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

• Jason06841
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.
Jason06841
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?

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.

Jason06841 and DaveE
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.

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.

DaveE
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.

Jason06841 and scottdave
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...

Baluncore
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.

Jason06841
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

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.

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.

Delta Prime, Jason06841 and scottdave
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.

scottdave
This (in the book you showed) is the kind of Algebra one uses when trying to obtain a desired result. Avoid the book.

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!

Okay, after cleaning up an off-topic dangerous discussion, this thread is now reopened. Have a nice day.

Baluncore

## 1. What is a half-wave rectifier with an inductor and a DC source?

A half-wave rectifier with an inductor and a DC source is an electrical circuit used to convert alternating current (AC) into direct current (DC). It typically consists of a single diode that allows current to pass during one half of the AC cycle, effectively blocking the other half. The inductor is added to the circuit to smooth out the output voltage by reducing the ripple and improving the efficiency of the voltage conversion.

## 2. How does an inductor improve the performance of a half-wave rectifier?

In a half-wave rectifier, the inductor acts as a smoothing component. It stores energy in the form of a magnetic field when the diode is conducting and releases this energy when the AC supply is in its non-conductive phase. This process helps to maintain a more constant voltage level across the load by reducing the fluctuations in the output, commonly known as ripple. The inductor's ability to limit changes in current also helps in reducing the sharpness of the voltage spikes and smoothing the output waveform.

## 3. What are the key considerations when designing a circuit with a half-wave rectifier, inductor, and DC source?

Key considerations when designing such a circuit include selecting the appropriate values for the inductor and the diode based on the desired output characteristics. The diode must be able to handle the peak inverse voltage and forward current without failure. The inductor's value must be chosen to adequately smooth the output while not overly delaying the response time of the circuit. Additionally, the load resistance should be considered as it affects the overall voltage output and the efficiency of the rectification process.

## 4. What are the limitations of using a half-wave rectifier with an inductor?

The primary limitation of using a half-wave rectifier with an inductor is its inefficiency in energy conversion compared to full-wave rectifiers. Since it only uses one half of the AC cycle, the output DC is not as stable or high in power. Additionally, the inclusion of an inductor can lead to a larger and potentially more costly design due to the physical size and material cost of inductors suitable for high current applications. There is also a risk of saturation of the inductor under high current conditions, which can affect performance.

## 5. How can the efficiency of a half-wave rectifier with an inductor be optimized?

Efficiency of a half-wave rectifier with an inductor can be optimized by carefully selecting components that match the desired output and load requirements. Using diodes with lower forward voltage drop can help reduce power losses. Choosing an inductor with a higher quality factor and appropriate inductance value can minimize losses due to resistance and magnetic hysteresis. Additionally, implementing filter capacitors at the output can further smooth the DC output, thus improving the overall efficiency of the rectifier circuit.

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