How Does a T-Type LCR Filter Affect AC Ripple in DC Output?

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Discussion Overview

The discussion revolves around the analysis of a T-type low pass filter used in conjunction with a rectifier to reduce AC ripple in the DC output. Participants explore the circuit's behavior under a 50 Hz alternating supply, focusing on phasor diagrams, impedance calculations, and the relationship between input and output voltages.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks help with calculating the current in the capacitor and understanding the phasor diagram for the T-type filter.
  • Another participant prompts for the phasor impedances of the circuit components and how to combine them in the phasor diagram.
  • A participant calculates the inductive reactance and asserts that it is much greater than the load resistance, suggesting the circuit can be treated as reactive.
  • Confusion arises regarding the phasor diagram, particularly in placing voltage and current phasors correctly.
  • One participant describes their progress in drawing phasor diagrams and calculating various voltages and currents, seeking validation of their approach.
  • Concerns are raised about the high input voltage required to achieve a specified output voltage, prompting questions about potential errors in calculations.
  • Another participant questions whether the rectifier is half-wave or full-wave and discusses the implications of the T-filter on AC ripple and DC output.
  • Clarifications are made regarding the assumptions needed for idealized calculations, particularly concerning the average DC value and the harmonic content of the waveform.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the calculations and assumptions involved. There is no consensus on the correct approach to the problem, and multiple viewpoints on the implications of the T-filter and rectifier type remain unresolved.

Contextual Notes

Participants note the lack of information regarding the type of rectifier used and the source impedance, which may affect the analysis. The discussion also highlights the complexity of applying phasor analysis to a waveform with harmonic content.

copresh1
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hi
if anyone can help me solve this problem (T- type low pass filter)it will be appreciated. i have managed to find the current in the capacitor and form there on i don know what to do.

thanks

circuit of a resistive load is fed from a rectifier, via
a filter. The filter is designed to remove unwanted a.c. that might be
present in the output of the rectifier.

To test the performance of the filter, a 50 Hz alternating supply was
connected directly to its input. The resulting
a.c. voltage appearing across the load was then measured.

(a) Sketch a phasor diagram for a circuit
(Hint: begin at the output and work back towards the input.)

(b) Estimate the magnitude of the a.c. input voltage required to give a
voltage of 1 V across RL.

(Hint: show that | XL2 | >> RL and therefore that the entire circuit can
be regarded as reactive.)
both L=1henry
C=100microfarads
RL=10 ohms
 
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Welcome to PF. Homework questions should generally be posted in the Homework forums, so this thread may get moved by a moderator. In the mean time, when we give help on homework problems, we try to ask some questions to guide you a little. It helps us to help you if you show what work you are able to do so far.

So, what are the phasor impedances of the inductors, capacitor and resistor? What would the phasor diagram look like for part (a)? How do the series and parallel combinations of the phasor impedances get combined? Can you show the inequality in the hint?
 
since i the values of RL and VL are given i have calculated IL
IL=0.1A
using IL i have found voltage of second capacitor ieVX2=XL * IL=314.2V
therefore VX2 is >> RL hence the reactance can be considered wholly reactive.

as for the phasors diagrams, i am totally confused bcoz i have to put voltage and current on the same phasor diagram and do not know where to start
 
"therefore VX2 is >> RL"

I'm not sure what that means. They want you to show that the reactive impedance of the inductor is much larger than the load resistance. Both sides of the inequality should have units of Ohms. What is the expression for the phasor impedance of an inductor? (include the phase angle in all of your phasor expressions)
 
i made a typing error. here is how i have progressed on the problem.letme knoe if i am wrong.

i have found XL>>RL, XL=314.2 ohms & RL=10 hms
... next i have drawn a phasor diagram with IL as the ref phasor, VL is inphase with IL and VX2 is at 90deg(inductive).VX2=31.4v
...next i have found that VX2=VX3(by adding their phasor sums, ie their inphase and quadrature voltages)
...next i ahve drawn the phasor diag for the capacitor with VX3 as ref phasor. i have then calculated Ic,where Ic=VX3/Xc.
... i have added the above phasor diagram to the forst diagram wherby i get Ic on the LHS.
... next i have drawn the phasor diag for the the first inductor and added this diag to the second diag, whereby VX1 is in the downward direction.
...to find Is, i have resolved IL &Ic into its components and i get Is=0.9A
using this Is i have found VX1=282.74
...next to find Vs i have resolved VX1 & VX2 into is components.
therefore Vs=VX1-VX2=250.33

so... is my answer right,or have i messed it up?
 
It looks like you are doing the right things. What did you get for the input waveform amplitude to get 1V across the load?
 
output voltage=1v ,was given as part of the question
 
RL = 10 Ohms and it takes 250Vac input to get 1Vdc output? If I'm reading your answer correctly, it seems kind of high...
 
hmmm...so where do you think i might have made a mistake?is the procedure i followed correct?
 
  • #10
Is the input rectifier half wave or full wave? Do they give you any source impedance for the voltage source? After the rectifier, the circuit is two inductors in series with a cap to ground between them (T-filter), and then the resistive load to ground, right?

The 50Hz input sine wave gets converted to a sharp 100Hz waveform by a full-wave bridge, so to a first approximation you can calculate how much that 100Hz waveform is reactively attenuated by the 1H & 100uF L-C filter effect. The output L-R should give you about a 10:1 divider on the AC ripple, but not on the DC Vout. And the input L-C will mostly smooth out the AC ripple as well.

So maybe as an experiment, just put in a 6Vpp 50Hz sine wave into the full-wave bridge, that gives you a 1.6Vpp output at 100Hz (half sines above ground, assuming about 0.7V for each diode in the rectifier). What happens to that waveform as it goes through the T filter? The average DC doesn't change much, does it?


EDIT -- I'm assuming that part (b) is asking about a 1Vdc across RL.
 
  • #11
the question doesn't say anything about full wave or half wave nor is the impedance of the voltage source given.
 
  • #12
Sorry if I'm not being of much help here, I'm not used to working with idealized questions very much. First of all, phasors only apply to single-frequency problems. A full-wave rectified sine wave (I'll assume it's full-wave) has a lot of harmonic content, so phasors generally would not apply. However, assume that the question is looking for phasors at 100Hz, and I guess you can draw phasor diagrams for the filtering that goes on with the T-filter into the load resistor.

But at least question -b- to me does not involve phasors at all, since it is asking about the resulting DC value. The T-filter only shapes the ripple, it does not directly affect the final DC value, IMO. The average DC value going into a T-filter equals the average DC value coming out (neglecting the DC R of the inductors), so what is the average value of a full-wave rectified sinusoid? And since this seems to be an idealized question, you can probably assume ideal diodes with zero forward voltage drop, as long as you state that as an assumption.

So what is the average value of a half-cycle of a sinusoid?
 
Last edited:

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