REALLY basic question on circuit analysis.

[HackedFridgeMagnet]

I haven't done this for a while but :

Write an expression of the circuit in ohms law.
Substitute in the expression for Vdiode in terms of Idiode.
You should have something like  Idiode = A * log(Idiode) + C

ie it has only one variable but you cannot simplify.
so you solve it like a computer.
put in a start value of Idiode on the log side and get the LHS value.

after 3 or 4 iterations it should converge.

They should have explained this in your course.

[M0BSW]

I haven't done this for a while but :

Write an expression of the circuit in ohms law.
Substitute in the expression for Vdiode in terms of Idiode.
You should have something like  Idiode = A * log(Idiode) + C

ie it has only one variable but you cannot simplify.
so you solve it like a computer.
put in a start value of Idiode on the log side and get the LHS value.

after 3 or 4 iterations it should converge.

They should have explained this in your course.

Cricky

[M0BSW]

Well, Kirchoff's Current and Voltage Law, probably also nodal and mesh analysis. That is what we studied in Circuits I. Apparently it seems to be very useful!

Thank You , I read up on these,as they have not mentioned it in the course I'm attending, if I learn enough I may not be bewildered when he starts on it, we just done Ohms law in depth, bit at a time I think, this week we are going to mess around with  Oscilloscopes and function generators, then the following week it's theory again , they mix it up  to keep you from falling asleep , something I can easily do, in a warm enviroment.

[IanB]

I haven't done this for a while but :

Write an expression of the circuit in ohms law.
Substitute in the expression for Vdiode in terms of Idiode.
You should have something like  Idiode = A * log(Idiode) + C

ie it has only one variable but you cannot simplify.
so you solve it like a computer.
put in a start value of Idiode on the log side and get the LHS value.

after 3 or 4 iterations it should converge.

They should have explained this in your course.

As observed earlier in the thread there is not enough information provided in the question to proceed this way. There are too many unknown values in the problem.

[zorthgo]

I don't wan't abuse your good will but can you help me on something else. I am having a huge problem understanding the variable on this problem I found online:

Half-wave Rectifier with Resistive Load For the half-wave rectifier of Fig. 3-1a, the source is a sinusoid of 120V rms at a frequency of 60Hz. The load resistor is5W. Determine (a) the average load current, (b) the average power absorbed by the load, and (c) the power factor of the circuit.

I thought that 120v rms was the V_rms. But apparently it isn't because the formula that I have to V_rms is V_rms=V_m/2, but in order to find V_m on part a instead of having V_m=2*V_rms=120V*2 they have V_m=120sqrt(2). Where did the square root come from since this is a half-wave rectifier. I know that you have a square root on full-wave rectifiers. Pleas help!

[alm]

This has nothing to do with the bridge rectifier, the same would apply to just the voltage source. It is about the relation between the amplitude of the sinusoidal signal (peak height) and the RMS voltage. Note that amplitude indicates the V_peak, which is half of V_peak-to-peak. V_rms is the root of the mean of the squared voltage, or the square root of 1 over the period times the definite integral of the sine squared over the period (see the bottom equation for V_rms in this Wikipedia entry). If you go through the calculus, you'll find that the result is that V_rms = 1/sqrt(2) times the amplitude of the sine. This is where the sqrt(2) factor comes from.

Now draw the sine input (indicate zeros and maxima/minima) and the output from a half bridge rectifier (leave away the smoothing cap). Indicate over which portions of the wave the diode is conducting (forward biased). Do the same for a full bridge rectifier. How does the peak amplitude of the output of the rectifier relate to the RMS voltage? What is the difference between the output from a full and half wave rectifier?

[IanB]

I don't wan't abuse your good will but can you help me on something else. I am having a huge problem understanding the variable on this problem I found online:

Half-wave Rectifier with Resistive Load For the half-wave rectifier of Fig. 3-1a, the source is a sinusoid of 120V rms at a frequency of 60Hz. The load resistor is5W. Determine (a) the average load current, (b) the average power absorbed by the load, and (c) the power factor of the circuit.

I thought that 120v rms was the V_rms. But apparently it isn't because the formula that I have to V_rms is V_rms=V_m/2, but in order to find V_m on part a instead of having V_m=2*V_rms=120V*2 they have V_m=120sqrt(2). Where did the square root come from since this is a half-wave rectifier. I know that you have a square root on full-wave rectifiers. Pleas help!

The half wave rectifier with a purely resistive load is a special case where you can apply some intuitive insight and you don't necessarily have to work out integrals.

If you do what alm suggested and plot out the input voltage wave form and the output voltage wave form, you will see that the half-wave rectifier has (as its name suggests) only allowed half of the sine wave through to the load. In effect, since only half of the cycles have got through to the resistor, the power delivered to the resistor has been cut in half compared to what would be delivered by the full wave supply voltage.

This is where the sqrt(2) comes from. If the power is reduced by 2 the RMS voltage is reduced by sqrt(2) since power in a resistor is given by V²/R. Therefore the RMS voltage across the resistor is 120 / sqrt(2) = 84.9 V.

[zorthgo]

Thanks for the explanation guys! You guys rock! I understand a little better!