Trying to analyze a half wave rectifier with inductor and DC source

In summary, the author is trying to explain how the equations work for a half wave rectifier with an inductor and DC source load, but does not understand how they got from the first equation to the second equation. The final equation seems correct, but the current will not be sinusoidal.
  • #1
Jason06841
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TL;DR Summary
I am trying to analyze a half wave rectifier with an inductor and DC source load.
I am trying to analyze a half wave rectifier with an inductor and DC source load. I understand the circuit but I guess I do not get the math. I am reading a book and this is the circuit and equations they came up with. I understand how they got from the first equation to the second equation but I do not understand their algebra to get from the second equation to the third. Shouldn't it be w/L instead of 1/wL? How do they get to the third equation?

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  • #2
Looks as if the final equation is correct but I do not understand Eq 2.
It also looks to me that the current will not be sinusoidal, as the DC voltage biases the diode beyond cut off for part of the cycle.
 
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  • #3
Jason06841 said:
TL;DR Summary: I am trying to analyze a half wave rectifier with an inductor and DC source load.

I am reading a book and this is the circuit and equations they came up with.
Give us a link and page number please.
 
  • #5
tech99 said:
Looks as if the final equation is correct but I do not understand Eq 2.
It also looks to me that the current will not be sinusoidal, as the DC voltage biases the diode beyond cut off for part of the cycle.
These are just the equations for when the diode is forward biased. Starting from Vdc+ to partly into the negative half cycle due to the energy stored in L.

He does a few things there from equation 1 to equation 2. First he redefines the domain from t to wt. Because that would make the lower differential d(wt). But w is constant (in this case) so he is able to pull it out and multiply it by dt and continue to solve the equation in terms of t. However, that puts the w in the denominator so when you rearrange you get w/L instead of what he got which was 1/wL.
 
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  • #6
Jason06841 said:
I am reading a book and this is the circuit and equations they came up with.
Some of us want to know the title, author and ISBN, so we can peruse that chapter.

It seems;
To get from eqn(1) to eqn(2), multiply the rhs by w/w = 1;
Then subtract Vdc from both sides of eqn(2).
Knowing; ZL = w⋅L ;
Eqn(3) is Ohms law, with i = v / ZL.
 
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  • #7
Baluncore said:
Some of us want to know the title, author and ISBN, so we can peruse that chapter.
Page 5 of the pdf he linked to...
 
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  • #8
Baluncore said:
Some of us want to know the title, author and ISBN, so we can peruse that chapter.

It seems;
To get from eqn(1) to eqn(2), multiply the rhs by w/w = 1;
Then subtract Vdc from both sides of eqn(2).
Knowing; ZL = w⋅L ;
Eqn(3) is Ohms law, with i = v / ZL.
The units in equation 2 and 3 do not balance.

Since i(t) and i(wt) are both the same dimension (elec. current) You don't put an omega inside the argument of function i, then just divide by omega to "cancel out". So yes that's correct to get the derivative to work out, but you need to divide the other side of the equation by omega as well.
 
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  • #9
Baluncore said:
Some of us want to know the title, author and ISBN, so we can peruse that chapter.

It seems;
To get from eqn(1) to eqn(2), multiply the rhs by w/w = 1;
Then subtract Vdc from both sides of eqn(2).
Knowing; ZL = w⋅L ;
Eqn(3) is Ohms law, with i = v / ZL.
Power Electronics, Daniel Hart, ISBN 978-0-07-338067-4
 
  • #10
scottdave said:
The units in equation 2 and 3 do not balance.

Since i(t) and i(wt) are both the same dimension (elec. current) You don't put an omega inside the argument of function i, then just divide by omega to "cancel out". So yes that's correct to get the derivative to work out, but you need to divide the other side of the equation by omega as well.
I do not think he is putting that w in the denominator simply to "cancel out". When he replaces t with wt he ends up with di(wt)/d(wt). Because w is a constant, it becomes di(wt)/wdt but because w is in the denominator on the rhs of equation (2) after rearranging, it should turn up in the numerator in equation (3) but for him it ends up in the denominator.
 
  • #11
Jason06841 said:
I do not think he is putting that w in the denominator simply to "cancel out". When he replaces t with wt he ends up with di(wt)/d(wt). Because w is a constant, it becomes di(wt)/wdt but because w is in the denominator on the rhs of equation (2) after rearranging, it should turn up in the numerator in equation (3) but for him it ends up in the denominator.
Yes, definitely a typo.
 
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  • #12
Turns out if I do not substitute wt and just integrate w.r.t. time it all comes out when I substitute wt at the end after I get my final result. Thanks everyone for your help. If anyone knows the proper way he should have done that substitution please post a reply. I have added a LTSpice plot and circuit if anyone is interested. The green is the voltage at the top of the inductor the red is the current through the inductor.

ltspice.png
 
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  • #13
This (in the book you showed) is the kind of Algebra one uses when trying to obtain a desired result. Avoid the book.
 
  • #14
Holy crap! I just clicked into this thread randomly and read the last few posts, and I, um, need to close it temporarily for, um, time for Mentor review. Lordy! :oops:
 
  • #15
Okay, after cleaning up an off-topic dangerous discussion, this thread is now reopened. Have a nice day.
 
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FAQ: Trying to analyze a half wave rectifier with inductor and DC source

What is the purpose of a half wave rectifier in a circuit?

A half wave rectifier is used to convert alternating current (AC) into direct current (DC). It allows current to pass through during one half of the AC cycle (either positive or negative), effectively blocking the other half, thus producing a pulsating DC output.

How does adding an inductor affect the performance of a half wave rectifier?

Adding an inductor to a half wave rectifier can improve the performance by smoothing the pulsating DC output. The inductor resists changes in current, which helps to reduce the ripple and produce a more stable DC voltage. This is because the inductor stores energy when the current increases and releases it when the current decreases.

What role does the DC source play in a half wave rectifier circuit with an inductor?

The DC source in a half wave rectifier circuit with an inductor provides a reference voltage or biasing. It can be used to maintain a certain level of voltage in the circuit, ensuring that the rectified output remains within a desired range. This can be particularly important in applications where a stable DC voltage is critical.

How can the efficiency of a half wave rectifier with an inductor be calculated?

The efficiency of a half wave rectifier with an inductor can be calculated by comparing the DC power output to the AC power input. The formula for efficiency (η) is η = (P_DC / P_AC) * 100%, where P_DC is the power delivered to the load and P_AC is the power supplied by the AC source. Factors such as the inductor's resistance and losses should also be considered in the calculation.

What are the common issues encountered when analyzing a half wave rectifier with an inductor and DC source?

Common issues include high ripple voltage, inductor saturation, and voltage drops across the inductor and diode. Ripple voltage can be mitigated by choosing an appropriate inductor value. Inductor saturation can be avoided by ensuring the inductor's current rating is sufficient for the application. Voltage drops can be minimized by using components with low resistance and forward voltage drop.

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