# Electronic circuit with Op-Amp, Switch and Diodes

1. Feb 16, 2012

### Femme_physics

1. The problem statement, all variables and given/known data

http://img689.imageshack.us/img689/6668/mycircuit.jpg [Broken]

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

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?

2. Relevant equations

3. 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 :)

Last edited by a moderator: May 5, 2017
2. Feb 16, 2012

### 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

Last edited: Feb 16, 2012
3. Feb 16, 2012

### likephysics

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

4. Feb 16, 2012

### technician

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

5. Feb 16, 2012

### 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?

6. Feb 16, 2012

### likephysics

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

7. Feb 17, 2012

### 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!

8. Feb 17, 2012

### 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

Last edited: Feb 17, 2012
9. Feb 17, 2012

### likephysics

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?

10. Feb 19, 2012

### Femme_physics

If I reverse, it's -1.5v.

Agreed.

I see! :)

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

Last edited by a moderator: May 5, 2017
11. Feb 19, 2012

### 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

12. Feb 19, 2012

### I like Serena

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.

13. Feb 27, 2012

### Femme_physics

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?

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 [Broken]

Last edited by a moderator: May 5, 2017
14. Feb 27, 2012

### I like Serena

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.

Yep! :)

Last edited by a moderator: May 5, 2017
15. Feb 27, 2012

### 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?

16. Feb 27, 2012

### 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)?

17. Feb 27, 2012

### Femme_physics

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

18. Feb 27, 2012

### 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. ;)

19. Feb 29, 2012

### Femme_physics

:) Agreed.

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 [Broken]

Last edited by a moderator: May 5, 2017
20. Feb 29, 2012

### 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

#### Attached Files:

• ###### op amp.jpg
File size:
76.6 KB
Views:
157
Last edited: Feb 29, 2012