# Type A Chopper

## Homework Statement

I am having quite a bit of difficulty with the problem I have attached. We need to determine the period of operation and the critical inductance. Solving for the period is not as difficult. However, determining $τ$ has proven to be rather tedious. Any help is appreciated.

The output voltage can be expressed as $V_o = \frac{t_{on}V}{T} + \frac{V_C(T-t_x)}{T}$ where $t_x$ represents the point that the load current falls to zero. In this instance, this value falls to zero at the boundary where the current is just discontinuos ($t_x=T$). Therefore, $T=\frac{t_{on}V}{V_c}$.

I then can do a bunch of manipulations to the current equations to come up with $t_x = τln[1+3(e^{t_{on}/τ} - 1)]$

At this point, I cannot seem to solve this equation by hand or with my TI-89. Any ideas?

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rude man
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Not sure of your equations or reasoning.

Two situations:
1. Q turns on, how high does the current i build up over T1 = 0.350 us? Let i(T1) = i0.
2. Q turns off, how much time T2 does it take for i to decay to zero?

Your frequency is then obviously 1/(T1 + T2).
Both situations require the solving of 1st order diff. eq. with known initial conditions. Together they allow you to solve for L. L appears in both equations. The first equation does not allow solving for L by itself since it also contains a second unknown, to wit, the current at the end of T1.

Unfortunately I see a problem here. We wind up with 2 equations but 3 unknowns: T2, L and i0, the current at the end of T1. This seems to indicate that there is no unique solution of T2 and L. I have to think about this some more.

I am going to take another go at it around 2 today.

EDIT: I peaked at a previous solution for a Type A chopper and a simplifying assumption that he made was that the current curves were approximately linear.

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rude man
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The problem as stated does not have a unique solution.

You could pick any L arbitrarily as long as L/R << T1 (= 0.350 μs). Then the current builds up to (V-Vc)/R very quickly when Q turns on, and when Q turns off the current goes back to zero in time T2 = (L/R) ln(V/Vc). The frequency would be arbitrarily close to 1/T1, just a bit lower.

The problem should give a value for L or T2, then a unique solution would exist.

Yes, you are 100% correct. The professor indicated to me that in this instance you arbitrarily choose one of the values. Which is good to know as if i had not encountered this i would have been stumped on the exam trying to solve for a unique solution when it doesnt even exist. Thanks again!

rude man
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In "real life" the load would not be a constant voltage generator but a capacitor with a resistive load in parallel. Would have made a much more interesting problem!

If we want to determine the energy output from the input source of the type A chopper for one cycle I can determine the real power supplied by the source and multiply that by the period, correct?

There is a particular wording that has me a little bit confused: "the power output of the chopper." Power output, as in the power at at the output (ie. at the load in the image above)?

One other thing. I have encountered one other question for a 3-phase full-wave controlled rectifier that states the following: Assuming a very large L, determine the RMS value of the fundamental component of the line current. The firing angle is not given to us. However, since L is rather large, that means the time constant is as well. Harmonics for full-wave controlled rectifiers are of n=6,12,18,..., and since L is large, that causes $z_n$ to get large rather fast, and therefore the nth peak amplitude of the current gets small rather quickly. In this instance, the RMS output current can be approximated as merely the average output current, and the harmonics can be neglected. The output current can therefore be approximated as a constant value with zero slope. Hence, the output voltage waveform does not depend on the firing angles. Since $z_n$ gets rather large, $I_{RMS} = \sqrt{I_o^2+I_{6}^2+....}\Rightarrow I_{RMS}≈I_o$

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rude man
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If we want to determine the energy output from the input source of the type A chopper for one cycle I can determine the real power supplied by the source and multiply that by the period, correct?

There is a particular wording that has me a little bit confused: "the power output of the chopper." Power output, as in the power at at the output (ie. at the load in the image above)?
I would integrate load current times the 200V load voltage over one cycle (T1 + T2).

One other thing. I have encountered one other question for a 3-phase full-wave controlled rectifier that states the following: Assuming a very large L, determine the RMS value of the fundamental component of the line current. The firing angle is not given to us. However, since L is rather large, that means the time constant is as well. Harmonics for full-wave controlled rectifiers are of n=6,12,18,..., and since L is large, that causes $z_n$ to get large rather fast, and therefore the nth peak amplitude of the current gets small rather quickly. In this instance, the RMS output current can be approximated as merely the average output current, and the harmonics can be neglected. The output current can therefore be approximated as a constant value with zero slope. Hence, the output voltage waveform does not depend on the firing angles. Since $z_n$ gets rather large, $I_{RMS} = \sqrt{I_o^2+I_{6}^2+....}\Rightarrow I_{RMS}≈I_o$
Oh dear, you've got me this time. I totally forgot what little I learned of 3-phase systems. If you could describe your system in more detail maybe I could help.

rude man,

Alright, so the energy output of the input source is $\int i(t)*Edt$. The power output of the chopper is the power consumed the the resistor and the source $V_C$. It is otherwise known as the load power.

In regards to the last question, the output waveform for the current normally looks like the image I have attached. However, we have been told L is really large and not given a firing angle. Normally, all of the calculations depend on the firing angle as it depicts the shape of the output waveform. However, we do not have it this time. Hence, I concluded that the harmonics can be neglected (n=6,13,18,...) and the output waveform is approximately constant (zero slope). Which is how I ended up here: $I_{RMS}≈I_{average}$

My exam is in two hours so hopefully I can clear this up.

EDIT: my exam was horrible. I felt as though I had a pretty good grasp on most of the concepts, however, he asked some rather challenging questions. For example, given a single phase full-wave controlled rectifier, what is the firing angle such that the current is on the boundary of continuity. Was too complicated to solve by hand or even with my solver. I wasn't about to waste any more time on the question guessing values. Not only did we not know the firing angle, but you also did not know the constant as a result of first order circuit (amplitude of the forced component).

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rude man
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rude man,

Alright, so the energy output of the input source is $\inti(t)*Edt$. The power output of the chopper is the power consumed the the resistor and the source $V_C$. It is otherwise known as the load power.
The R in your circuit is part of the inductor L and is not the load resistance. So as shown the output energy per cycle is just Vc∫i(t)dt over (T1 + T2). Actually, it's that even if the load is a parallel R-C network with R the load and C the filkter capacitor (this is the usual setup). Of course, for smaller R (larger load) you get a larger current and thus a larger energy dissipation per cycle.

In regards to the last question, the output waveform for the current normally looks like the image I have attached. However, we have been told L is really large and not given a firing angle. Normally, all of the calculations depend on the firing angle as it depicts the shape of the output waveform. However, we do not have it this time. Hence, I concluded that the harmonics can be neglected (n=6,13,18,...) and the output waveform is approximately constant (zero slope). Which is how I ended up here: $I_{RMS}≈I_{average}$

My exam is in two hours so hopefully I can clear this up.

EDIT: my exam was horrible. I felt as though I had a pretty good grasp on most of the concepts, however, he asked some rather challenging questions. For example, given a single phase full-wave controlled rectifier, what is the firing angle such that the current is on the boundary of continuity. Was too complicated to solve by hand or even with my solver. I wasn't about to waste any more time on the question guessing values. Not only did we not know the firing angle, but you also did not know the constant as a result of first order circuit (amplitude of the forced component).
Sorry to hear that. That does not sound like a question that should be asked in a supervised, time-limited exam. It sounds too long to do even if you're on top of all the theory.

Do you still want to pursue the 3-phase system question?

I better focus on my other finals for now. I still have 6 to go. I will be back!!! Thanks again!