Engineering Finding Output Voltage: Maths & Graphs

AI Thread Summary
The discussion focuses on determining the output voltage (Vo) in a circuit with multiple diodes based on varying input voltages (Vi). For Vi less than 9.3V, all diodes are on, resulting in Vo equaling Vi. However, when Vi exceeds 9.3V, the assumption that all diodes remain on leads to contradictions, as it causes voltage levels that prevent current flow. Participants suggest evaluating the currents through the diodes to confirm their states and explore methods to predict diode behavior without current calculations. The conversation emphasizes understanding the relationships between input, output, and diode states in nonlinear circuits, particularly in the context of a diode-bridge modulator.
erece
Messages
70
Reaction score
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 ?
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    4.9 KB · Views: 452
Physics news on Phys.org
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.
 
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 ?
 
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[/color] ?
According to me, Vi=4V[/color] so all diodes can be ON & in that case VA = Vi+0.7 = 4.7V[/color] 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 ?
✔ [/size][/color]
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.
 
Last edited:
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 ?
 
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
 
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.
 
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 ?
 
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).
 
Last edited:

Similar threads

Back
Top