Backing to the thread topic.

I was trying to write a few more equations, to practice. I need to try to eliminate the issue when voltages are 'upside down'!

So, for instance, to know how much voltage is delivered to the load I wrote:

-V_z + V_R_Load - V_be = 0 V

V_R_Load = 8.2 V - 0.82 V = 7.38 V

 

ok so have u goting it

 

Yes, If the probes are attached to the same ends of the resistor, but you have now connected your meter backwards so whilst it might read minus something it is telling you positive something - in the old days it would also be telling you that your meter is now dead.

It's not the metter or the probes. Ok, they can also matter... But the question here is everything is in the same position but the resistor, which is flipped by 180º. That will shift the sign!

Edited;

 

Well I can't tell whether you are getting anywhere with this, so I am off now.

Don't worry about the rail being negative, it is at the bottom where negative should be, and any current flowing downhill is a positive current. Do not get confused by thinking it must be a negative current as it is coming from a negative rail - IT IS NOT - it is a positive going into the negative rail.

 

ok so have u goting it

I hope so!

Well I can't tell whether you are getting anywhere with this, so I am off now.

Don't worry about the rail being negative, it is at the bottom where negative should be, and any current flowing downhill is a positive current. Do not get confused by thinking it must be a negative current as it is coming from a negative rail - IT IS NOT - it is a positive going into the negative rail.

Ok, I'll keep that in mind!

 

and the voltage at the emitter is approx -8V2

 

Oops, -7V2

 

if it is any consolation I have met many qualified electricians who could not grasp the concept of a negative supply. They kept on on poking the black probe into the black wire and so measuring wrt to negative rail. Got into all sorts of tangles on telephone systems operating on -48V of course when the batteries started to go down all their measurements changed. 0v on the other hand is always 0V.

 

I got V_R_Load = 7.38V. I did the calcs at post #21.

 

I got V_R_Load = 7.38V. I did the calcs at post #21.

I think you got PLUS 7V38

 

Use the MOVE function to swap ends of the resistor in-place, and then resim. What was formerly a negative current will become a positive current, and vice-versa. Devices with multiple pins are defined with positive current flowing into the component.
In LTSpice, when you reverse a component, the current probes also reverses.

Notice the red arrow of the current probe: if you get a positive value, it means the current flows in the same direction of the arrow, if you get a negative value, the current flows in the other direction.

 

I think you got PLUS 7V38

Yes, positive value because I was sweeping the net loop clockwise, which would meet the voltage drop at R1 in the same direction of the sweep!

 

OK previously you apparently calculated the voltage at +7V38 - an absolute value. Absolutely (aka wrt to zero) the voltage at the emitter is -7V38.

Now you say a voltage ***drop*** is 7V38 - well a voltage drop implies downhill, negative. The voltage drop in RL is indeed 7V38 and it drops from 0V to -7V38 (wrt to zero).

 

Looking at R1 you might wish to say the voltage across R1 is 7V38 as the top is 7V38 more postive than the bottom, I might say the voltage across R1 is -7V38 as the bottom is 7V38 more negative than the top. This is not only true but avoids mentioning a positive voltage in a circuit where there can be no positive voltages. I probably should not call it a voltage drop.

 

Use the MOVE function to swap ends of the resistor in-place, and then resim. What was formerly a negative current will become a positive current, and vice-versa. Devices with multiple pins are defined with positive current flowing into the component.
In LTSpice, when you reverse a component, the current probes also reverses.

Notice the red arrow of the current probe: if you get a positive value, it means the current flows in the same direction of the arrow, if you get a negative value, the current flows in the other direction.

Yes, that's why I said that if the component is flipped, the sign is also swapped to the opposite polarity!

But I always try to not mess with that so that I can have a consistent reference! If I switch components orientations back and forth I'll easily loose track of my references and I'll mess up all the circuit analysis! Also LTSpice sometimes, by convenience or not, plots negative values without me asking for it! That also confuses me!

OK previously you apparently calculated the voltage at +7V38 - an absolute value. Absolutely (aka wrt to zero) the voltage at the emitter is -7V38.

Now you say a voltage ***drop*** is 7V38 - well a voltage drop implies downhill, negative. The voltage drop in RL is indeed 7V38 and it drops from 0V to -7V38 (wrt to zero).

Yeah, that can turn pretty *$&%# up to understand if we loose track of the words meanings and the circuit references which may differ from person to person!

[MOD NOTE: Please refrain from obscenities.]

 

Yeah, that can turn pretty *$&%# up to understand if we loose track of the words meanings and the circuit references which may differ from person to person!

I am not absolutely sure whether we are disagreeing or violently agreeing.

 

Yeah, that can turn pretty *$&%# up to understand if we loose track of the words meanings and the circuit references which may differ from person to person!

I am not absolutely sure whether we are disagreeing or violently agreeing.

I absolutely agree with you. And sometimes I get frustrated because different persons may have different wording to say the same thing and that may get pretty confusing, at least to me! (I don't want to be violent in any way! Sorry if I have!)

Ok, let me try to keep things insight in the thread. We are discussing two subjects at the same time.

Let me re group the values I already worked out:
I_R1= 8.48 mA
V_R_Load = -7.38 V (I'm using @grahamed method. I think it's more logic!)
V_R1 = 3 V

 

Ok, I'm re-doing this calcs because I was wrong about the Zener power dissipation. It's 500mW
So, using the problem data and not LTSpice values, I get:

P_z = V_z * I_z
0.5 V*A = 8 V * I_z
I_z_max = 62.5 mA

I_z_min = 10% of I_z_max = 6.25 mA

I_R1 = (V_z - V_in) / R1 = (-8 V - (-11 V) ) / 330 Ω = 9.1 mA

I_B + I_z = I_R1
I_B = 9.1 mA - 6.25 mA = 2.85 mA

Then I think I know how to answer the question which is (just to recall): "What is the voltage at the load?"

Write a net loop equation that includes the Zener, E-B junction and the load resistor.

That is what I'm doing as we speak!

Edited;

-V_z + V_out - V_BE = 0
V_out =8 V + 0.67 V = 8.67 V

Is this correct?

The other question was: "What is the max current that can be 'delivered' to the load???"

In other words this means what is the max current that can flow through the load that doesn't drops the current on the zener below the previously calculated I_z min!

This would be I_E for the I_z min = 6.25 mA

As we have I_B, we can evaluate I_E = I_B * (β + 1) = 2.85 mA * (50 + 1) = 145.35 mA

 

Well, it depends on how I place the resistor. If I place it just like I pick it up from the list, it will have one direction, though, current will be either positive or negative. But If I rotate the resistor by 180º, the current will have the opposite sign/direction! So, it's kinda hard to answer that. Also, sometimes, not sure why, maybe by convenience, LTSpice plots negative values without me asking for!

By convention (and I'm not aware of any simulators that don't follow it), current into a pin, any pin, is positive.

For a two terminal device, the current through the device is defined to be the current into pin 1.

That explains why the polarity of the current depends on the orientation of the part.

With most libraries there is no way to easily tell which pin is pin 1. That's why I go in and edit the library symbols to put a small tick or other small mark at pin 1.

The problem is if I, at the middle of the problem/calcs, miss consistency in analyzing the circuit. I'll end up with wrong calcs!

Which is why it is important to clearly define what you are doing and be methodical and consistent is setting up and in working your equations from start to finish. No more being sloppy!

 

PsySc0rpi0n said:
-V_z + V_out - V_BE = 0
V_out =8 V + 0.67 V = 8.67 V
it is error.
-Vz=Vout+V_BE
Vout=-Vz-V_BE
V_BE= -0.67
V_out=-8 V -(- 0.67)= -7.33V