
#1
Mar409, 07:26 PM

P: 99

So now that I'm taking Electronics 1, the course right after Linear Circuits (DC/AC), the whole parallel/series schematic thing is kicking my ***. Pic 1 (idealdiode) is an example of just a tricky way of drawing something in parallel: namely, a voltage drop. With the (ideal) diode off = open circuit, we can say that Vo = the drop across the rightmost 1k Ohm resistor. There is a voltage division going on for sure, that's how I solved the problem, but knowing the current, would we get the same answer if we just did I*1k ohm?
When the (ideal) diode is on, it acts as a short (we can ignore the resistor). I have a hard time convincing myself that Vi = Vo when I tell myself that they are the same because the "voltage travels through the wire". Are we assuming Vi here to be connected to the same node as Vo, and also to ground? Is that why Vo = Vi, because they are both connected to the same 2 nodes? Again, I'm sure choosing a wiser set of words to express this would make it much easier to understand. In idealdiode2, with a <0 value for Vi, D1 is on = short, and D2 is off = open circuit. Here is where I have trouble. What would be a good way to express the voltage at the node above D2 and below the 1k ohm resistor? I have already solved these problems, with some good thought put along with the help of the solutions manual. I'm not asking for homework help, in other words; I don't want to get penalized. Basically, what I want to get out of this thread is a better way of explaining, and convincing myself, of the things explained above. 



#2
Mar509, 12:33 AM

Mentor
P: 39,600

I'd suggest thinking in terms of real diodes instead (after all, that's what you will use in the real world). First understand the diode equation for real diodes, and then reduce the Vf voltage to a very small number if you want to do an "ideal diode" circuit question.
So, what would your answers be to the above queries if the Vf were, say, 0.7V? 



#3
Mar1009, 07:49 AM

P: 13

Congratulations on taking Electronics 1! I am sure you will enjoy it but it will give many days of sore heads and head scratching. Just keep in mind that it will make sense in the end no matter how hard you are struggling.
To answer you first question. I think you are confusing yourself a bit with the description. Let us say that Vi is always a positive voltage. Therefore D1 will conduct and if it is an ideal diode it will become a short circuit. You have to accept the term short cicuit means that it is if there are no components there anymore. If you redraw the circuit it would be as if the first 1k resistor and D1 were not there so you would draw a line from Vi to Vo and have the other 1k resistor going down to ground. Joining the line from Vi to Vi means that Vi is at the same potential as Vo. This does not make these points also connect to ground. The other 1k resistor is still between Vi/Vo and ground. If you removed the diode from the circuit you would have a potential devider. this is a very important building block for electronic circuits. If you imagine that Vi was 10V. Because both of the resistors are the same value, the same amount of voltage will be dropped across them. As there are two resistors that means half the voltage is dropped across them ie 5V. that would make Vo 5V. If the first resistor was 8k and the second resistor 2k then it acts as a ratio. Therefore the two resistor in series add to 10k. Therefore the top resistor has a ration of 8k/10k. This means that 8/10 of the voltage will be dropped across it ie 8V (8/10 * Vi). This means that the second resistor has 2/10 of the voltage ie 2V. Therefore Vo would be 2V. I hope this hasn't confused you more and that it has explained things a bit better. http://www.powerups.co.uk 



#4
Mar1009, 08:50 AM

P: 99

General query: Parallel, series, voltage division, etc 



#5
Mar1009, 11:33 AM

P: 13

No you wouldn't ignore the diode.
In the first case the ideal diode is a short circuit. Current will always try to find the path in a circuit with the least resistance. The least resistance you can get in an ideal circuit is a short circuit so all of the current will flow through the short circuit path and none through the resistor therefore taking the resistor out of the circuit. Using ohms law will prove this but that is something you will have to work out yourself. If the diode was in series ie R1 then the diode then R2 then the diode would still be a short circuit if Vi is positive. Therefore you can still "ideally" draw the same circuit without the diode ie. R1 then R2 which once again gives you a potential divider circuit with half the voltage dropped across R1 and half across R2 making Vo=Vi/2. All of the theory above can be proved using ohms and Kirchoff's laws and I am sure that this will become more apparant as your course goes on. Do not get disheartened. Most of us struggled to get our heads around these problems when we started out in electronics. I found that some of the lecturers at my university did not explain things very well at all but others made things very easy to understand. If you get bogged down and can't understand something  take a break for a while and clear your head. It's amazing how things become clearer after a break! http://www.powerups.co.uk 



#6
Mar1009, 01:11 PM

P: 303

No they are not connected to the same node. There is a diode between vi and vo (obviously). Even though there is a diode between Vi and Vo, Vi still equals Vo because there is no energy lost in an ideal diode when it conducts. In this case the diode acts like a wire but by definition of a node, vi and vo is not a node. If you treat the diode as non ideal, then there would be a .7V drop across it. Just let the ideal simmer and marinate in your head. 


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