

Ideally, galvanic input voltage sources act as such "natural jumpers". The problem is when capacitors are inserted in the input circuit (your case study). Then we need to close the path of the biasing current by additional "artificial jumpers" (resistors)... and this brings new problems...

To control the transistor from the side of the emitter, the base voltage should be fixed (common-base configuration). But if the "jumper" has relatively high resistance, the base will "move" when the emitter source tries to "move" the emitter. As though, in this case, the base resistor acts as a kind of a "base regeneration"?

Let's finally see where bias currents flow. Accoding to Cyril's Golden Rules about currents:))), every current returns to where it has started... and should be presented by a fully closed path (loop). Look at the attached picture (from a similar question) to see where bias base currents flow (pay no mind to the input source VIN).

As you can see, the bias current (e.g., IB1) "departs" from the positive terminal of the negative power supply V-... enters into the right ground... leaves the left ground... passes through the "jumper" RB1... base-emitter junction BE1... emitter "current source" (the humble resistor RE here but you can imagine a perfect current source instead)... and finally, successfully returns to where it has started - the negative terminal of the negative power supply V-. So, the bias base current is created only by the negative power supply.

I would comment that "real world" op-amps are vastly more complex even in their equivalent datasheet schematic representations, whereas Josef's is a simplified minimal circuit to demonstrate basics of the situation (and yes could hold true if being built out of discrete components, for example in a power amplifier input stage).

The bottom line is that ac coupling an op-amp's non-inverting input without a resistive reference to some fixed point is going to end in tears.

Hi Susan,

Simple things are sometimes useful. The wiring by Annex I tried a "big boost" in 1972. And it worked :-))

Ooo ... dynamic load... another great idea...

http://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper1.htm

Hi Cyril,

I prefer current source simply (in collector) - not dynamic load. It enables to sever value of "collector resistance" and quiescent current (thus supply voltage, too). Ideal curent source has infinity output resistance - and we can use small supply voltages. This idea enables to construct modern OPAs. Certain problem is DC quiescent voltage of the T1. But in "full structure" all solves negativ feedback (DC).

Josef

Then what is a dynamic load?

Well, strictly speaking, the upper part of your circuit ("current source") is exactly a dynamic load since it changes its static (chordal) resistance when the lower part tries to change the common current. But, as a whole, this is not a pure dynamic load arrangement because of the parallel negative feedback introduced by RB.

IMO it is not correct to set the quiescent current from the side of the collector; it is more correct to set it from the side of the emitter as it is usually done in the long-tailed pair. But this is another story... here it is not possible...

https://www.researchgate.net/post/Can_we_apply_an_input_current_to_the_collector_of_a_BJT_whose_base_is_held_at_a_constant_voltage_and_to_take_the_collector_voltage_as_an_output#share

Hi Cyril

and what about new situation in the annex?   :-)

Josef

Yes, this is the case (cascode configuration)  where "a  current source sets the quiescent current from the side of the emitter"... The problem is only that the emitter is "movable" (due to the 100% serial negative feedback) and there is no gain. This problem is successfully solved in the differential pair where another transistor (emitter follower) keeps the emitter "immovable":

https://www.researchgate.net/post/Can_we_reveal_the_basic_idea_behind_the_long-tailed_pair_and_explain_its_operation_in_an_intuitive_way

But I can agree that if we introduce a 100% parallel negative feedback in the first configuration (RB = 0, e.g., by connecting the collector to the base), they both become the same cascode configuration where two dual electrical sources interact - a current source (T2) sets the current through a voltage source (T1)... and v.v.,  a voltage source (T1) sets the voltage across a current source (T2)...

Hi Cyril

And what about this circuit :-))

And next possibility.

A real constellation of tricky circuit ideas!

In the first circuit diagram above I see a common-emitter stage (T1) drawing a current from a common-base stage (T3)... and this composed stage is comprised by a parallel negative feedback (through RB)... and it is driven by the upper current source (T1).

Usually, such a cascode (T1 + T3) is used to prevent from the Early effect...

In the second picture, the network R1-R2 is bootstrapped... and thus acts as a dynamic load (current "source").

The output is boosted by T2 (emitter follower).

So, the common basic idea in all these circuit solutions is to set the quescient current by a current source included in the collector? And, in order to "soften" the conflict between the two current "sources", a parallel negative feedback is applied?

Hi Cyril, in these simple situations is negative feedback from the collector the simplest solution. For AC signals we are easily able remove the feedback. See below.

Sophisticated solutions created with a lot of imagination and fantasy...

As I can see, in regard to the AC signals, the final circuit solution is already a real dynamic load arrangement where two current "sources" connected in series "fight" each other... As a result, the voltage of the middle point between them vigorously changes (high gain).