Electronic circuit with Op-Amp, Switch and Diodes

[Femme_physics]

Homework Statement

http://img689.imageshack.us/img689/6668/mycircuit.jpg

Given a circuit that includes Operational Amplifiers, double-switch - S, for two situations, 1 and 2. Type of switch - DPDT - Double pole double throw. Two diodes and two lightbulbs, L1 and L2.

Given:

Voltage of each diode = 0.7V during conductance.
Lightbulb L1 is red, and its resistance is 120 ohms
Lightbulb L2 is green and its resistance is 150 ohms.
The rest of the values are as they appear in the circuit


Answer the following questions:

A) Explain how the circuit works. What is the name of type of Op-Amp connection, in this type of circuit? At which condition of switch S is every lightbulb turns on?
B) Calculate Va, Vb, Vc, at points a, b, c, when S is at condition 1.
C) Calculate Va, Vb, Vc, at points a, b, c, when S is at condition 2.
D) Calculate the current at the red lightbulb when it's turned on. Calculate the current at the green lightbulb when it's turned on. Which lightbulb shines brighter?

Homework Equations

The Attempt at a Solution

A) (I'm really not sure what do they mean by explain the way the circuit works. I'm tempted to answer with "Read and electronics book, *******!")
But, basically, it's about current flowing and depending on the condition of the switch, and the amplification, the lightbulbs either light up or don't.

The Op-Amp in this type of connection is a NON-INVERTER op-amp.

When the switch is at case 1, we see that Vout is negative, so current flows from minus to plus... therefor L1 lights up.

When the switch is at case 2, current still flowing from minus to plus, though this time L2 lights up due to the connections.

B) They're all zero because the current skips the op-amp entirely.

C) Vc is basically Vout.

So, as per the formula I pasted above:

Vout = -R2/R1 x Vin = -40k/10k x 2 = -8V
Vc = -8V
Va = 0 (because there is no current source or anything between V+ and the ground point)
Vb is a bit more complicated

I use KVL to get:

-8 + (40k x It) + (10k x It) -2 = 0

It = 0.2 [mA]


Vb = It x R1 = 0.0002 x 10000 = 2V
Vb = 2V


I'll answer the last question once I see the rest of it is correct :)

 

[technician]

Hi FP... good old negativr feedback opamp ! It is an INVERTING opamp
You are correct to recognise how the circuit works and you get Vout = -8V when switch in position 1 and Vout = +8V when in position 2.
Can you see how the diodes + lamp branches will respond to one output being -ve then being +ve?
Part(B) voltage at c will not be zero

 

[likephysics]

Hi FP... good old negativr feedback opamp ! It is an INVERTING opamp
You are correct to recognise how the circuit works and you get Vout = -8V when switch in position 1 and Vout = +8V when in position 2.
Can you see how the diodes + lamp branches will respond to one output being -ve then being +ve?
Part(B) voltage at c will not be zero

Vout is +10V when switch is at position 2.

 

[technician]

Well spotted 'I like physics'
Apologies...careless

 

[technician]

Sorry 'I like physics' I think Vout =+8V when the switch is in position 2. In both switch position Va and Vb = 0
Why do you think it is +10V?

 

[likephysics]

Sorry 'I like physics' I think Vout =+8V when the switch is in position 2. In both switch position Va and Vb = 0
Why do you think it is +10V?

Sorry, I overlooked the "Earth" connected to non-inv terminal.
Good trick question though.

 

[Femme_physics]

How come Vout is once positive and once negative?

The way I see it it's always the same formula

Vout = -R2/R1 x Vin

Therefor no matter the switch condition Vout must always be negative!

 

[technician]

with the switch in position 1 can you see that the + of the battery is connected to V-
via R1
And with the switch in position 2 the - of the battery is connected to the V- via R1

PS
The - in the equation tells you the amplifier is INVERTING so a + at the input produces a - at the output and a - at the input produces a + at the output

 

[likephysics]

How come Vout is once positive and once negative?

The way I see it it's always the same formula

Vout = -R2/R1 x Vin

Therefor no matter the switch condition Vout must always be negative!

It has to do with the ground connection or reference potential.

Say for example, you are using a multimeter to measure battery voltage.
One terminal of the multimeter is reference, which you always connect to -ve terminal of the battery, the other connects to +ve terminal and you get 1.5v. What happens if you reverse the connection?

 

[Femme_physics]

If I reverse, it's -1.5v.

with the switch in position 1 can you see that the + of the battery is connected to V-
via R1
And with the switch in position 2 the - of the battery is connected to the V- via R1

Agreed.

PS
The - in the equation tells you the amplifier is INVERTING so a + at the input produces a - at the output and a - at the input produces a + at the output

I see! :)

BTW - is this the way the current flows in the circuit for S1?
http://img828.imageshack.us/img828/5405/tukgn.jpg

 

[technician]

No, it goes the other way. Remember Vout is -8V when Vin is +2V
When the switch is in position 2 the 2V battery is the other way round (upside down)
Then the current will have the direction you have shown

 

[I like Serena]

BTW - is this the way the current flows in the circuit for S1?

I think you have drawn the situation for S2, except that battery is the wrong way around.
This would be correct!

In S1 the battery would be as you drew it, but then the current would flow from the plus pole at +2V to the opamp at 0V.

 

[Femme_physics]

In S1 the battery would be as you drew it, but then the current would flow from the plus pole at +2V to the opamp at 0V.

Wait, but the battery coordinance never changes. If the longer line of the plus side is up, it stays up, unless someone physically changes the circuit, no?

No, it goes the other way. Remember Vout is -8V when Vin is +2V

You mean to tell me it's like this? (I drew it just focusing on the +2 and -8) ->

http://img851.imageshack.us/img851/9973/opopu.jpg

 

[I like Serena]

Wait, but the battery coordinance never changes. If the longer line of the plus side is up, it stays up, unless someone physically changes the circuit, no?

In S2 the wires from the battery are crossed, so the plus pole (the longer line) connects to the bottom of the circuit.
Schematically you can indicate this by drawing the longer line on the bottom while not crossing the wires.

You mean to tell me it's like this? (I drew it just focusing on the +2 and -8) ->

http://img851.imageshack.us/img851/9973/opopu.jpg

Yep! :)

 

[Femme_physics]

Ah, perfectly understood :) Though, one thing is still confusion to me. When asked "Explain the action of the circuit"-- what the hell do they expect me to write? To start explaining about op-amps and switches?

 

[I like Serena]

Hmm... the action of the circuit.
I'd interpret that as saying what "you" can do to the circuit (input), and how "you" would see the result of the circuit (the output).

So what does the circuit do if the switch is in S1 (one possible input)?
And what in S2 (the other possible input)?
Can you light up both LED's simultaneously (possible output)?
Or switch them both off simultaneously?

Furthermore, what voltage is used for the switch (the input) and what voltage is used on the LED's (the output)?

 

[Femme_physics]

Well, in that case then they should ask this question last, not first! I got to do some calculations first!

 

[I like Serena]

Okay, since it's the first question, perhaps you should only say that you can put the switch either in S1 or S2.

Or just make up a nice story, since these type of open questions are not really hard science. ;)

 

[Femme_physics]

:) Agreed.

In both switch position Va and Vb = 0

In any op-amp, whenever you get that either Va, or Vb = 0, that means the other one is necessarily 0, as well?

I'm referring to the locations of a and b as they are in my current exercises near V- and V+

Edit: For S2, this is the case right? ->

http://img444.imageshack.us/img444/692/feely.jpg

 

[technician]

In this diagram the - of the battery is connected to the - input. The amplifier is inverting so Vout will be + which means that current will flow from Vout to the right to go through the diode on the far right... just as you have drawn.
BUT the current goes to the left through the feedback resistor so your current through the battery part of the circuit is the wrong way round
edit
drawing attached

 

[NascentOxygen]

In any op-amp, whenever you get that either Va, or Vb = 0, that means the other one is necessarily 0, as well?

Yes...and no.

Va and Vb are equal providing the OP-AMP is operating as an amplifier and in the linear region. But sometimes OP-AMPs are used as comparators, where the output sits at either +Vcc or -Vcc and nowhere in-between, in which case it is not operating as a linear amplifier, and it therefore cannot make Va=Vb. (I'm sure you encountered this with an earlier OP-AMP circuit.)

So you have to be confident (or else assume) that it is operating as an amplifier in order to say Va=Vb.

 

[Femme_physics]

Let me see if I got the full picture right...

http://img19.imageshack.us/img19/4117/fullpic.jpg

 

[NascentOxygen]

Not there yet! Consider the top sketch. The battery's positive voltage is pushing current into the input resistor, as you show. With the feedback resistor arrangement here, this OP-AMP is inverting.

So what does that tell you should be the polarity of the OP-AMPs output voltage?

 

[Femme_physics]

Not there yet! Consider the top sketch. The battery's positive voltage is pushing current into the input resistor, as you show. With the feedback resistor arrangement here, this OP-AMP is inverting.

Agreed.

So what does that tell you should be the polarity of the OP-AMPs output voltage?

I'm not sure I understand what you mean by the polarity of the OP-AMP output voltage. You mean Vout? It shoud be -8 [V] as we agreed.

Simply based on the formula

Vout = -Rf/Rv x VIn

 

[I like Serena]

Let me see if I got the full picture right...

http://img19.imageshack.us/img19/4117/fullpic.jpg

You mean that in both cases the rightmost LED lights up?
I thought you had found that in one of those cases the left LED would light up...

 

[NascentOxygen]

So if the OP-AMP's Vout is negative, which of those diodes is going to be conducting? What will be the direction of the current through that diode?

(We are still on the top diagram of your latest sketch.)

 

[Femme_physics]

You mean that in both cases the rightmost LED lights up?
I thought you had found that in one of those cases the left LED would light up...

That's what I thought so initially, but you guys corrected me in terms of how current flows-- that's the conclusion I came to...

So should it be like this?

http://img864.imageshack.us/img864/262/s2s2.jpg

But that means the op-amp current doesn't split-- how come?

So if the OP-AMP's Vout is negative, which of those diodes is going to be conducting? What will be the direction of the current through that diode?

(We are still on the top diagram of your latest sketch.)

At the top diagram? The furthest one will be conducting.

 

[NascentOxygen]

Now you have them both wrong.

Let's try again, staying with the top circuit, not jumping to the second yet.

If the OP-AMP's Vout is negative, which of those diodes is going to be conducting? Carefully examine the circuit (ignore all your arrows) and decide, then explain why.

 

[technician]

If you are happy that the output of the op amp goes -8V or +8V then you can picture the output to behave like a battery as far as the diodes and lamps are concerned.
I have attached a drawing showing how the op amp appears to the diodes and lamps. Can you see how current flows.
(I emphasise that this battery I have drawn is how the output behaves...it is not a real battery !)
Hope this helps

 

[Femme_physics]

If the OP-AMP's Vout is negative, which of those diodes is going to be conducting? Carefully examine the circuit (ignore all your arrows) and decide, then explain why.

I still only see the furthest one conducting :( I don't know what's wrong with my S1 current loop sketch?..

technician-- i haven't read your reply yet. Sec.

 

[Femme_physics]

Oh! Oh! I think I got it...

http://img62.imageshack.us/img62/6664/s1s2c.jpg

Yes?

 

[NascentOxygen]

No. Both wrong!

You fixed some errors, but introduced more. You keep jumping ahead before your error is sorted out.

Look what you've done to the inputs: you have current going the wrong way from the battery. Current should enter the external circuit from the long side (+) of the cell symbol.

 

[NascentOxygen]

Let's go back to the sketches here, and just focus on the top one https://www.physicsforums.com/showpost.php?p=3792186&postcount=22

No need for any more sketches. You correctly figured out which diode is conducting. So you know that the circuit is operating, you just need to account for why it does.

 

[Femme_physics]

Let's go back to the sketches here, and just focus on the top one https://www.physicsforums.com/showpos...6&postcount=22

No need for any more sketches. You correctly figured out which diode is conducting. So you know that the circuit is operating, you just need to account for why it does.

Because there's voltage?

Look what you've done to the inputs: you have current going the wrong way from the battery

That's because it's an inverter

 

[I like Serena]

Because there's voltage?

That's because it's an inverter

Uhh ?

Current flows from + to -.
So it (almost) always flows away from the plus pole of a battery (the long line) and it (almost) always flows toward the minus pole of a battery (short line).
Certainly in this case it does.
However, in some of your drawings it does, but in others it doesn't...