Hi !
I want to design an op-amp with the following properties :
- open-loop gain = 10,000 (minimum)
- input impedance = 100 kilo ohm (minimum)
- output impedance = 1 kilo ohm (maximum)
- Vout = 0 for V+ (non-inverting input) = V- (inverting input)
( power supplies must be +/- 6 volt )
actually this is my final project for " principles of electronics " . I've designed following circuit which satisfies first , second and third condition , but I have no idea about fourth .



could anybody help me how to satisfy the fourth condition ?
( I've checked uA741 ( one of the most successful op-amps ) , it doesn't have this property ! )

 

The polarity of the -6V battery was wrong.

 

It's OK

 

The polarity of the -6V battery was wrong.

It's fine. It just looks unusual because we commonly see a +6V source with the positive terminal connected to ground and the negative terminal connected to the rail.

 

Hi !
I want to design an op-amp with the following properties :
- open-loop gain = 10,000 (minimum)
- input impedance = 100 kilo ohm (minimum)
- output impedance = 1 kilo ohm (maximum)
- Vout = 0 for V+ (non-inverting input) = V- (inverting input)
( power supplies must be +/- 6 volt )
actually this is my final project for " principles of electronics " . I've designed following circuit which satisfies first , second and third condition , but I have no idea about fourth .

View attachment 79305

could anybody help me how to satisfy the fourth condition ?
( I've checked uA741 ( one of the most successful op-amps ) , it doesn't have this property ! )

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.

 

It's fine. It just looks unusual because we commonly see a +6V source with the positive terminal connected to ground and the negative terminal connected to the rail.

If that is fine then the voltage of bjt will be reversed, as Q1, Q2, Q3, Q4, Q6, Q7, Q11, is that really ok?

 

If that is fine then the voltage of bjt will be reversed, as Q1, Q2, Q3, Q4, Q6, Q7, Q11, is that really ok?

Well, maybe I'm looking at it wrong.

As drawn, the top rail will be at +6V and the bottom rail will be at -6V.

Why will that result in any of the BJTs being reverse-biased?

Why do you want both rails to be at +6V?

Or, as drawn, what voltage do YOU think the bottom rail with be at?

Remember, a voltage source with a negative output voltage means that the positive terminal is at a negative voltage relative to the negative terminal.

 

Well, maybe I'm looking at it wrong.

As drawn, the top rail will be at +6V and the bottom rail will be at -6V.

1 . Why will that result in any of the BJTs being reverse-biased?

2. Why do you want both rails to be at +6V?

3. Or, as drawn, what voltage do YOU think the bottom rail with be at?

Remember, a voltage source with a negative output voltage means that the positive terminal is at a negative voltage relative to the negative terminal.

As I lebeled 1,2,3 those three lines, are you asking me those question or to the TS?

 

As I lebeled 1,2,3 those three lines, are you asking me those question or to the TS?

As you labeled WHAT three lines? I don't see ANY diagram that you've posted here at all?

I'm asking YOU (ScottWang).

 

As you labeled WHAT three lines? I don't see ANY diagram that you've posted here at all?

I'm asking YOU (ScottWang).

Three lines was labeled from your posts.
What I said the reversed voltage were shown in the yellow circles as below.

 

Why would any of those voltages be as you indicate?

Again, please answer this very simple question: What voltage do YOU think the bottom rail of that circuit is going to be at?

 

Why would any of those voltages be as you indicate?

Again, please answer this very simple question: What voltage do YOU think the bottom rail of that circuit is going to be at?

My viewpoint is -6V from ground, that's why I felt the polarity of battery was wrong.

 

My viewpoint is -6V from ground, that's why I felt the polarity of battery was wrong.

So, if the bottom rail is at -6V (which it is), why do you think that the emitter of, say Q4 is going to be more positive than the base? That means that the voltage on the base is <-6V. How is that happening?

 

So, if the bottom rail is at -6V (which it is), why do you think that the emitter of, say Q4 is going to be more positive than the base?
That means that the voltage on the base is <-6V. How is that happening?

That's from the circuit of TS in first posts, because the positive of battery was connected to e then you can along the wiring to check that its another side was connected to ground, can't you see they was reversed?

 

That's from the circuit of TS in first posts, because the positive of battery was connected to e then you can along the wiring to check that its another side was connected to ground, can't you see they was reversed?

How is it reversed? You yourself acknowledged that the voltage on the bottom rail, as shown in the OP's diagram in Post #1 (and the one you marked up) is at -6V.

The positive TERMINAL of the battery is connected to the bottom rail -- but the specified voltage of the battery (which is just a DC voltage source in a simulator) is MINUS 6V.

If he turns that battery around THEN the voltage on the bottom rail would be +6V.

If Vbatt is the voltage specified for the battery (in this case it is -6V), then that means that

Vbatt = Vpterm - Vnterm

where Vpterm is the positive terminal and Vnterm is the negative terminal.

Here the positive terminal is connected to the bottom rail, let's call that Vbrail, and the negative terminal is connected to ground, which is by definition 0V.

So we have

Vbatt = Vbrail - 0V

Solve for Vbrail

Vbrail = Vbatt = -6V

QED

Now let's do it the way you seem to want, which is to connect the positive terminal of the batter to ground and the negative terminal of the battery to the bottom rail:

Vbatt = 0V - Vbrail

Vbrail = -Vbatt = -(-6V) = +6V

 

How is it reversed? You yourself acknowledged that the voltage on the bottom rail, as shown in the OP's diagram in Post #1 (and the one you marked up) is at -6V.

The positive TERMINAL of the battery is connected to the bottom rail -- but the specified voltage of the battery (which is just a DC voltage source in a simulator) is MINUS 6V.

If he turns that battery around THEN the voltage on the bottom rail would be +6V.

If Vbatt is the voltage specified for the battery (in this case it is -6V), then that means that

Vbatt = Vpterm - Vnterm

where Vpterm is the positive terminal and Vnterm is the negative terminal.

Here the positive terminal is connected to the bottom rail, let's call that Vbrail, and the negative terminal is connected to ground, which is by definition 0V.

So we have

Vbatt = Vbrail - 0V

Solve for Vbrail

Vbrail = Vbatt = -6V

QED

Now let's do it the way you seem to want, which is to connect the positive terminal of the batter to ground and the negative terminal of the battery to the bottom rail:

Vbatt = 0V - Vbrail

Vbrail = -Vbatt = -(-6V) = +6V

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

 

Please tell me why the battery for the op amp connected that way, that's how the TS drawn.
View attachment 79314

I don't know where you are getting that schematic from.

This is what is in Post #1 (which has not been edited since the TS posted it -- at least there's no edit notification, which should mean that any edits that were made were finished before anyone else viewed the post):



And this appears to be the schematic that you annotated in Post #10.

What you have in Post #16 has lots of indications that it is NOT a schematic the TS would have posted. Their schematic looks like a direct screen capture from a schematic capture program of some flavor. Yours has black traces where all of the OP traces are red, it has black square connection dots where the OP connection dots are red circles, and it has missing grid dots in the area where the supplies are. Finally, it has a -6V battery and yours does not, despite the fact that YOU referred to the -6V battery in Post #2.

 

I don't know where you are getting that schematic from.

This is what is in Post #1 (which has not been edited since the TS posted it -- at least there's no edit notification, which should mean that any edits that were made were finished before anyone else viewed the post):

View attachment 79315

And this appears to be the schematic that you annotated in Post #10.

What you have in Post #16 has lots of indications that it is NOT a schematic the TS would have posted. Their schematic looks like a direct screen capture from a schematic capture program of some flavor. Yours has black traces where all of the OP traces are red, it has black square connection dots where the OP connection dots are red circles, and it has missing grid dots in the area where the supplies are. Finally, it has a -6V battery and yours does not, despite the fact that YOU referred to the -6V battery in Post #2.

I didn't change any wiring except the battery, and I just moved them from middle position to the right side, it's more easily to check, the problem was that you saw the battery was fine(-6), but I saw the symbol of battery was wrong.

 

The polarity of the lower battery is incorrect.

 

The polarity of the lower battery is incorrect.

This times you discuss with WBahn from the first post, because he said that was fine ...