# How opamp work with Feed back?

Discussion in 'Homework Help' started by EDFG, May 14, 2016.

1. ### EDFGNew Member

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Hello,
How op amp work with Feedback?
i need help in this circuit how it is working ad what will be output?

2. ### ronsimpsonWell-Known MemberMost Helpful Member

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Assuming that you know how op amps work and how feed back works:

First look at how Q works. It is a grounded base amp. The gain is set by two resistors. In your case R-emitter is 0 ohms so the gain is very high. (is a non inverting amp)
What ever current is pulled out of Q_E is also goes through Q_C and through the C-resistor. (1000)
[note +5V battery in my schematic is the same as your +5V input. My 1000 resistor is the same as your 1k resistor)

In you circuit. The +5V and 1k resistor pulls op amp(-) input up above above ground and op amp(+) input. This will cause op amp output to go down. As soon as op-amp out reaches -0.7V current is pulled out of Q_emitter. This causes Q_Collector to have current. As op amp output current reaches 5mA then there is 5 volts across the 1k resistor. Now op amp(-) is at zero volts, and all balances.

If the op amp output pulls too low, say 5.1mA then op amp(-) will be at -0.1mA this will cause the op amp out to go up a little to find balance again.

3. ### EDFGNew Member

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please tell i only know this..
1. The opamp multiply the input
2. if there is negative feedback the gain will decrease and positive the circuit will behave like oscillator.

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5. ### ronsimpsonWell-Known MemberMost Helpful Member

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A op amp has two inputs. (-) and (+)
The op amp looks at the difference between (-) and (+). If (-) equals (+) then the output stays at the voltage where it is.
If (+) is higher the output will go up.
If (-) is higher the output will go down.

6. ### ronsimpsonWell-Known MemberMost Helpful Member

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You started out with a too complicated circuit. Here is a simple one.
5V and 1k pull up on (-) input. That makes the output go down.
Balance is when the output is at -5volts.

The boy on the left is "5V" in weight.
The distance from boy_left to center is "1k ohms"
The distance from boy_right to center is the other "1k ohm" resistor.
To balance how much must boy_right weigh?

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5V

8. ### ronsimpsonWell-Known MemberMost Helpful Member

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Yes

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Then?
What to do?

10. ### EDFGNew Member

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I think we need +10 at right to balance

11. ### MikebitsWell-Known Member

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Your very good at teaching Ron. The see saw analogy is great

12. ### audioguruWell-Known MemberMost Helpful Member

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I wonder why the student comes to a forum for the answer to the simple circuit instead of being properly taught about electronics by the teacher. Maybe the student missed some classes about transistors and opamps?

1) What is the voltage of the (-) input of the opamp? The open loop gain is extremely high so it is the same as its (+) input, 0V.
2 The opamp is inverting then does its output go positive or negative? Negative, since the input is positive.
3 When the output of the opamp goes negative then it drives the emitter of the transistor negative by how much voltage? -0.7V. That is the answer.

13. ### ronsimpsonWell-Known MemberMost Helpful Member

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I have tough (not English seeking) engineers how to be engineers. (I can use hardware or the chalk board but not words)
One example I use is to get a meter stick out. (yard stick if you are in a back ward country)
Put Jim on the "0" end of the meter stick. Joe holds the "1" end of the stick. Jill looks at he 0.5 m center of the stick.
Jim--------Jill-----------Joe
Jim is the source, of in this case, the "5V" input.
Jill is the (-) input.
Joe is the output.

Jim holes his end of he stick up 5cm off the desk. (5 volts)
Jill reports that her pint is 2.5cm up. Because she is "negative" she reports "down".
Joe, being the output, then drops his end below the desk some until he reaches -5cm.
At this point Jill reports "0" so Joe holds at -5cm.

We have balance. +5cm on one end, -5cm on the "output" end which causes the center to be 0cm.

If the two resistors are not 1:1 but 9:1 you move Jill over to 0.1m point and see what gain does.

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14. ### KerimMember

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This is how I analyse the circuit:

In the linear region, V(in-)=V(in+) at the opamp inputs, practically speaking. Actually, they are very close (not equal) due to the high gain of the opamp.
Since the 5V input tries to increase the voltage at (in-), the opamp output has to go down; since V(in-) > V(in+).
But Vout cannot be lower than -Vbe (Vbe of the transistor Q). As seen on the circuit, the opamp output is shorted to ground via the base-emitter junction of Q.
So the problem answer is close to B(-0.7).

Now let us see the currents.

The opamp output sinks, from the Q emitter, the current (Ic + Ib) where Ic = 5V/1K to let V(in-)=0, and Ib= Ic/beta.
Since the Q collector has to be close to ground, V(in-)=0, beta is the current gain of Q at Vce = Vbe (close to its saturation region) which is lower than its beta at a rather high voltage of Vce (in the linear region).

Hope this helps.

Kerim

15. ### EDFGNew Member

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Ok, if there is no bjt at output then the op amp out put will be -15v right due to high gain.

Is this part right?

Then bjt after bjt the base is 0v and enitter is -15v so
All
We need 5v or -5v ?
Because you have mention -5v

16. ### EDFGNew Member

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Ok, i got it will -0.7 V

17. ### audioguruWell-Known MemberMost Helpful Member

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No. The negative feedback causes a gain of -1. Then with the +5V input there will be a -5V output. With no negative feedback then the output will be -13V to -14V.
No. The base-emitter of the transistor is a 0.6V to 0.7V forward biased silicon diode with a current in it of about 0.5mA in it and can never go more than 0.6V to 0.7V in this circuit.

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18. ### EDFGNew Member

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The negative feedback causes a gain of -1. Then with the +5V input there will be a -5V output.

How you have calculated gain of -1?

19. ### ronsimpsonWell-Known MemberMost Helpful Member

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Input = +5, output=-5, so the gain is -1.
+5 X -1 = -5

If the output was +5 then the gain would be +1.

20. ### EDFGNew Member

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We know input is 5v but from where we got -5v output?

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