LINEAR VOLTAGE STABILIZER

rjenkinsgb

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• Sep 12, 2022

• #2

The opamp compares the output voltage to the zener reference voltage and the opamp output will go high or low if the voltage is low or high (as the voltage sensing is on the inverting input).

That acts to adjust the transistor base drive and hold the regulator output at a constant voltage, within a fairly limited range of current.

eg. It may be OK for a few hundred mA with an appropriate transistor, but not high currents as most opamps can only give fairly small output currents, something probably in the 10mA to 50mA range depending on the type.

The opamp would also have to be suitable for the voltage range involved; some can't take above 3 - 5V, other can operate on 40V or more.


Have a read through this:

https://www.electronics-tutorials.ws/opamp/opamp_1.html

 

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S

sxy

Member

• Sep 12, 2022

• #3

Thanks a lot!
What is the output of the op-amp?
Is there any effect on the output of the op-amp; from the non-inverting side?
and why the voltage to the Base of the transistor decreases when VL increase?
So actually you say that when the voltage in the inverting side increases;
the output voltage of the op-amp will decrease in order to make equal V(+) and V(-)?
thanks a lot

 

Last edited: Sep 12, 2022

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rjenkinsgb

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• Sep 12, 2022

• #4

sxy said:

So actually you say that when the voltage in the inverting side increases;
the output voltage of the op-amp will decrease in order to make equal V(+) and V(-)?

Click to expand...

Exactly!

 

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S

sxy

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• Sep 12, 2022

• #5

Thanks a lot.
Is there any effect on the output of the op-amp; from the non-inverting side?

 

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danadak

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• Sep 12, 2022

• #6

Is there any effect on the output of the op-amp; from the non-inverting side?

Click to expand...

Yes. The over simplified model of an OpAmp is -



Ignore Ro, Ri. So output is internal A x (V+ - V-), but then thats modified by feedback.


Regards, Dana.

 

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crutschow

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• Sep 12, 2022

• #7

sxy said:

Is there any effect on the output of the op-amp; from the non-inverting side?

Click to expand...

Yes, the op amp will try to keep the voltage at the inverting input equal to the voltage at the non-inverting input.

 

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S

sxy

Member

• Sep 12, 2022

• #8

Thanks a lot!
So why: VL=Vz-Vi
and why if A=infinite (A=output voltage gain);
then Vi=o.

 

Last edited: Sep 12, 2022

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danadak

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• Sep 12, 2022

• #9

Vl = Vz

Vi =~ 0

Let R2 go to infinity in this example, which replicates your simple regulator.

Voltage Regulator using Op Amp and Transistor

Regards, Dana.

 

Last edited: Sep 12, 2022

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S

sxy

Member

• Sep 12, 2022

• #10

Thanks a lot!
What is Vi; in my circuit?
So why: VL=Vz-Vi in my circuit?
Why, in a op-amp, whose voltage gain is infinite, the voltage Vi tends to zero?
Thanks

 

Last edited: Sep 12, 2022

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danadak

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• Sep 12, 2022

• #11

Sorry, I just realized Vi was not Vin.

Vi, for an OpAamp with - fdbk does go to 0 when its internal open
loop gain, Aol, goes to infinity.

Take a look at this -

Inverting Op-Amp and the Concept of Virtual Ground

In this article, the inverting op-amp configuration and the concept of virtual ground is discussed and the output expression is derived.

www.allaboutelectronics.org

If you take this simple model and add the feedback Rs and do the analysis you find

1) As internal Aol approaches infinity R0 will approach 0.

2) As internal Aol approaches infinity V1 - V2 approaches 0, eg. virtual ground.

3) And several other important affects in the AC realm.



Note Vd = V1 - V2 in the above, the differential voltage at the inputs.


Regards, Dana.

 

Last edited: Sep 12, 2022

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S

sxy

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• Sep 12, 2022

• #12

Thanks a lot!
So why: VL=Vz-Vi in my circuit?
Thanks

 

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rjenkinsgb

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• Sep 12, 2022

• #13

sxy said:

So why: VL=Vz-Vi
and why if A=infinite (A=output voltage gain);
then Vi=o.

Click to expand...

Anything other than zero, *infinity = infinity.

In reality, Vi would likely be something from microvolts to millivolts, as opamps are not perfect in the real world.

 

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S

sxy

Member

• Sep 12, 2022

• #14

Thanks a lot!
Still the same question haven't been answered:
why: VL=Vz-Vi
Thanks

 

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rjenkinsgb

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• Sep 12, 2022

• #15

sxy said:

Still the same question haven't been answered:

Click to expand...

Yes it has, repeatedly, if you study the answers given....

 

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S

sxy

Member

• Sep 12, 2022

• #16

Thanks a lot!
But it should be:
VL=A(Vz-Vi)
(A=output voltage gain).
Thanks

 

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danadak

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• Sep 13, 2022

• #17

Vl = A ( Vz - Vl) = AVz - AVl

so Vl (1 + A) = AVz or Vl = AVz / (1 + A) = Vz / (1 + 1/A )

if A goes to infinity Vl = Vz


Regards, Dana.

 

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Tony Stewart

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• Sep 13, 2022

• #18

let A= 10^5 (typ BJT Op Amp) and VL=Vz since closed loop gain = 1 as there is no attenuation to Vi- negative feedback. . Non-inverting gain = 1+ |inverting gain=0|=1. But we hope Vz is constant, but italso has some series resistance so it varies much less than Vin.


If VL-Vz= small error,E then Vi= E/10^5 thus Vi~0 however if OpAmp has input offset, in uV or mV that too will be added to output error, E. In datasheets input offset error= Vio

 

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danadak

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• Sep 13, 2022

• #19

And of course there is error due to PSRR and Vbe and OpAmp Rout but for the
simple circuit and crappy tolerances of Vz most designs, I would posit, would not
care given the approach used. Not counting AC effects and noise as well.

Regards, Dana.

 

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S

sxy

Member

• Sep 13, 2022

• #20

Thanks a lot!
VL=A(Vz-Vi)
VL-is the output of the circuit
Whereas Vi is the input of the negative side of the op-amp.
But you assumed that VL and Vi is the same thing
and it is not correct.
IL*RL=VL is the output of the circuit
Thanks

 

Last edited: Sep 13, 2022

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