REALLY basic question on circuit analysis. - Page 1
Forum section: electronics / beginners

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 07:42:59 pm
content: 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 ID, 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:ID=Vth(R1/(R1+R2))But the formula only hasID=Vth/(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!

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 07:46:22 pm
content: Also... If it is just the basic ohms law, doesn't the diode have any resistance?

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 07:51:49 pm
content: It's simple Ohm's law. The current IDthrough the diode is the current in the circuit, given byID= VTh/ (RTh+ RL)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.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 08:21:07 pm
content: 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 RL. I am positive that I need to use the formula ID=IseVf/.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.

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 08:32:18 pm
content: Did you miss a "-1"? That is to say,ID= IS(eVf/0.026- 1)

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 08:35:34 pm
content: (OK, I guess the "-1" might not really matter in many cases.)

author: Unknown
author_rank: Unknown
timestamp: October 20, 2012, 08:35:46 pm
content: Vfrepresents the voltage across the diode. Why would this have to be equal to VThin your bottom circuit? What about the voltage across the other components in that loop?IanB: Assuming Vfis much larger than VT, the -1 term can be neglected. This is often done in introductory courses to simplify the math.

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 08:42:18 pm
content: 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 Vs, R1and R2given, so the left half of the circuit is fully defined and can be made into a Thevenin equivalent.On the right, we have ISgiven, 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 RL? Without this information, I don't see how to calculate a value for RL?

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 08:44:07 pm
content: Thanks alm, I see what you are saying!... but without knowing the current through RthI can't calculate the voltage drop through it. Now I have a chicken or the egg problem.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 08:52:54 pm
content: Well, after realizing that Vfis not VthI 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!

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 09:17:04 pm
content: Question... I was looking at my notes, and now I am not sure about Vf. Is that the voltage drop across the diode (which is usually .7V), or the voltage going through the diode?

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 09:20:50 pm
content: I think the .7 is called the Vcut in

author: Unknown
author_rank: Unknown
timestamp: October 20, 2012, 09:26:32 pm
content: 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 IS, since 1.5 uA seems awfully high for a saturation current.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 09:34:27 pm
content: Because I am a little confused as to what Vfis. Because on one of the examples, for Shockley's equation ID=IseVf/.026, the Vfis actually the voltage drop across the diode. Which we were taught to just assume it is 0.7V. But I was thinking that the Vfwas actually the voltage coming into the diode before the voltage saturation. Apparently I was wrong. Can you confirm that for Vfon that formula they are actually talking about the voltage saturation and not the voltage applied to the diode. Thanks!

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 09:36:25 pm
content: You are absolutely right about the Isthere is a correction that the teacher put out and the value is actually 25.69X10-15A.

author: Unknown
author_rank: Unknown
timestamp: October 20, 2012, 10:01:35 pm
content: 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.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 10:07:47 pm
content: 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 (ID) through the resistor be the same as that of the load resistance (Ith=ID=IL)? Because in circuits I, resistors in series had the same current running through them.

author: Unknown
author_rank: Unknown
timestamp: October 20, 2012, 10:17:00 pm
content: 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).

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 10:26:51 pm
content: 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 IDwe would know the current through RLand could find the voltage drop across the diode, which would allow the problem to be solved in a meaningful way.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 20, 2012, 10:39:30 pm
content: alm: We studied Kirchoff's law in Circuits I.IanB: Shouldn't I be able to calculate IDby using the formula ID=(25.69x10-15A)e.7V/.026?

author: Unknown
author_rank: Unknown
timestamp: October 20, 2012, 10:52:22 pm
content: Yes, assuming Vf= 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 Vfor temperature will cause huge variations in ID. 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 Vf= 0.7 V, calculate the voltage across RThand RL, 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 ID, like 1 mA, and go from there.

author: IanB
author_rank: Super Contributor
timestamp: October 20, 2012, 11:05:01 pm
content: Quote from: zorthgo on October 20, 2012, 10:39:30 pmOn paper, yes. In reality, no. Consider:Vf= 0.68 V, ID= 5.9 mAVf= 0.72 V, ID= 27 mAAnd 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?

author: zorthgo
author_rank: Regular Contributor
timestamp: October 21, 2012, 12:43:13 am
content: Good point!... Not reliable at all.

author: M0BSW
author_rank: Frequent Contributor
timestamp: October 21, 2012, 11:46:04 am
content: Quote from: alm on October 20, 2012, 10:17:00 pm,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 ".

author: zorthgo
author_rank: Regular Contributor
timestamp: October 21, 2012, 12:07:39 pm
content: 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!

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: October 21, 2012, 12:57:55 pm
content: 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) + Cie 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.

author: M0BSW
author_rank: Frequent Contributor
timestamp: October 21, 2012, 04:21:45 pm
content: Quote from: HackedFridgeMagnet on October 21, 2012, 12:57:55 pmCricky

author: M0BSW
author_rank: Frequent Contributor
timestamp: October 21, 2012, 04:26:15 pm
content: Quote from: zorthgo on October 21, 2012, 12:07:39 pmThank 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.

author: IanB
author_rank: Super Contributor
timestamp: October 21, 2012, 04:27:44 pm
content: Quote from: HackedFridgeMagnet on October 21, 2012, 12:57:55 pmAs 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.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 21, 2012, 07:10:45 pm
content: 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 Vrms. But apparently it isn't because the formula that I have to Vrmsis Vrms=Vm/2, but in order to find Vmon part a instead of having Vm=2*Vrms=120V*2 they have Vm=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!

author: Unknown
author_rank: Unknown
timestamp: October 21, 2012, 07:43:55 pm
content: 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 Vpeak, which is half of Vpeak-to-peak. Vrmsis 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 Vrmsinthis Wikipedia entry). If you go through the calculus, you'll find that the result is that Vrms= 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?

author: IanB
author_rank: Super Contributor
timestamp: October 21, 2012, 09:39:02 pm
content: Quote from: zorthgo on October 21, 2012, 07:10:45 pmThe 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.

author: zorthgo
author_rank: Regular Contributor
timestamp: October 22, 2012, 02:54:37 am
content: Thanks for the explanation guys! You guys rock! I understand a little better!