I'm still not comfortable will the circuits with negative voltages. I need to go through this again!

 

I'm still not comfortable will the circuits with negative voltages. I need to go through this again!

It still has to follow Kirchoffs voltage Law around the loop, the sum of the voltage drops equaling zero . Vz + Vbe + Vload = 0. -8.2 + 0.7 + 7.5 = 0

Attached are the meter readings following KVL. Note the meter polarity:

 

Ok, guys...

Back to this problem...

I want to start this problem all over again! I'm not sure if it wouldn't be better to start a new thread, as this one is already a bit long and confusing!

I can start a new thread entitled "Problem 2 - take 2" or so!

 

Ok, I'm going to restart this same problem in the same thread!
Hope this won't create any confusion!

I'm going to try to write the equation for the V_RLoad, assuming the zener is ON (working as zener) and also the PNP transistor is also ON (Vbe <= -0.67 V or Veb >= 0.67 V).

So, as Vbe = -Veb, I have 2 options and I'm not sure, but I think I'm doing something wrong in one of the options!

-Vz + V_RL - Vbe = 0 V

or

-Vz + V_RL + Veb = 0 V

If I'm assuming that the PNP transistor is ON, then Vbe = -0.67 V or Veb = - Vbe = - (-0.67 V) = 0.67 V, right?

So, both those equations MUST return the same, right?

-8 V + V_RL - (-0.67 V) = 0 V

or

-8 V + V_RL + 0.67 V = 0
V_RL = 7.33 V

I think this is correct, no? Please, check attached image!

 

Happy New Year for everyone!

After these small period of holidays and festivities days, I'm back... My exam will be 01/02/2016 so I need to keep going with practising.

Ok, I have a question about my last post!
I'm starting again this problem and I have calculated the load voltage drop as being 7.33V. This is a positive value because I'm considering the voltage drop and not the voltage at the Load wrt to gnd, right? And also because I was using a net loop equation that, by chance, I considered the sweep direction, the same as the voltage drop across the resistor which is from the higher potential to the lower potential.

This is a bit tricky because it is easy to loose track of directions and signs to use when it comes to use that value of +7.33 V in other equations, for instance!

 

-Vz-Vbe-Vrl=0 -->
Vrl=-Vz-(-0.67)=-7.33V

 

Do the same calculation for the positive regulator. The signs are the same.
V(z)-Vbe-Vrl=0
This equation is the same for all cases. Only the signs are replaced. It is a sign of V (z) and the sign of Vbe.

 

Do the same calculation for the positive regulator. The signs are the same.
V(z)-Vbe-Vrl=0
This equation is the same for all cases. Only the signs are replaced. It is a sign of V (z) and the sign of Vbe.

Yeah finally I got the whole picture after a friend of mine in IRC has helped me!

I'm already solving a new problem!

Thanks to all!