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Problem understanding how a Bias Limiter works.
- Thread starter saadishere
- Start date
Is this homework? Then it should have been posted in the homework section.
Ignore the battery (remove from the circuit). Calculate the voltage at point A. Replace the voltage at point A by the battery's voltage minus the diode's drop if the voltage as calculated in the first step is lower than the voltage from the battery (minus the diode's drop).
With some caution you could also use this approach:
Apply the superposition theorem to the circuit.
First "turn off" the AC source by replacing it with a short circuit. Calculate the output.
Second "turn off" the battery by replacing it with a short circuit. Calculate the output.
The sum of both calculations is the circuit's behaviour.
Since the superposition theorem is only valid for linear circuits but you have a diode (nonlinear element) in your circuit, you will have to make different calculations for the diode being conductive or non-conductive. For the final result superimpose the different results piecewise. The graph on the right of the schematic you provided gives you the necessary clues (linear - sinusoidal - linear).
Ignore the battery (remove from the circuit). Calculate the voltage at point A. Replace the voltage at point A by the battery's voltage minus the diode's drop if the voltage as calculated in the first step is lower than the voltage from the battery (minus the diode's drop).
With some caution you could also use this approach:
Apply the superposition theorem to the circuit.
First "turn off" the AC source by replacing it with a short circuit. Calculate the output.
Second "turn off" the battery by replacing it with a short circuit. Calculate the output.
The sum of both calculations is the circuit's behaviour.
Since the superposition theorem is only valid for linear circuits but you have a diode (nonlinear element) in your circuit, you will have to make different calculations for the diode being conductive or non-conductive. For the final result superimpose the different results piecewise. The graph on the right of the schematic you provided gives you the necessary clues (linear - sinusoidal - linear).
Sorry i didn't understand
Ignore the battery (remove from the circuit). You then have only the AC source and the resistive divider. Calculate the voltage at point A for this circuit.
Now look at this voltage:
- If the voltage is higher than the battery's voltage minus the diode's drop, then the diode is reverse biased (non conducting). No current flows through a reverse biased diode. Therefore the voltage at point A follows the calculated voltage. This is the sinusoidal part of the curve in the diagram you posted.
- If the voltage at point A is lower than the battery's voltage minus the diode's drop, then the diode is forward biased (conducting). Current will flow from the battery through the diode into point A. For the sake of this example assume the voltage drop across the diode to be constant (0.7 V). Since the battery's voltage is constant, too, the voltage at Point A will be constant at V(battery)-V(diode).This is the flat part of the curve in the diagram you posted.
Now look at this voltage:
- If the voltage is higher than the battery's voltage minus the diode's drop, then the diode is reverse biased (non conducting). No current flows through a reverse biased diode. Therefore the voltage at point A follows the calculated voltage. This is the sinusoidal part of the curve in the diagram you posted.
- If the voltage at point A is lower than the battery's voltage minus the diode's drop, then the diode is forward biased (conducting). Current will flow from the battery through the diode into point A. For the sake of this example assume the voltage drop across the diode to be constant (0.7 V). Since the battery's voltage is constant, too, the voltage at Point A will be constant at V(battery)-V(diode).This is the flat part of the curve in the diagram you posted.
Ok thank you very much 
Thank you , I understand better the dynamic now however I have a question
After the AC goes on the resistance , we should have a drop in the voltage at A right?
I am trying here to link the AC voltage and the potential of A so I can know if the diode is forwared biased or no.
Basicaly how can I know when the AC provide a voltage that make potential A bigger than Vbias - 0.7 V.
Thank you for your help , I appreciate it.
After the AC goes on the resistance , we should have a drop in the voltage at A right?
I am trying here to link the AC voltage and the potential of A so I can know if the diode is forwared biased or no.
Basicaly how can I know when the AC provide a voltage that make potential A bigger than Vbias - 0.7 V.
Thank you for your help , I appreciate it.
In this circuit the diode acts as a switch. The diode switches ON to ensure that Va can never be less than Vbias-0.7 volts. Because the diode has two states (ON & OFF) the circuit operates in two different states, i.e., there are two different circuits. I would suggest that you draw the two circuits, one where Va=Vbias-0.7 (diode ON) and the other where Va>Vbias-0.7 (diode OFF). Note that for the (diode OFF) case, you can just remove the diode and Vbias from the circuit in order to calculate Va.
thank you Laplace , that makes more sens.
