REALLY basic question on circuit analysis.

REALLY basic question on circuit analysis.
Posted by zorthgo on 20 Oct, 2012 19:42
This is kind of a simple question. But I can't put my finger on it...

I am studying for my electronics test next week, so I am going over stuff that we did in class. One of the examples, although very easy has gotten me stumped. On the attached file under case a and b, I need to find I_D, which is the current that will go through the load resistance. My problem is that I don't recognize the equation that gives me the 3mA for the ideal case and the 2.94mA for the actual diode case. At first I thought it was voltage/current division but for that I would need one of the two resistance on the numerator like:

I_D=V_th(R1/(R1+R2))

But the formula only has

I_D=V_th/(R1+R2)

It is missing resistance on top.

Can anyone tell me what formula that is? Is it a weird form of current/voltage division that I don't recognize?

Thanks!

EE 3300 - Electronics (Work Notebook) 20121020.pdf

#1 Reply
Posted by zorthgo on 20 Oct, 2012 19:46
Also... If it is just the basic ohms law, doesn't the diode have any resistance?

#2 Reply
Posted by IanB on 20 Oct, 2012 19:51
It's simple Ohm's law. The current I_D through the diode is the current in the circuit, given by

I_D = V_Th / (R_Th + R_L)

The diode is "ideal" which means it passes current with no voltage drop. Therefore the diode behaves just like a piece of wire in this case.

In the second case, the diode is "real" and has an (assumed) voltage drop of 1.2 V. So the voltage driving current round the circuit is now reduced by the diode drop. So subtract 1.2 V from 60 V and repeat the above calculation.

#3 Reply
Posted by zorthgo on 20 Oct, 2012 20:21
Thanks IanB!

I have another question if you wouldn't mind. I am trying to solve a similar exercise but this time I have to find the resistance of R_L. I am positive that I need to use the formula I_D=I_se^V_f/.026. But the problem is that my number is coming out way too large. So large that my calculator can't even handle it. So, there must be something wrong. I would really appreciate if anyone could give a quick look at it.

EE 3300 - Electronics (Work Notebook) 20121020.pdf

#4 Reply
Posted by IanB on 20 Oct, 2012 20:32
Did you miss a "-1"? That is to say,

I_D = I_S(e^V_f/0.026 - 1)

#5 Reply
Posted by IanB on 20 Oct, 2012 20:35
(OK, I guess the "-1" might not really matter in many cases.)

#6 Reply
Posted by alm on 20 Oct, 2012 20:35
V_f represents the voltage across the diode. Why would this have to be equal to V_Th in your bottom circuit? What about the voltage across the other components in that loop?

IanB: Assuming V_f is much larger than V_T, the -1 term can be neglected. This is often done in introductory courses to simplify the math.

#7 Reply
Posted by IanB on 20 Oct, 2012 20:42
On looking at the problem statement, it doesn't seem to me that there is enough information to solve the problem. At least one further item seems to be needed.

We have V_s, R₁ and R₂ given, so the left half of the circuit is fully defined and can be made into a Thevenin equivalent.

On the right, we have I_S given, which is a constant parameter of the diode. But it seems to me we need to know one other item, such as a voltage or current associated with the diode or R_L? Without this information, I don't see how to calculate a value for R_L?

#8 Reply
Posted by zorthgo on 20 Oct, 2012 20:44
Thanks alm, I see what you are saying!... but without knowing the current through R_th I can't calculate the voltage drop through it. Now I have a chicken or the egg problem. $:-\$

#9 Reply
Posted by zorthgo on 20 Oct, 2012 20:52
Well, after realizing that V_f is not V_th I am completely stomped. Maybe it is a typo or something. I think I am going to move on and do another problem. This one seems to be a dud!

Thanks guys for the help though!

#10 Reply
Posted by zorthgo on 20 Oct, 2012 21:17
Question... I was looking at my notes, and now I am not sure about V_f. Is that the voltage drop across the diode (which is usually .7V), or the voltage going through the diode?

#11 Reply
Posted by zorthgo on 20 Oct, 2012 21:20
I think the .7 is called the V_{cut in}

#12 Reply
Posted by alm on 20 Oct, 2012 21:26
What do you mean with voltage going through the diode? You can have a voltage across a device, or a current through it.

I concur that the problem appears to have insufficient information. Even if you approximate the diode as a resistor or a constant voltage source, there is insufficient information to solve the equations in the Thevenin circuit (you only have one voltage, one resistance and one resistance/voltage to solve the current and three voltages). I wonder if something got mixed up with the I_S, since 1.5 uA seems awfully high for a saturation current.

#13 Reply
Posted by zorthgo on 20 Oct, 2012 21:34
Because I am a little confused as to what V_f is. Because on one of the examples, for Shockley's equation I_D=I_se^Vf/.026, the V_f is actually the voltage drop across the diode. Which we were taught to just assume it is 0.7V. But I was thinking that the V_f was actually the voltage coming into the diode before the voltage saturation. Apparently I was wrong. Can you confirm that for V_f on that formula they are actually talking about the voltage saturation and not the voltage applied to the diode. Thanks!

#14 Reply
Posted by zorthgo on 20 Oct, 2012 21:36
You are absolutely right about the I_s there is a correction that the teacher put out and the value is actually 25.69X10^-15A.

#15 Reply
Posted by alm on 20 Oct, 2012 22:01
The diode has only two terminals, so the only voltage that it can sense is the potential difference between these two terminals. It has no memory of past voltages. This voltage determines its behavior (current) in the Schockley model. If solve the Schockley equation for Vf for reasonable currents (say 1uA to 1A), you'll get values fairly close to 0.7 V. If you would assume that the forward voltage is exactly 0.7 V, this would define the current through the diode and allow you to solve the problem, but I would expect this to be explicitly stated in the problem.

#16 Reply
Posted by zorthgo on 20 Oct, 2012 22:07
I would assume the teacher just forgot to state it. Because he has been using the .7 for all of his examples. So, would the current (I_D) through the resistor be the same as that of the load resistance (I_th=I_D=I_L)? Because in circuits I, resistors in series had the same current running through them.

#17 Reply
Posted by alm on 20 Oct, 2012 22:17
Did they discuss Kirchoff's circuit laws? Because you should be able to derive this from Kirchoff's current law. This law applies to any circuits you'll encounter in introductory circuit analysis courses (exceptions include antennas).

#18 Reply
Posted by IanB on 20 Oct, 2012 22:26
Using the voltage across a diode to determine the current through it is a very unreliable procedure. As alm mentioned, a range of currents from microamps upwards will all produce a similar voltage drop. The usual way of solving the diode equation is to find the voltage drop given the current through it (and the junction temperature).

If we had a stated value of I_D we would know the current through R_L and could find the voltage drop across the diode, which would allow the problem to be solved in a meaningful way.

#19 Reply
Posted by zorthgo on 20 Oct, 2012 22:39
alm: We studied Kirchoff's law in Circuits I.

IanB: Shouldn't I be able to calculate I_D by using the formula I_D=(25.69x10^-15A)e^.7V/.026?

#20 Reply
Posted by alm on 20 Oct, 2012 22:52
Yes, assuming V_f = 0.7 V and applying Shockley's model would allow you to solve this circuit. The issue with this is that no EE would do this. The exponential relation between current and voltage means that any tiny variation in V_f or temperature will cause huge variations in I_D. The very same property that allows you to assume it's approximately 0.6-0.7 V regardless of the current makes it useless to derive the current from.

You either assume it's approximately 0.7 V and that the current is purely determined by the other devices (assume V_f = 0.7 V, calculate the voltage across R_Th and R_L, derive current from Ohm's law) or use Shockley's diode equation and accept that it's a function of the current (and not constant). Both solutions would require information beyond what's given in this problem. I would probably pick some I_D, like 1 mA, and go from there.

#21 Reply
Posted by IanB on 20 Oct, 2012 23:05
Quote from: zorthgo on 20 Oct, 2012 22:39
IanB: Shouldn't I be able to calculate I_D by using the formula I_D=(25.69x10^-15A)e^.7V/.026?

On paper, yes. In reality, no. Consider:

V_f = 0.68 V, I_D = 5.9 mA
V_f = 0.72 V, I_D = 27 mA

And that's assuming the value of 0.026 is accurate for the given diode (it isn't).

How reliable do you think your estimated current will be given such uncertainty?

#22 Reply
Posted by zorthgo on 21 Oct, 2012 00:43
Good point!... Not reliable at all.

#23 Reply
Posted by M0BSW on 21 Oct, 2012 11:46
Quote from: alm on 20 Oct, 2012 22:17
Did they discuss Kirchoff's circuit laws? Because you should be able to derive this from Kirchoff's current law. This law applies to any circuits you'll encounter in introductory circuit analysis courses (exceptions include antennas).
,
This is interesting, as I'm fairly NEW to this side of the hobby I should read this Kirchoff's current law is there any other , laws I should be reading , I've done the ohms law so far.guess there's lots more "He Gulped ".

#24 Reply
Posted by zorthgo on 21 Oct, 2012 12:07
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!

#25 Reply
Posted by HackedFridgeMagnet on 21 Oct, 2012 12:57
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.

#26 Reply
Posted by M0BSW on 21 Oct, 2012 16:21
Quote from: HackedFridgeMagnet on 21 Oct, 2012 12:57
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

#27 Reply
Posted by M0BSW on 21 Oct, 2012 16:26
Quote from: zorthgo on 21 Oct, 2012 12:07
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.

#28 Reply
Posted by IanB on 21 Oct, 2012 16:27
Quote from: HackedFridgeMagnet on 21 Oct, 2012 12:57
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.

#29 Reply
Posted by zorthgo on 21 Oct, 2012 19:10
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!

#30 Reply
Posted by alm on 21 Oct, 2012 19:43
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?

#31 Reply
Posted by IanB on 21 Oct, 2012 21:39
Quote from: zorthgo on 21 Oct, 2012 19:10
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.