Diode incremental model help

In summary, you are having trouble calculating the diode current with the incremental model. You might consider letting the 10V source be represented by a variable and then finding the Thevenin equivalent for the network as "viewed" from the diode D2's point of view.
  • #1
asdf12312
199
1

Homework Statement


upload_2014-9-21_15-24-8.png

upload_2014-9-21_15-25-2.png

Homework Equations


r_d= V_T/I_D, V_T=25mV

The Attempt at a Solution


i alreayd find the operation of diodes, D1 off and D2 on (2nd circuit). the current through this diode i also calculate as 34mA, also r_d=0.7353 ohm. I am having trouble using incremental model, in my book it gives voltage divider like V(r_d/(r_d+R)) where V is ripple in voltage (+1 here) and R is I guess the eq. resistance. except I don't think I can voltage divider here, I try to change the V source to 3.7 and 5.7V (+1) and I get the diode current change by 1mV each time.. but not sure how to prove this. Using node voltage I find V at node is -20V, the diode current is (4.7+1+20)/1k + (20)/1.93k this also give change in diode current +1mA for +1V of the 10V source, but is this the right way to do it?
 
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  • #2
Your operating point current and diode resistance look about right. You might consider letting the 10V source be represented by a variable and then finding the Thevenin equivalent for the network as "viewed" from the diode D2's point of view. So the Thevenin voltage will be an expression that depends upon ##V_s##.

With the incremental model diode attached as its load you'll have a series circuit with ##V_{th}, R_{th}, r_d, ## and ##v_d##. Writing an expression for the diode current with respect to ##V_s## will be simple then.

You might even consider finding ##\frac{dI}{dV_s}## for that expression. Being a linear circuit, it should be equivalent to ##\frac{\Delta I}{\Delta V_s}##.
 
  • #3
I'm not sure if this is right but I get Vth = 3.096V (or 23.096V?) and Rth = 1k||1.93k= 658.703. so with Vth and Rth in series with Rd, do I need to include the -20V or not? If I just calculate first change in diode voltage then current from ripple, +1V*(Rd/(Rd+Rth)) = 1.11mV, Id=Vd/Rd = 1.11mV/0.7353 =+1.51mA... something like that?
upload_2014-9-22_1-22-18.png
 
  • #4
asdf12312 said:
I'm not sure if this is right but I get Vth = 3.096V (or 23.096V?) and Rth = 1k||1.93k= 658.703. so with Vth and Rth in series with Rd, do I need to include the -20V or not? If I just calculate first change in diode voltage then current from ripple, +1V*(Rd/(Rd+Rth)) = 1.11mV, Id=Vd/Rd = 1.11mV/0.7353 =+1.51mA... something like that?
View attachment 73450
Ah, you left the 10V source as 10V and didn't make it a variable. The problem is, any variation in that 10V will be scaled by the resistor network in the same fashion that the 10V gets scaled when it gets transformed into its contribution to the Thevenin voltage. I think what you've done above is apply your +1V to the Thevenin voltage itself. That will result in a larger effect on the current than if it were the 10 V source varying by 1V. I was thinking to carry a variable for the 10V source through the Thevenin model in order to retain the network scaling factor. But no matter...

Another approach is to take advantage of superposition to determine the effect on the current of just the noise source of the 10V. Consider your circuit re-drawn as follows:

Fig1.gif

Now, ##V_s## represents the noise on the 10 V supply. If you suppress ALL sources except for ##V_s## then calculating your Thevenin model should be straightforward.
 
  • #5


The diode incremental model is a simplified representation of a diode that is used to analyze small-signal variations in a circuit. It is based on the assumption that the diode is a linear device and its characteristics can be approximated by a small-signal resistance (r_d) and a constant voltage drop (V_T). The equation you have mentioned, V(r_d/(r_d+R)), is used to calculate the voltage across the diode using the voltage divider rule.

In order to use the incremental model, you need to first determine the operating point of the diode. This can be done by finding the current through the diode using the diode equation and the given voltage source. Once you have the operating point, you can then use small-signal analysis techniques, such as the voltage divider rule, to calculate the variations in voltage and current due to small changes in the circuit parameters.

It seems like you have correctly calculated the diode current for different values of the voltage source, which is a good start. To prove your results, you can use the incremental model to calculate the expected changes in diode current for small variations in the voltage source. You can also verify your results by simulating the circuit in a software like LTSpice.

Overall, the incremental model is a useful tool for analyzing small-signal variations in a circuit, but it is important to remember that it is only an approximation and may not be accurate for large variations or non-linear behavior of the diode.
 

1. What is the diode incremental model?

The diode incremental model is a mathematical representation of the behavior of a diode in an electronic circuit. It takes into account the small-signal behavior of the diode and is used to analyze and design circuits that use diodes.

2. How is the diode incremental model different from the ideal diode model?

The ideal diode model assumes that a diode is either completely on (with zero voltage drop) or completely off (with infinite resistance). The incremental model takes into account the small changes in diode behavior when subjected to small changes in voltage or current.

3. What are the parameters used in the diode incremental model?

The diode incremental model uses three parameters: the forward voltage drop (VF), the reverse current (IR), and the reverse resistance (RR).

4. How is the diode incremental model used in circuit analysis?

The diode incremental model is used in conjunction with other circuit analysis techniques, such as Kirchhoff's laws, to calculate the behavior of a circuit with diodes. It allows for a more accurate analysis of circuits that use diodes compared to the ideal diode model.

5. Are there any limitations to the diode incremental model?

Yes, the diode incremental model is only accurate for small changes in voltage or current. It also does not take into account other factors that can affect diode behavior, such as temperature or manufacturing variations.

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