Ok once again here for a little guidance after hours of pondering.

question states "For the circuit shown in Figure 3, Zener diodes have the following characteristics. Determine and Draw voltage characteristics of the circuit."

First things first im assuming the voltage characteristics is a graph of Vi vs Current in the series circuit.?

the Zener diode characteristics can be found in the attached image of the workings and also the circuit diagram. i drew the zener diode charecteristics for visual aid to help me try to understand whats going on.

so what i have done is began to test voltages random Vd voltages that satisfy the top Id and Vd relationship (found in the image attached). and then substituted the Id value into the corresponding equation for the next zener diode in the series circuit. No matter what Vd i use (satisfying the vd condition) in the first equation and then substitute the corresponding current into an equation for the second diode (because it is a series circuit current must be the same through all compnents. but no matter what initial assumed Vd1 value the second diode will always be closed?

sort of a bit confused at the moment ? am i using the right id relationships for the corresponding diodes and voltages?(see working)

I have been using the top/1st Id Vd relationship for the first zener diode as a postive voltage would just result in the diode being a standard forward bias diode? and i have been using the third relationship for the 2nd diode with a + vi input voltage (same as D1)

Thanks/ ideas/ what am i missing here?

 

hi alex,
Could you post an image of the original Fig #3, which shows the Zener characteristics.?

E

 

hi alex,
Could you post an image of the original Fig #3, which shows the Zener characteristics.?

E

This is all the information given. copied word for word. im thinking maybe theirs an issue with the current polarity through the diodes? do i have to make it negative through on of the diodes due to the diodes polarity?

 

hi alex,
The image shows ambiguous Vd and Vo symbols, where did the question come from.?
The io = 0 when -3v < Vo < 0.5v, suggests that the Vin test range is say -5V thru to say +5v.

E

 

hi alex,
The image shows ambiguous Vd and Vo symbols, where did the question come from.?
The io = 0 when -3v < Vo < 0.5v, suggests that the Vin test range is say -5V thru to say +5v.

E

Sorry this may be my bad hand writing. all V symbols are Vd in the Id characteristics of the diodes. so Id= (Vd-0.5)/15 when Vd>0.5v. Id=0 when -3v<Vd<0.5v. Id= (Vd+3)/10 when Vd<-3.

I really get the feeling that the lecturer just does not really care. I have no idea where he got the question but i believe he has re written it and made it difficult to understand for no purpose other then to inconvenience? it doesnt realy help learning when the question structure sucks. had a discussion on the phone with a fellow student today to try understand the question and an hour of brain storming got us no were.

How do you draw the voltage characteristics for a circuit when we are only given an input and no specified point to which we must asses the voltage.????

The whole voltage characteristics part has me baffled? I can find the current through a diode then solve if other diode is on or off and then find the voltage drop accross the resistor. sum the voltage drops the find a V in value and graph that against current?. then graph that for different Vd values?

 

hi alex,
With two zener diodes connected in opposing series, one zener can be considered a regular diode, depending upon the polarity of the applied Vin voltage.
I will look thru it again.
E

 

hi alex,
With two zener diodes connected in opposing series, one zener can be considered a regular diode, depending upon the polarity of the applied Vin voltage.
I will look thru it again.
E

Yea so if we look at the circuit from a positive voltage applied to the top buss. the voltage must be enough to have a drop of 0.5v across the first diode( acting as a normal diode) after that there must be enough potential to zen er breakdown the second diode ei( more then 3 volts of potential) and then solve the current through the resistor for its drop. the circuit is a series circuit so current is constant. Vi = Vd1 + Vd2 +Vresistor.

and graph accordingly
?

 

hi,
For checking assume one Z1 is 3V and Z2 =15.5v.
Does io = 0 apply to all 3 conditions.???
ie: conduction points, then you could calc the current thru the 50R outside these non conduction limits.
E

 

Ok so the lecturer got back to me and said take V out as the voltage across the resistor? therefor yes you can draw the voltage characteristics of V in vs V out.

Without that information given in the question does this not make the question ambiguous?

 

Morning alex,
I still believe the original image showing the io=0 conditions has been either presented or copied incorrectly.
It does not make any sense to have all the variables as Vd.???

E