Ive just begun to begin to learn about diodes and their applications at school, and my question should be very simple to answer...Basically i cant figure out my professor's answer for the question (ive attached a picture)...
i couldnt understand his explanation either when i inquired, perhaps someone here can help me.

Based on my redrawing...V out should be 20v-0.7v=19.3...
However my teachers soloution was 12v-.7v=11.3v

Why didnt my professor factor both sources befor answering?
Any help is greatly apreciated...

 

billion_boi@hotmail.com,

Ive just begun to begin to learn about diodes and their applications at school, and my question should be very simple to answer...Basically i cant figure out my professor's answer for the question (ive attached a picture)...
i couldnt understand his explanation either when i inquired, perhaps someone here can help me.

Based on my redrawing...V out should be 20v-0.7v=19.3...
However my teachers soloution was 12v-.7v=11.3v

Why didnt my professor factor both sources befor answering?
Any help is greatly apreciated...

Because the bottom voltage is irrelevant. Anyway, you will be happy to know your teacher is correct. You show 8 volts feeding the resistor, but the voltage can be any value that does not starve the diode or burn it out. The voltage will still be 12 volts minus the voltage across the diode (approximately 0.7 volts for a silicon diode). Different voltages supplied to the resistor will change the current through the diode, but as long as the diode is conducting, the voltage will still be 12-0.7 = 11.3 volts.

Ratch

 

Ratch, from what I can understand in your response it's that this circuit is linear...it is not as I had redrawn it where I had assumed the sources were actually in series with one another...So now obviously since this diode operates in forward bias only the 12V source's current will get past it...
Am i correct?

 

Who did make the redrawn, you or the professor?

If you measure the voltage across the 1K resistor then you will measure 19.3V.

If you measure the voltage between Vo and ground (0V) then you will measure 11.3V.

 

When calculating or measuring a voltage, it is important to keep in mind that such a measurment must have a reference. In this circuit, the reference point or ground is assumed to be at the junction where the negative terminals of both voltage sources are connected.

As ratch and your professor have stated, if you place the ground probe of a voltmeter at the reference point in the circuit and the measurement probe on the cathode of the diode, you should get a reading of 12v minus the forward voltage drop of the diode. This will be true even if the -8V voltage source is replaced with a voltage source of a different voltage as long as the voltage source is of a value sufficient to forward bias the diode.

hgmjr

 

Oh man, i think im more confused than ever... mik3, the redrawn circuit was by me, and you seam to imply that the Vsources are in series...So 19.3v across the resistor makes sense to me

hgmjr, ive redrawn the circuit and added a ground refrence... but if you take a look at my drawing can you see why the Vout looks to me like 20V - Forward Vdrop of Diode(0.7)...?

 

Split the 20V battery into a 12V and an 8V battery in series and measure the voltage between the connection of the batteries (connection between the batteries) and Vo. The connection between the batteries is your ground reference.

 

Oh man, i think im more confused than ever... mik3, the redrawn circuit was by me, and you seam to imply that the Vsources are in series...So 19.3v across the resistor makes sense to me

hgmjr, ive redrawn the circuit and added a ground refrence... but if you take a look at my drawing can you see why the Vout looks to me like 20V - Forward Vdrop of Diode(0.7)...?

Take heart, things are darkest just before the dawn.

hgmjr

 

Have your original circuit. And think that you have a voltmeter across you 12V supply and your Vo . Your voltmeter will measure 12V -0.7V voltage. Do not redraw the circuit since it was already the simplest circuit it could be.

 

All you needed to do is add the power supplies and think like hmgjr about where the common node was between the supplies.

Consult the attached diagram ...

 

Here are some statements about the circuit.

a) There is no zero volt ground in this circuit, so we won't talk about it.

b) You are told that the bottom of the 1k resistor is at -8V. Don't argue with that, because it says so.

c) You are told that the voltage at the top of the diode is +12V. Again, no point arguing, because it's written right there, black on white.

d) You know that a diode which is conducting will have 0.7V across it. What your professor failed to explain is what that actually means - here it comes:
One end of that diode (the anode end) is 0.7V higher in voltage than the bottom.

e) In other words, the bottom end of your diode is 0.7V lower in voltage than the top. We don't care where 0V is. We don't need to know. We know that the bottom of the diode is 0.7V lower than the top. What's the top voltage? 12V. What's the voltage Vo at the bottom of the diode? 0.7V lower, = 11.3V.

f) If the top of the diode was 2850V, the the bottom of the diode (Vo) would still be 0.7V less, at 2849.3V

g) If the negative supply at the bottom of the resistor was negative 8 gazillion volts, we still know that the diode's cathode (Vo) is 0.7V lower in voltage than it's anode. Whatever the anode voltage is, subtract 0.7V and you have the cathode voltage.

h) How much higher is the top of the resistor (Vo) than its bottom end? Vo is 11.3V, we worked that out already. The bottom of the resistor is -8V because it says so on the diagram. We can say then that the top end of the resistor is 11.3 - (-8) = 19.3V higher than the bottom.

i) What we mean when we say there is 19.3V across the resistor is this: One end of the resistor is 19.3V different from the other end. This is what a voltmeter will read if you connect it across the resistor. A voltmeter doesn't tell you what the voltage is at some point is - it tells you what the voltage difference is between two points in a circuit.

j) If we shifted all the voltages up by 1000V: Voltage at the top is 1012V. Voltage at the bottom is 992V. There would STILL be 0.7V across the diode, and there would STILL be 19.3V across the resistor. But because we know there is 0.7V across the diode, Vo would now be 1012 - 0.7 = 1011.3V. Yes, Vo shifted up by 100V too.

k) This illustrates very nicely the distinction between the phrase "the voltage AT point X is blah" and the phrase "the voltage ACROSS component Y is blah".