Electric circuit with negative reference voltages and diodes

In summary, you should assume D2 is not conducting when trying to figure out the voltages for nodes Va and Vb. If D2 is on, then Va > Vb.
  • #1
canadiansmith
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0
Hi I am having trouble figuring out what to do with negative reference voltages in my circuit. I attached my problem(Problem 3) and I know that some current will be drawn out of the ground and I think all of the diodes will be on but I am unsure how to approach this problem.
 

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  • #3
Sorry Here it is right side up
 

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  • #4
OK, you are right about the diodes all being on. You need to show some work here. What have you tried so far? Have you read the forum rules about how to post homework problems?
 
  • #5
Yes I have read the rules regarding homework. So far I have redrawn the circuit with batteries in place of the diodes each = to 0.7 volts. The 0.7 V batteries all oppose their assigned current ( ie the negative end of the battery is the same side as the cathode on the diode.
So first I tried to figure out each node using KVL starting with Va. But quickly I have figured out this cannot work. So right now I am sort of suck because I am not sure if what I am trying to do is to get each negative potential voltage to zero or if I should somehow find a way to sum up the voltage sources.
 
  • #6
Oh and the current from the +5 Voltage source across R1 is 1mA
 
  • #7
phinds said:
OK, you are right about the diodes all being on. You need to show some work here. What have you tried so far? Have you read the forum rules about how to post homework problems?

You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.
 
  • #8
gneill said:
You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.

Hm ... I worked the whole thing under that assumption and got good and consistent answers. Are you sure I need to redo it?
 
  • #9
ok so I have Va=7.15V and the current on the left side = 1.43 mA
Vb= 4.3V and current right side = 0.86mA. But how do I know that I can make the assumption to turn D2 off?
 
  • #10
Like with D2 off I get Va>Vb which would mean that my assumption would not turn D2 off.
 
  • #11
Oh wait. Va should be >Vb so the assumption is correct. Thanks for your help. One last question: when choosing diodes to be on or off do you simply take your best guess or is there a more systematic approach?
 
  • #12
canadiansmith said:
Like with D2 off I get Va>Vb which would mean that my assumption would not turn D2 off.

Is Va > Vb always a sufficient condition to have D2 on?
 
  • #13
phinds said:
Is Va > Vb always a sufficient condition to have D2 on?

I believe so. The polarity on the open switch that is used to replace D2 would be such that the diode would be turned off. Assuming my Va and Vb are correct
 
  • #14
canadiansmith said:
I believe so. The polarity on the open switch that is used to replace D2 would be such that the diode would be turned off. Assuming my Va and Vb are correct

Well, you need to think about this some more. What does "on voltage" on a diode mean to you?
 
  • #15
Well its the voltage required to turn on the diode to allow current through it. Usually 0.7V for silicon.
 
  • #16
So is it always turned on if that voltage is greater than zero?
 
  • #17
no. If the voltage is less than the "on" voltage then the diode will still be off.
 
  • #18
gneill said:
You might want to check that assumption about the diodes all being forward biased.

Assume for a moment that D2 is not conducting. What would be VA? Compare it to VB.

OOPS --- that IS a bad assumption isn't it.

Good catch.
 
  • #19
canadiansmith said:
no. If the voltage is less than the "on" voltage then the diode will still be off.

Good.

Also, note the post I just made about D2 being on.
 
  • #20
OK. So if d2 is on does this change my value for va and vb? I don't think it would.
 
  • #21
canadiansmith said:
OK. So if d2 is on does this change my value for va and vb? I don't think it would.

I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.
 
  • #22
phinds said:
I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.

What? I am confused. What post about D2 being on? Can you please explain.
 
  • #23
canadiansmith said:
What? I am confused. What post about D2 being on? Can you please explain.

Read through the posts again in order. You'll get it.
 
  • #24
oh no. I see what you are saying now. So is everything that i have done wrong now? Like the way I calculated Va and Vb is this still valid.
 
  • #25
Suppose that you remove D2 from the circuit (excise it... open circuit it). What would be the potential VA? What would be the potential VB? What's VA - VB?
 
  • #26
Va-Vb would be 7.15-4.3 = 2.85V
 
  • #27
canadiansmith said:
Va-Vb would be 7.15-4.3 = 2.85V

Where would a 7.15V potential come from? The highest potential source I can see is +5V.
 
  • #28
That would be the potential with respect to ground.
 
  • #29
canadiansmith said:
That would be the potential with respect to ground.

What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.
 
  • #30
gneill said:
What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.

Ok I found Va by (R1/R1+R2)*(5-(-100-0.7) = 7.15V since the bottom of the left side is -10. Likewise for Vb i found 0-0.7-(-5) = 4.3 V
 
  • #31
Is this wrong? If so please help me. I have been working on this all day.
 
  • #32
canadiansmith said:
Ok I found Va by (R1/R1+R2)*(5-(-100-0.7) = 7.15V since the bottom of the left side is -10. Likewise for Vb i found 0-0.7-(-5) = 4.3 V

Your formula for VA cannot yield 7.15V. It can't! VA must be less than 5V, since he current is flowing down through R1 (so there's a voltage drop across R1), and the top of R1 is at only +5V. It might be convenient to find the current in the branch first, then find the voltage drop on R1. +5V minus that voltage drop will give you VA.

The drop across D3 is 0.7V, so VB can only be 0.7V below ground potential. That makes VB -0.7V.
 
  • #33
ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off
 
  • #34
canadiansmith said:
ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off

Show your calculation for Va if D2 is not conducting.
 
  • #35
gneill said:
Show your calculation for Va if D2 is not conducting.

Exactly what I was going to say :smile:
 
<h2>1. What is a negative reference voltage in an electric circuit?</h2><p>A negative reference voltage is a voltage that is used as a reference point in a circuit. It is usually set at a lower value than the positive voltage in the circuit and is used as a baseline for measuring the voltage at other points in the circuit.</p><h2>2. How is a negative reference voltage created in a circuit?</h2><p>A negative reference voltage can be created in a circuit using a voltage divider, which is a combination of resistors that divide a positive voltage into a lower negative voltage. It can also be created using a voltage regulator or an operational amplifier.</p><h2>3. What is the purpose of using a negative reference voltage in a circuit?</h2><p>A negative reference voltage is used in a circuit to provide a stable and reliable reference point for measuring voltages. It can also be used to bias certain components, such as diodes, in a circuit.</p><h2>4. How do diodes work in an electric circuit with negative reference voltages?</h2><p>Diodes in an electric circuit with negative reference voltages act as one-way valves for current flow. They allow current to flow in one direction, from the positive to the negative reference voltage, but block current flow in the opposite direction.</p><h2>5. What are some common applications of electric circuits with negative reference voltages and diodes?</h2><p>Electric circuits with negative reference voltages and diodes are commonly used in power supplies, voltage regulators, and signal processing circuits. They can also be found in audio amplifiers, battery chargers, and electronic control systems.</p>

1. What is a negative reference voltage in an electric circuit?

A negative reference voltage is a voltage that is used as a reference point in a circuit. It is usually set at a lower value than the positive voltage in the circuit and is used as a baseline for measuring the voltage at other points in the circuit.

2. How is a negative reference voltage created in a circuit?

A negative reference voltage can be created in a circuit using a voltage divider, which is a combination of resistors that divide a positive voltage into a lower negative voltage. It can also be created using a voltage regulator or an operational amplifier.

3. What is the purpose of using a negative reference voltage in a circuit?

A negative reference voltage is used in a circuit to provide a stable and reliable reference point for measuring voltages. It can also be used to bias certain components, such as diodes, in a circuit.

4. How do diodes work in an electric circuit with negative reference voltages?

Diodes in an electric circuit with negative reference voltages act as one-way valves for current flow. They allow current to flow in one direction, from the positive to the negative reference voltage, but block current flow in the opposite direction.

5. What are some common applications of electric circuits with negative reference voltages and diodes?

Electric circuits with negative reference voltages and diodes are commonly used in power supplies, voltage regulators, and signal processing circuits. They can also be found in audio amplifiers, battery chargers, and electronic control systems.

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