Sorry, I just noticed the -6.

This is the first time I have ever seen it done like that. It is very confusing and leads to misinterpretation.

 

The polarity of the lower battery is incorrect.

It's OK.
Look at the bias point of the BJTs :

red boxes are branch currents and purple ones are node voltages . As you see , all the BJTs are in active region .
furthermore , using two batteries (plus and minus Vcc) is a common step in op-amp and specially differential amplifier design !

 

With no feedback it becomes more challenging.

In practice, an opamp's gain is so high that shorting the two inputs to each other will probably rail the output. Plus, there will be a common mode gain which will also affect the output. About the best you can do is design it so that nominally it has this property (and notice that the basic 741 has pins specifically for the purpose of nulling the amplifier).

It's been a long time (over 15 years) since I did this kind of stuff, but my first thought would be to ask what the effect would be of tweaking the 22.6kΩ resistor value on the quiescent output voltage.

If that doesn't do it, then look for a place where you can inject or extract current in order to move the output around (there should be numerous places where you can do this) without having too much effect on the gain.

changing the 22.6 k resistor , makes things complicated ( disturbs Widlar current source etc )
I think I should inject some current to output node to increase its voltage , But this may disturb output voltage swing .
I'll check it.
Thank you

 

In my opinion the polarity of the lower battery is shown incorrectly.
The long dash of the battery symbol is the +ve terminal and the short dash the -ve terminal.
You need to turn the symbol 180° and label it 6V (no sign). There is no -6V battery. Only the negative node should be labeled as -6V.

 

I've modified my circuit by adding a current source which injects current to base of Q10 :

As you see , output voltage is nearly 0 volt . but new problems appear :
- dissipation power increases ( cause of new current source )
- value of the corresponding resistor ( 5.35 k ) should be ascertained with high precision

I think it's better to use an internal feedback . I have no idea how to perform it . ( Because feedback decreases open-loop gain )
could you please help me ?

 

In my opinion the polarity of the lower battery is shown incorrectly.
The long dash of the battery symbol is the +ve terminal and the short dash the -ve terminal.
You need to turn the symbol 180° and label it 6V (no sign). There is no -6V battery. Only the negative node should be labeled as -6V.

you're right . there's no -6 volt battery . but in simulation it's OK.

 

Simulation can't show the real problem in the real world, so do the same thing as the real world as you can then the simulation will closing to the real world.

 

Simulation can't show the real problem in the real world, so do the same thing as the real world as you can then the simulation will closing to the real world.

battery polarity or its orientation is NOT my problem . my problem is how to fix output voltage at 0 volt when V+(non-inverting input) = V- (inverting input)

 

You can use a DC-servo circuit or add some circuit at the input stage to compensate a DC-offset. Because now without any closed loop feedback your amp will amplified his own DC-voltage. Which is a normal behavior, because every real world op amp without feedback network will go into saturation if we connect both of his inputs to GND.

 

Please tell me why the battery for the op amp connected that way, that's how the TS drawn.

View attachment 79314

The polarity of the lower battery is incorrect.

The polarity of the battery in which post? The original post or the modified one that ScottWang made. One is correct and one is incorrect (and note that ScottWang is claiming that both are incorrect!).

The one in the original post is correct since the positive terminal of the battery is at -6V relative to the negative terminal, hence the lower rail is at -6V which what is intended.

 

In my opinion the polarity of the lower battery is shown incorrectly.
The long dash of the battery symbol is the +ve terminal and the short dash the -ve terminal.
You need to turn the symbol 180° and label it 6V (no sign). There is no -6V battery. Only the negative node should be labeled as -6V.

The battery in a simulator is just a symbol for a DC voltage source that can have any voltage, positive, negative, or zero. You might PREFER to always have it be positive and to orient the battery symbol accordingly (as do I), but that does not make it incorrect and the simulator does not care. (as clearly shown by the DC bias point simulation done by the TS). Neither node of the battery is labeled as a voltage -- the battery voltage is labeled as being -6V which is, by definition, the voltage difference at the positive terminal relative to the voltage at the negative terminal.

 

You can use a DC-servo circuit or add some circuit at the input stage to compensate a DC-offset. Because now without any closed loop feedback your amp will amplified his own DC-voltage. Which is a normal behavior, because every real world op amp without feedback network will go into saturation if we connect both of his inputs to GND.

I agree with you . I don't know any commercial op-amp with this feature ( Vout = 0 for V+ = V- ) . but this feature is necessary for my op-amp because It's my final project and my designing should satisfy this condition .
offset voltage needed to set output at 0 volt is -383.068 uV ! (from DC sweep analysis) obviously we can not generate this voltage with this high precision in real world and real circuits .

 

I agree with you . I don't know any commercial op-amp with this feature ( Vout = 0 for V+ = V- ) . but this feature is necessary for my op-amp because It's my final project and my designing should satisfy this condition .
offset voltage needed to set output at 0 volt is -383.068 uV ! (from DC sweep analysis) obviously we can not generate this voltage with this high precision in real world and real circuits .

Is the requirement that you have to meet merely is simulation? Does it have to meet if over a range of conditions (temperature, common mode voltage, supply voltage, etc.), or just at one fixed operating point? If the latter, then you can probably use a resistor to bleed current from the first stage just enough to balance it. It's a kluge, but it might be acceptable for your purposes. The better way is to use feedback (which I was assuming wasn't an option and that you had chosen or been given a particular topology).

 

Sorry, I just noticed the -6.

This is the first time I have ever seen it done like that. It is very confusing and leads to misinterpretation.

I agree -- and schematics are about communicating the circuit to other humans as much as they are about technically describing a circuit for a simulator. I prefer to vertically stack supplies and to keep DC supplies, whether they use the battery symbol or not, with a positive voltage. But in a text simulation file it is cleaner (to my mind) to use positive and negative voltages with the negative terminals tied to the reference. This can be shown pretty intuitively in a schematic by putting both supplies side-by-side and tying the negative terminals together and to the common and connecting the positive terminals to labels for the two rails (or routing explicit wires if that doesn't clutter things up too bad).

 

Is the requirement that you have to meet merely is simulation? Does it have to meet if over a range of conditions (temperature, common mode voltage, supply voltage, etc.), or just at one fixed operating point? If the latter, then you can probably use a resistor to bleed current from the first stage just enough to balance it. It's a kluge, but it might be acceptable for your purposes. The better way is to use feedback (which I was assuming wasn't an option and that you had chosen or been given a particular topology).

I'm supposed to DESIGN and SIMULATE and IMPLEMENT an op-amp with four properties . (I mentioned in my first post )
The fourth property ( Vout = 0 for V+ = V- ) , as you know , means common mode gain must be zero ( or at least very low)
power supplies are plus and minus 6 volt . and temperature is between 20 and 30 celsius degrees . ( normal temperature )

in that way (using resistor to balance differential stage) , value of the resistor must be ascertained with high precision ( for example 153.764 k ) and it's not practicable.
could you please explain how to use feedback ?

 

The polarity of the battery in which post? The original post or the modified one that ScottWang made. One is correct and one is incorrect (and note that ScottWang is claiming that both are incorrect!).

The one in the original post is correct since the positive terminal of the battery is at -6V relative to the negative terminal, hence the lower rail is at -6V which what is intended.

I already said my viewpoint, if you and TS say ok then it's ok.