# Power Electronics: Choppers RLE Load

• jaus tail
In summary, the conversation discusses a question involving finding the average source current and voltage for a load with RLE properties. The conversation also mentions a possible printing error in the given options for the answer. There is a suggestion to use exponential waveform instead of the triangular approximation, but it is not necessary for the homework. The conversation ends with a discussion about the assumptions and unknown variables in the problem.
jaus tail

## Homework Equations

I rms for triangular waveform = I(max)/ (1.732)
R = V/I

## The Attempt at a Solution

Since there is RLE load, current waveform is triangular. It is

Source current is

Average of I source. I used integration. Assumed Time period = 10 seconds. So T on = 5 seconds. (question says duty = 0.5)

So integrating I got average source current = 14.7 A
Average source voltage = 200 V
So input resistance = 200 / 14.7 = 13.6A

But I'm not able to get the first one. I get answer as 26.33 A But that is not even in the option.
I used formula square I(load) = square dc component (26.05) + square (rms of triangular waveform)

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It says minimum is 32.75A and maximum is 26.05A. Something seems wrong, since we expect max > min.

So I'd say there's a printing error, maybe two, and in either the data or the answer options. Would (A) 26.17A satisfy you? If not, you could play around with various changes in the answers to see whether you can stumble across consistent values.

jaus tail
Yes the printing error occurred to me as well. I guess only the max n min words should be exchanged. Cause I'm getting right answer for second part of question.
How do you get 30.17A as answer?

I looked for a single digit to change that would turn what's printed into something close to what you're looking for.

Is this correct method?

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Your calculations are probably right. You answered while I was still adjusting my comment. I settled on 26.17 as a likely option.

Maybe if you worked it out exactly, using the exponentials instead of approximating to straight lines, it might match the 26.17A answer exactly?

jaus tail
So instead of triangular ur saying exponential waveform? e-Rt/L?

jaus tail said:
So instead of triangular ur saying exponential waveform? e-Rt/L?
Well, only if you want to test whether my hunch is right. The homework would not expect you to, the triangular approximation is fine. Though in the real world you should show that the triangular approximation is justified and not simply assume it is, i.e., not just hope it is.

I got something like this:

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I guess triangular waveform is justified when the inductance is very big. If resistance is very big then waveform of current = waveform of voltage = square.
I'm getting integrating value as 27.6A
How did you integrate so fast?

I haven't worked out an answer. I see insufficient information given to do so, in this instance.

The rising exponential, A(1-e-t/T) has magnitude A = (200-E)/R – 26.05
and T = L/R = unspecified
and passes through the point 6.7 at t=?? (5ms your nomination)

Too many unknowns to evaluate to a number, I think.

jaus tail
I assumed T to be 10 seconds and since duty cycle is 0.5 we get inductor storing energy for 5 seconds and releasing energy for 5 seconds.

## 1. What is a chopper in power electronics?

A chopper is a DC to DC converter that is used to control the output voltage of a DC power source. It works by switching the input voltage on and off to produce a desired output voltage.

## 2. What is an RLE load in power electronics?

RLE stands for "Resistance-Inductance-Electrolytic" and is a type of load that is commonly used in power electronics for testing and evaluation purposes. It consists of a resistor, inductor, and electrolytic capacitor connected in series.

## 3. How does a chopper control the output voltage?

A chopper uses a technique called pulse width modulation (PWM) to control the output voltage. It switches the input voltage on and off in a controlled manner using a high frequency signal, and the average voltage delivered to the load is determined by the duty cycle of the switching signal.

## 4. What are the advantages of using choppers in power electronics?

Choppers offer several advantages, including high efficiency, precise control of output voltage, and the ability to step up or step down the input voltage. They also have a simple and compact design compared to other DC to DC converters.

## 5. What are some common applications of choppers in power electronics?

Choppers are commonly used in applications where precise control of output voltage is required, such as in motor speed control, battery charging, and voltage regulation in power supplies. They are also used in renewable energy systems, electric vehicles, and other industrial and consumer electronics devices.

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