I'm assuming the section near the battery. I need to reduce the whole thing to a equivalent circuit with just a battery, resistor and diode. So whichever would get me to that the easiest would be best, so I can apply the real properties on a simplified model of the diode's circuit.NascentOxygen said:
So if I broke it off before the diode. I would have the 10V source and a simplified 10K resistor.gneill said:
No, but that might work also depending upon what your end goal is.bnosam said:
Then everything on the left would simplify to the 10V source and just a single 10K resistor, no? The reason I want to simplify it is to apply the real diode model's properties to find current the compute output voltage of the diode using iterative analysis.gneill said:
No. You should "cut" the circuit where indicated and do the analysis to obtain the Thevenin model. The two 5k resistors are not in series. Hint: It's a voltage divider. Here's the circuit re-drawn to make it more obvious:bnosam said:
By "output voltage of the diode" do you mean the potential drop across the diode, or the potential drop across the 2.5k load resistor?
Ohh ok. I'm not used to drawing circuits in the manner that I presented it in, that's just the form my book uses. I struggle a lot with circuits in general. It makes sense that it is a voltage divider, though. So the voltage divider where R1 = R2 the output voltage is half of the input voltage. 10 * 1/2 = output from those two voltages would be 5 volts?gneill said:
Yes, the Thevenin voltage for the voltage divider would be 5 V. You still need to determine the Thevenin resistance.bnosam said:
Ah. So then leave the 2.5 k resistor in place. You'll have a Thevenin model consisting of a voltage source in series with a resistance (Rth) in series with a diode and 2.5 kΩ resistor.
So I would remove the voltage source and then with that voltage source set to short circuit, the two 5K resistors are in parallel and the Rth = 2.5k Ohms. Is that correct?gneill said:
Yes.bnosam said:
Then I have the source, diode and two resistors in series in the circuit?gneill said:
Thank you very much for you assistance!gneill said:
Thevenin equivalent resistance refers to the equivalent resistance of a circuit when viewed from two terminals. It is used to simplify complex circuits and analyze their behavior.
Finding thevenin equivalent resistance with diode allows us to analyze the behavior of a circuit that contains a diode. This is important in understanding the overall functionality and performance of electronic devices.
Thevenin equivalent resistance with diode is calculated by first finding the open-circuit voltage and short-circuit current of the circuit. The equivalent resistance is then calculated by dividing the open-circuit voltage by the short-circuit current.
Using thevenin equivalent resistance with diode allows for easier analysis and design of electronic circuits. It also helps to simplify complex circuits and make calculations more efficient.
One limitation of using thevenin equivalent resistance with diode is that it assumes the behavior of the diode is linear. In reality, diodes have a non-linear behavior which can affect the accuracy of the calculations. Additionally, thevenin equivalent resistance may not work for circuits with multiple diodes or complex non-linear components.
