Electronic circuit with Op-Amp, Switch and Diodes

by Femme_physics
Tags: circuit, diodes, electronic, opamp, switch
 PF Patron P: 2,544 1. The problem statement, all variables and given/known data 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? 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 :)
 P: 1,508 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
P: 591
 Quote by 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
Vout is +10V when switch is at position 2.

P: 1,508

Electronic circuit with Op-Amp, Switch and Diodes

Well spotted 'I like physics'
Apologies.....careless
 P: 1,508 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?
P: 591
 Quote by 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?
Sorry, I overlooked the "Earth" connected to non-inv terminal.
Good trick question though.
 PF Patron P: 2,544 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!
 P: 1,508 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
P: 591
 Quote by 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!
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?
PF Patron
P: 2,544
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?
 P: 1,508 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
HW Helper
P: 6,164
 Quote by Femme_physics 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.
PF Patron
P: 2,544
 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) ->

HW Helper
P: 6,164
 Quote by 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?
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! :)
 PF Patron P: 2,544 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?
 HW Helper P: 6,164 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)?
 PF Patron P: 2,544 Well, in that case then they should ask this question last, not first! I gotta do some calculations first!
 HW Helper P: 6,164 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. ;)

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