Morning...

I need help urgently...

Please see the attached circuit. My teacher is asking to calculate the efficiency of the regulator for 3 different R_loads. I'm not sure where to measure values to calculate efficiency.

I need to calculate (P_out/P_in)*100% = ((V_out*I_out)/(V_in*I_in))*100% but I'm not sure where to measure V_in and I_in... Can you guys help?

 

Morning,
Did you understand the explanation in my last post.?

calculate the efficiency of the regulator

The teacher is asking about the 'regulator', not the overall efficiency of the 'circuit'.

So where do you think the power 'losses' [wasted as heat, etc] are in the regulator circuit.?
This should give you a hint on how to calculate the regulator efficiency.

E

EDIT:
I would suggest that for LTS sim, you initially fix the value of R_load to say 560R.

Use the .meas statement to calculate the Power dissipation[V * I ] in the regulator components.

 

Morning,
Did you understand the explanation in my last post.?



The teacher is asking about the 'regulator', not the overall efficiency of the 'circuit'.

So where do you think the power 'losses' [wasted as heat, etc] are in the regulator circuit.?
This should give you a hint on how to calculate the regulator efficiency.

E

I didn't see that post I guess...

Well, I think there will be losses mainly in resistors. So, maybe in R_s and R_load!

And by the regulator, I understand it will probably be after the rectifier, right? From V_Rs to the right???

Edited;

If my thoughts are correct:

P_in = V_Rs*I_rs
and
P_out=V_zener*I_R_load

Is this correct?


Edited1;
This is what I get from LTSpice for currents in regulator components!

 

Try my suggestion for fixing R_load at 560R,,,, what is the Power dissipation in the R_load. [ this is the 'useful' power ].
Lets take it step by step.

Post your answer.
E

 

Try my suggestion for fixing R_load at 560R,,,, what is the Power dissipation in the R_load. [ this is the 'useful' power ].
Lets take it step by step.

Post your answer.
E

~56mW

 

OK.
You know the Vavg of V_Rs and the value of Iavg thru R_s [from using the LTS sim] , so what is the power dissipation in R_s.?

 

OK.
You know the Vavg of V_Rs and the value of Iavg thru R_s [from using the LTS sim] , so what is the power dissipation in R_s.?

It will be 6.9244*38.231=256mW

 

Is there really 6.9244Volts across R_s.??? [check with LTS and the thermometer cursor for the actual power]

 

Is there really 6.9244Volts across R_s.??? [check with LTS and the thermometer cursor for the actual power]

Tat was one of my questions... V_Rs is regarding GND or V_out...

If is regarding V_out I need to drag & drop the LTSPice pen to plot that voltage drop!
Just a sec!

Edited,

Voltage across Rs is 1.491.
Current across Rs is 38.231mA

Power dissipation in Rs would be 1.491*38.231mA = 57mW (avg)
LTSpice says 62mW (avg).

This would be P_in, right?

 

Ok, I have calculated P_out and P_in but I think results are not OK, because the efficiency in an half-wave rectified filter is lower than in a full wave rectified filter, right? It is giving me around 92% of efficiency...


But shouldn't be this value much lower?

And it's even worse if I use the 330Ω resistor! It's abou 154% efficiency! Not sounding right!

 

Hi,
If I understand correctly, the teachers said the regulator efficiency, not the overall efficiency of the circuit.??
So why would you consider the rectification.?
E

 

Hi,
If I understand correctly, the teachers said the regulator efficiency, not the overall efficiency of the circuit.??
So why would you consider the rectification.?
E

I think I'm not considering the rectification... If I understand "rectification", is the diode D1 and the capacitor C1, right?

My last calcs were somehow wrong.

But the corrected ones are not better!!



I've made the calcs for 330 R_load just because that's what is asked to us! But with 560, probably the results will be of the same kind!

 

Check the last part of your equation, [numerator and denominator ], how can the efficiency be greater than 100%.?

The results for 560R and 330R will not be the same, do a check calculation, using 330R.

Also Vout should be 5.6V when doing calculations.

 

Hi,
I have simplified the circuit, it should be easier for you to understand.
As you can appreciate I will not give you the answer.

Consider this.
You have a known input Power1 = Vin * Iin.
The usable output Power2 is Vz * Iload.
The 'wasted power' is due to the 39R and the zener current

Peff ratio = Power2/Power1 , is always less than 1

E
EDIT.
simplified the plot

 

Yes I understand that Peff can't be grater than 1 but that's what calcs said!

P_out = V_Zener*I_load = 5.6*16.425mA = 91.98mA

Why you say V_in is 7.1V??? That is the same as I told in post #27 which you said that was not correct in post #28.
Yesterday I said that V across Rs was 6.9244V and you found it weird. I don't understand where to measure V_in to calculate P_in.

 

Why you say V_in is 7.1V??? That is the same as I told in post #27 which you said that was not correct in post #28

No I did not say that.

I asked you, so what is the power dissipation in R_s.?

You replied, It will be 6.9244*38.231=256mW
So posted, asking , Is there really 6.9244Volts across R_s.???, which is something entirely different.!

I think it would be best if some other member/s helped you with your questions.

 

No I did not say that.

I asked you, so what is the power dissipation in R_s.?

You replied, It will be 6.9244*38.231=256mW
So posted, asking , Is there really 6.9244Volts across R_s.???, which is something entirely different.!

I think it would be best if some other member/s helped you with your questions.

I'm sorry if I was harsh somehow. I need to apologize. I'm really sorry. I think I misunderstood you.

I've learned quite a lot with you and I would love to learn more...

In the meantime I spoke to my teacher and he helped me out and I think I managed to find some acceptable results.

Hope I still can get help from you...

Once more I apologize.
Thanks
Psy

 

hi Psy,
Life is too short to bear grudges, let us move on, I enjoy helping anyone.

My intention in breaking your problem down in to simple steps, was to take each component of your circuit and measure the power dissipated in that component.
In that way you would understand which components were causing the 'losses' of Input power, also the 'usable' power in the load.
From those two values you could calculate the efficiency of the regulator.

Knowing the power 'wasted' as heat dissipation in any regulator is important.

As I understand the group policy of this Forum we are not to give the final solution to any Home work questions, only Hints or Clues to help you understand.
Sometimes it is difficult in posting a reply and wording it in such a way as not to give the answer but to help you understand.

Eric

 

hi Psy,
Life is too short to bear grudges, let us move on, I enjoy helping anyone.

My intention in breaking your problem down in to simple steps, was to take each component of your circuit and measure the power dissipated in that component.
In that way you would understand which components were causing the 'losses' of Input power, also the 'usable' power in the load.
From those two values you could calculate the efficiency of the regulator.

Knowing the power 'wasted' as heat dissipation in any regulator is important.

As I understand the group policy of this Forum we are not to give the final solution to any Home work questions, only Hints or Clues to help you understand.
Sometimes it is difficult in posting a reply and wording it in such a way as not to give the answer but to help you understand.

Eric

I appreciate that...

But I was in a rush to send the task completed to our teacher within the deadline or we would have a penalty due to send the task off of the deadline. And I really was trying to do what you told me but calcs keep going crazy above 100% when I expected low values under 30%.

After I speak to my teacher he agreed to calculate Power_in from the regulator to the right side of the circuit. But probably his intent was to us to calculate the power at the very beginning of the whole circuit because when I said him that the question was talking about the "regulator" he said "ok, you're right. So calculate power_in as being in the beginning of the regulator", which would be at R_s.
After a few calcs, finally i got values around 34% for 330Ω (my colleagues, last year, achieved 37% for a full-wave rectifier), which makes sense because I've read in the internet that full-wave provides better efficiency than half-wave ones.

Thanks
Psy

 

It is sometimes difficult to know what the teacher is exactly asking you to show in your solutions,.
You were correct to point out to him that he had asked for the 'regulator' and not the overall efficiency of the half wave circuit.

Your 34% for 330R, have you thought of a way to improve the efficiency? by changing a component, assume Iz= 10mA.??

E