Op-Amp Circuit Help: Voltage Follower & Multisim Guide for Homework

[moe619]

Homework Statement

in the attachment

Homework Equations

The Attempt at a Solution

for q.1) its a voltage follower non-inverting
Vcc has to be greater than 5V for the zener diode work?
formulas?

q2) does anybody know how to use multisim ?
if i copy the figure in multisim would i be able to work out a,b,c,d

 

[NascentOxygen]

Looking at Fig. 1, if you have studied op-amps you should be able to figure this out. Assuming an ideal op-amp, what are its important characteristics?

 

[moe619]

positive feedback?
i read that the input draws no current.. but output has a value

 

[NascentOxygen]

positive feedback?

There is feedback. What input terminal does the op-amp's feedback connect to?

i read that the input draws no current.. but output has a value

Good. That, together with the op-amp's differential gain being "very high" is all the op-amp detail that is needed to analyze the circuit.

Since the inputs of the op-amp itself draw no current, how much current will be flowing through the "horizontal" 10kΩ resistor?

 

[moe619]

its midnight so ill get back to you in a couple of hours.thanks.

 

[moe619]

the input it connects to is( - ) so its negative feedback.
i don't think there's current flowing in the horizontal 10 K. because of no voltage drop?
how do we find Vcc

 

[NascentOxygen]

the input it connects to is( - ) so its negative feedback.

Negative, yes. What percentage of the output voltage is fed back to the inverting input?

i don't think there's current flowing in the horizontal 10 K. because of no voltage drop?

What will be the voltage on the left hand side of that horizontal 10kΩ resistor?

how do we find Vcc

Usually we don't. Vcc is given, or we simply assume it is appropriate for the circuit to work properly. I don't think you are asked to find a suitable Vcc, are you?

You are asked to describe how the circuit works---or to explain why it does what it does.

 

[moe619]

Negative, yes. What percentage of the output voltage is fed back to the inverting input?

What will be the voltage on the left hand side of that horizontal 10kΩ resistor?

Usually we don't. Vcc is given, or we simply assume it is appropriate for the circuit to work properly. I don't think you are asked to find a suitable Vcc, are you?

You are asked to describe how the circuit works---or to explain why it does what it does.

it has to balance to reach a point of equilibrium 99.99% ?

0 Volts?

true.

 

[NascentOxygen]

The correct answer is 100% of output is fed to the inverting input.

Do you understand how the zener diode is being used here?

 

[moe619]

not really i know what it does like, it won't work if it doesn't get 5V or something

 

[NascentOxygen]

Well, let's assume it does get sufficient voltage to develop 5V here, otherwise if we consider the circuit is not being powered there is not going to be much to explain.

Perhaps you are thinking Vcc could be any old time-varying waveform? I'd say it can't be. It is a respected convention in electronics to denote only a fixed DC voltage as Vcc. (If you want to put a figure on it, think something anywhere in the range 9V - 18V. The particular value should not be critical.)

 

[moe619]

if we don't put a figure on it how are we going to find the output current for part b and is Rf important to note down is Rf the horizontal one on the left 10K?

 

[NascentOxygen]

if we don't put a figure on it how are we going to find the output current for part b

That would be important only if the output current were dependent on Vcc.

and is Rf important to note down is Rf the horizontal one on the left 10K?

Rf is your feedback resistance, I presume? Rf connects the feedback (in general) from output to inverting input. In your circuit, the resistance in this position (viz., from output to inverting input) is a piece of wire, and nominally 0 ohms.

Instead of worrying about component values, at this stage you should be trying to figure out what the circuit does and how it works. These details typically don't require specific component values, usually.

 

[moe619]

is part a done ?

 

[NascentOxygen]

is part a done ?

You have worked out the formula for I_L?

 

[moe619]

i=v/r

we makeup v (9-18)
and how about resistance we just use the 10K before ground?

 

[moe619]

please reply

 

[NascentOxygen]

You may be under a misunderstanding that the only voltage source in the circuit is Vcc. That would be quite wrong. Not shown in many schematic diagrams are the + and - DC supplies powering the op-amp itself.

The Vcc shown serves to supply current to the zener diode. What voltage do you expect to find at the junction of the two 10k resistors and the zener diode?

Bearing in mind what you said about the horizontal 10k resistor, deduce the voltage that you would expect to measure on the right hand side of that horizontal 10k resistor (i.e., as measured at the inverting input (-) of the op-amp).

 

[moe619]

V1 and V2 my lecturer kind of gave us an idea that there both 15 V and Vcc is 10V now I am more confused

 

[NascentOxygen]

The voltages powering the op-amp do not dictate the output voltage of the op-amp, meaning you can largely ignore those voltages---which is why they are not even shown on some circuit diagrams. What is important is that the output voltage of the op-amp is controlled by the input voltages of the op-amp.

 

[moe619]

then let's calculate the current..?

 

[NascentOxygen]

I framed some leading questions in post #18. Still waiting to see your answers.

I'll be away, may not be back on for 3 or 4 days.  http://img9.imageshack.us/img9/9287/smileycar71.gif

 

[moe619]

5 V? we said the horizontal 10k would be 0 V

 

[NascentOxygen]

5 V? we said the horizontal 10k would be 0 V

Right. So how to determine the op-amp's output voltage? Remember, it's nothing to do with Vcc, providing Vcc is sufficiently high.

 

[moe619]

the output voltage of the non-inverting amplifier is in phase with the input voltage.
Vout=Vin

 

[NascentOxygen]

the output voltage of the non-inverting amplifier is in phase with the input voltage.
Vout=Vin

Meaning Vout would be how many volts?

 

[moe619]

5V? more or less?

 

[NascentOxygen]

5V? more or less?

Let's suppose the op-amp has an open loop low frequency differential gain of 10,000. Can you determine Vout exactly?

 

[moe619]

approximately V out can be determined
there is a formula but its for a summing/difference amplifier
Vout= -(V1 + V2) (for unity gain mode)

Vout= -(Rf/R) (V1 + V2) (where R1=R2=R)

......
Vout= V2 - V1 (for a unity gain amplifier)

Vout= (Rf/R1) (V2 - V1) (for an amplifier with gain where R1=R2 and Rf=R3

 

[NascentOxygen]

At seeing the word exactly I thought you may https://www.physicsforums.com/images/icons/icon3.gif have applied this formula:

Vout = A (V₊  ̶ V₋)

But 5V is close enough. So with an output voltage of 5V, how much current is flowing in the load resistor? I believe that is what you are asked find?

 

[moe619]

Vout = A (V₊  ̶ V₋) never seen this formula is it the same as Vout=Av*Vin

V=IR
5=I x 10k
I= 0.0005 A

 

[NascentOxygen]

Vout = A (V₊  ̶ V₋) never seen this formula is it the same as Vout=Av*Vin

The op-amp triangle shows two inputs. What is Vin?

V=IR
5=I x 10k
I= 0.0005 A

That looks right.