Answer my queston pls - Question 14(Op amp) of the worksheets.

wayneh

Joined Sep 9, 2010
16,164
The base-emitter voltage drop across a transistor is like a diode drop, roughly 0.7V. The old circuit controlled the base voltage to the zener voltage, so the output was zener-minus-BE drop.

The improved circuit holds the output at the zener voltage. It's regulation is tighter because the BE drop, which varies a little, has now been included in the feedback loop.
 

Thread Starter

srikar sana

Joined Feb 3, 2016
18
The base-emitter voltage drop across a transistor is like a diode drop, roughly 0.7V. The old circuit controlled the base voltage to the zener voltage, so the output was zener-minus-BE drop.

The improved circuit holds the output at the zener voltage. It's regulation is tighter because the BE drop, which varies a little, has now been included in the feedback loop.
i still didnt got the point ineed a detailed reply pls
 

Thread Starter

srikar sana

Joined Feb 3, 2016
18
Don't think you could get more detailed than that.
the problem with me i am not being able to see how it is getting added to it and how the input of the op amp is changing and so ....................
i request you to pls try explaining me once more. (moreover i think if i stay longer with a problem i will i understand that )
 

ronv

Joined Nov 12, 2008
3,770
the problem with me i am not being able to see how it is getting added to it and how the input of the op amp is changing and so ....................
i request you to pls try explaining me once more. (moreover i think if i stay longer with a problem i will i understand that )
The zener diode voltage is not perfect it varies a little with the current thru it.


So when you increase the load in the first circuit it needs more base current as well so there is some voltage change across the zener.
With the second circuit the current in the zener is constant because of the op amp "buffer". So the zener voltage stays more constant and the output is more constant.
 

Thread Starter

srikar sana

Joined Feb 3, 2016
18
The zener diode voltage is not perfect it varies a little with the current thru it.


So when you increase the load in the first circuit it needs more base current as well so there is some voltage change across the zener.
With the second circuit the current in the zener is constant because of the op amp "buffer". So the zener voltage stays more constant and the output is more constant.
yeah i understood that sir but what i need not understood was how did the drop 0.7v got eleminated after adding a feedback
(assume that i am the first student in the followup question )
 

Thread Starter

srikar sana

Joined Feb 3, 2016
18
The zener diode voltage is not perfect it varies a little with the current thru it.


So when you increase the load in the first circuit it needs more base current as well so there is some voltage change across the zener.
With the second circuit the current in the zener is constant because of the op amp "buffer". So the zener voltage stays more constant and the output is more constant.
i dont understand that how 0.7v gets added to the output .(the same doubt which was raised by the first student in the quesiton )
 

ronv

Joined Nov 12, 2008
3,770
i dont understand that how 0.7v gets added to the output .(the same doubt which was raised by the first student in the quesiton )
I see, the third picture.
The op amp wants to make the - terminal the same as the + terminal. So the op amp output will be the zener voltage plus the Vbe drop of the transistor.
So if the zener voltage is 10 volts the output of the op amp will be 10.7 volts.
 

wayneh

Joined Sep 9, 2010
16,164
The time to respond to a voltage change will be limited only by the op-amp, and even a "slow" op-amp is very fast. For practical purposes in a power supply circuit like this one, the op-amp is "infinitely" fast and adjusts its output to equalize the voltages on its inputs.

The simulator provided by ronv allows you to look at a very fast transition and see that it is not truly infinitely fast. The model's accuracy will depend on the accuracy of the parameters entered into the models of the components. It may only be an approximation, but as ronv notes it helps visualize a transition.
 
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