Finding Output Voltage: Maths & Graphs

In summary, the conversation discusses finding the output voltage for a circuit with diodes, where the diodes are on or off depending on the input voltage. The solution involves calculating the standing potential at points A and B and determining the changeover point where diodes will switch from being off to on. It is difficult to analyze nonlinear circuits, so following the shorts in the diodes and checking for conduction conditions can help determine the state of the diodes.
  • #1
erece
72
0

Homework Statement


I need to find the output voltage mathematically or graphically


Homework Equations





The Attempt at a Solution


My approach :
as long as the Vi is < 9.3 V , the diode D1 will be ON because in that case VA will be < 10 V and current can flow through 10k resistor. And if this is true then D2 will also be 0N and Vo = Vi.
Now second interval i.e Vi > 9.3 V, same reasoning for D3 i.e D3 will be ON as long as Vi > -9.3 V & D2 is ON & D4 is also ON. But in this case if we track from input to the output through D3 & D4 , then i get Vo = Vi and this is making D1 ON which i assumed to be OFF. Summarizing my results :

0<Vi<9.3 all diodes are ON & Vo = Vi
Vi>9.3 all diodes are ON & Vo=Vi... but this is not possible because this makes VA > 10 which shows current can't flow from +10V battery. So in short i know i am doing but the question is where ?
 

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  • #2
erece said:
Vi>9.3 all diodes are ON & Vo=Vi... but this is not possible because this makes VA > 10 which shows current can't flow from +10V battery. So in short i know i am doing but the question is where ?
If, for a diode to be forward biased, you need a voltage of a level which can't exist, then your assumption of that diode conducting is wrong.
 
  • #3
can you tell me what will be the output voltage Vo for Vi = 5V ?
According to me, Vi=5V so all diodes can be ON & in that case VA = Vi+0.7 = 5.7V and Vo = 5.7-0.7 = 5V, this will make Vc = 4.3 V again possible. So is this the correct output for Vi=5V ?
 
  • #4
EDIT: I've changed to Vi=4V for your example.
erece said:
can you tell me what will be the output voltage Vo for Vi = 4V ?
According to me, Vi=4V so all diodes can be ON & in that case VA = Vi+0.7 = 4.7V and Vo = 4.7-0.7 = 4V, this will make Vc = 3.3 V again possible. So is this the correct output for Vi=4V ?
For Vi=4V that sounds right, but you need to check the currents through the diodes to be sure — do this by determining the currents through all 3 resistors since you know the [expected] voltages across them.

If you go on increasing Vi much beyond 4V, the currents through D1 and D4 will be decreasing and at some point these currents will become zero. At that stage, Vo is no longer directly determined by Vi.

Evaluate the currents for Vi of 5V and you'll see what I'm talking about.
 
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  • #5
i got it nascent
is there a way to find that which diodes will be ON and which will be OFF for a particular input voltage without calculating the currents ?
 
  • #6
erece said:

The Attempt at a Solution


My approach :
as long as the Vi is < 9.3 V , the diode D1 will be ON because in that case VA will be < 10 V and current can flow through 10k resistor. And if this is true then D2 will also be 0N and Vo = Vi.


That's not what I got..

If Vi=-10V ...

Point A will be at Vi+0.7 = -9.3V D2 will be off.

Current will flow from ground, through the 10K near the output, through D4 and the 10K to -10V. Vo will be about -9.3/2 = -4.65V
 
  • #7
erece said:
i got it nascent
is there a way to find that which diodes will be ON and which will be OFF for a particular input voltage without calculating the currents ?
If you imagine D1 and D4 not conducting, you can calculate the standing potential at point A (and B). This defines the changeover point which, if Vi falls further, will forward bias D1.
 
  • #8
NascentOxygen said:
If you imagine D1 and D4 not conducting, you can calculate the standing potential at point A (and B). This defines the changeover point which, if Vi falls further, will forward bias D1.

this is the solution given in the textbook. Still i have a doubt that how do we come to know that we have to observe the state of D1 & D4 ?
 
  • #9
It's always harder to see what is happening in nonlinear circuits.

What I do is follow the shorts in the diodes (pretend they conduct) and see if some conditions for conduction arise.

For this circuit, you will see if Va > Vc (plus some diode drops), all the diodes conduct. So Va=Vc (plus some diode drops) is one transition point that will see one or more diodes shut off. If the diodes are on then there is a relationship between Vi, Vo, Va and Vc.

(This circuit is a variation of the diode-bridge modulator / diode-bridge switch. One use is in AM radio modulation).
 
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Related to Finding Output Voltage: Maths & Graphs

1. How do I calculate output voltage?

The formula for calculating output voltage is V = IR, where V is the voltage, I is the current, and R is the resistance. You can also use the Ohm's Law formula, V = I x R, to calculate output voltage.

2. How do I graph output voltage?

To graph output voltage, you will need to plot the voltage values on the y-axis and the corresponding current values on the x-axis. Make sure to label your axes and include a title for your graph.

3. Why is output voltage important?

Output voltage is important because it is the measure of the electrical potential difference between two points in a circuit. It is used to determine the amount of power that can be delivered to a load and is necessary for proper functioning of electronic devices.

4. What is the relationship between output voltage and resistance?

The relationship between output voltage and resistance is inverse. As resistance increases, output voltage decreases. This means that the higher the resistance in a circuit, the lower the output voltage will be.

5. How can I increase output voltage?

To increase output voltage, you can either increase the current or decrease the resistance in a circuit. This can be achieved by using a transformer, increasing the number of batteries, or using different components with lower resistance values.

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