Ok, so knowing the exact problem now:

Vin of rectifier = 230Vrms
Frequency = 50Hz

I consider that all info from Buck problem can be used here if needed, such as Po = 100W, Ro = 18Ω and Io is constant because problem says that buck acts as an highly inductive load and this means Io = const.

We want 80Vdc at Buck input and we need to evaluate the firing angle to provide this 80V at Buck input!

So from Mohan's book I read that:



Considering Id = Io = 2.36A from Buck problem, can we simply do:

\(\displaystyle{100W = 0.9\cdot \sqrt{2}\cdot 230\cdot 2.35\cdot cos(\alpha)}\)

\(\displaystyle{\alpha= arcos\left ( \frac{100\,W}{0.9\cdot \sqrt{2}\cdot 230\, V\cdot 2.35\, A}\right )}\)

\(\displaystyle{\alpha = 81.67^{o}}\)

to answer the first question?

 

But 2.36A *80V = 189W

I = 100W/80V =1.25A Isn't it?

 

But 2.36A *80V = 189W

I = 100W/80V =1.25A Isn't it?

Yes, indeed...

So
\(\displaystyle{100W = 0.9\cdot \sqrt{2}\cdot 230\cdot 1.25\cdot cos(\alpha)}\)

\(\displaystyle{\alpha= arcos\left ( \frac{100\,W}{0.9\cdot \sqrt{2}\cdot 230\, V\cdot 1.25\, A}\right )}\)

\(\displaystyle{\alpha =74.14^{o}}\)

 

Ok, so knowing the exact problem now:

Vin of rectifier = 230Vrms
Frequency = 50Hz

I consider that all info from Buck problem can be used here if needed, such as Po = 100W, Ro = 18Ω and Io is constant because problem says that buck acts as an highly inductive load and this means Io = const.

We want 80Vdc at Buck input and we need to evaluate the firing angle to provide this 80V at Buck input!

So from Mohan's book I read that:

View attachment 145421

Considering Id = Io = 2.36A from Buck problem, can we simply do:

\(\displaystyle{100W = 0.9\cdot \sqrt{2}\cdot 230\cdot 2.35\cdot cos(\alpha)}\)

\(\displaystyle{\alpha= arcos\left ( \frac{100\,W}{0.9\cdot \sqrt{2}\cdot 230\, V\cdot 2.35\, A}\right )}\)

\(\displaystyle{\alpha = 81.67^{o}}\)

to answer the first question?

Hi,

Where did that formula come from, can you provide the derivation?
I ask because they seem to be assuming a constant DC current.

When i do this for sine voltage and current, i get close to 90 degrees which agrees with half of 230*1.25/sqrt(2).

 

Hi,

Where did that formula come from, can you provide the derivation?
I ask because they seem to be assuming a constant DC current.

When i do this for sine voltage and current, i get close to 90 degrees which agrees with half of 230*1.25/sqrt(2).

I'm not sure if that current you're saying "constant" is the same or not than the one the problem says to be constant. Problem says to consider the Buck as a highly inductive load, meaning output current constant! I was reading the chapter about single phase controlled rectifiers in Ned Mohan's book and this formula comes up from there, which is the book taken as reference for our classes! I can post a link to the pdf I'm using if needed! Should I?

This is the page of that book:

 

I'm not sure if that current you're saying "constant" is the same or not than the one the problem says to be constant. Problem says to consider the Buck as a highly inductive load, meaning output current constant! I was reading the chapter about single phase controlled rectifiers in Ned Mohan's book and this formula comes up from there, which is the book taken as reference for our classes! I can post a link to the pdf I'm using if needed! Should I?

This is the page of that book:
View attachment 145482

View attachment 145483
View attachment 145484
View attachment 145485

Hi again,

From the looks of it and the area they show in that one diagram, it does look like that assumes that the load is inductive.
Just a few questions then...
1. Who is stating that the input to the buck looks like an inductive load, or is that wrong?
2. Does this buck have decent size input capacitors (like 1000uf)?
3. Does the problem specify using the idea that the input to the buck in inductive, or can you change that if you like?

The calculation with the 'constant' current i see allows for negative current so they call it constant. That's probably ok but i did not check it with that idea in mind. I checked out the expression for an AC voltage and an AC current, which is different.

You can post the link to the pdf and i'll take a look at that. Thanks.

 

Just a few points:

Using a phase-angle controller ahead of a buck regulator isn't particularly common, but can make an effective pre-regulator. I've actually done this to make the output of a permanent-magnet generator driven at variable speed managable with a compact buck circuit.

You can implement a fully controlled bridge using 2 SCRs and 2 conventional diodes for single phase, three of each for 3-phase. This can considerably simply the gate drive circuit (very often done with transformers). It can also simplify heatsinking - have a look at what the stud is connected to in stud-mount diodes and SCRs for a hint as to how to configure the bridge. There are lots of diode-SCR modules to make controlled bridges on the market.

Evaluating RMS current for a controlled bridge with capacitive filtering is definitely a job for simulation. It's ugly! (remember that the SCR current will drop to zero well before the zero crossing of the AC input for the half-cycle in which the SCR turned on & that the value of the cap has a big influence on RMS to average current ratio).

 

Who is stating that the input to the buck looks like an inductive load, or is that wrong?

The "average" input of a buck regulator looks like a negative resistance (i.e. negative slope, not negative absolute value).
The input of a buck, unless it is something rather bizarre, must be filtered. Buck regs work the input caps hard (high ripple current) because current is drawn from them in rectangular pulses.

"Knowing that the current at the rectifier output is 120% of the buck output current ..."
Huh?! If the buck is after the controlled bridge and there is filtering between the two, this is a nonsense statement unless the buck is obscenely inefficient. In a buck reg, the input voltage is higher than the output voltage and the input current is less than the output current.

 

@MrAl the problem statement is just as it is in post #18. And as this is an exam, I cannot change anything!

Just a few questions then...
1. Who is stating that the input to the buck looks like an inductive load, or is that wrong?
2. Does this buck have decent size input capacitors (like 1000uf)?
3. Does the problem specify using the idea that the input to the buck in inductive, or can you change that if you like?

1 - Problem states that the buck plays the role of an highly inductive load to the rectifier! That's what it states.
2 - No info is provided about capacitors. The problem is translated in that post #18.
3 - I'm not sure if this is supposed to be different from question 1? I'm not sure I understood!

What about what @Jony130 said in his calcs? Could it be answered like I did after his corrections?


@ebp this is an exam problem. Not a practical circuit to be built or anything like that. Theoretical stuff though...

 

@MrAl the problem statement is just as it is in post #18. And as this is an exam, I cannot change anything!



1 - Problem states that the buck plays the role of an highly inductive load to the rectifier! That's what it states.
2 - No info is provided about capacitors. The problem is translated in that post #18.
3 - I'm not sure if this is supposed to be different from question 1? I'm not sure I understood!

What about what @Jony130 said in his calcs? Could it be answered like I did after his corrections?


@ebp this is an exam problem. Not a practical circuit to be built or anything like that. Theoretical stuff though...

Hello again,

Ok no problem, that was one of the main questions too. If you cant change anything, then they dont want you to chance anything regardless how well or not well this represents a real life circuit of some kind. So we must stick with that then.

What i find interesting is that the input caps will stay charged for some time and so i think that will alter the conduction angle, but it's a moot point for now

 

The notion that the input of a buck regulator is highly inductive is complete nonsense.

The input to the switch in a buck will indeed look inductive when the switch is on. As soon as the switch turns off, the input looks like an open circuit. Opening the buck switch may or may not result in the SCR turning off due to forward current dropping below the required hold current - it depends on the switching characteristics of the SCR. If the gate drive to the SCR is DC (which is not usually the way), then when the buck switch turns on again the SCR may or may not turn on again - once more depending on the characteristics of the SCR and IF the voltages actually permit it to. Effectively this is a strange way to cathode switch the SCR, attempting to force it to switch at the same rate as the buck switch - which is complete nonsense. The results of any attempt to simulate such a circuit will depend greatly on the characteristics of the SCR chosen for the model. The control loop of the buck will be tormented - it will go to maximum duty cycle as long as the SCRs are off or the instantaneous voltage out of the SCR bridge is less than the voltage on the filter caps at the output of the buck. It will spend most of its time trying to cope with the large signal problem. This is a useless circuit. It becomes useful by adding capacitive or LC filtering between the SCR bridge and the buck regulator. The shape of the current waveforms in the SCR bridge, and hence the RMS value, vary dramatically with the values of the components in the filter. The average current from the bridge is easy to evaluate.

Assuming the questions have been translated properly, it is my contention that whoever ask the questions does not understand what he/she is asking.

Using four SCRs instead of 2 with 2 diodes isn't wrong, it just makes the model and simulation unnecessarily complex.

 

The notion that the input of a buck regulator is highly inductive is complete nonsense.

The input to the switch in a buck will indeed look inductive when the switch is on. As soon as the switch turns off, the input looks like an open circuit. Opening the buck switch may or may not result in the SCR turning off due to forward current dropping below the required hold current - it depends on the switching characteristics of the SCR. If the gate drive to the SCR is DC (which is not usually the way), then when the buck switch turns on again the SCR may or may not turn on again - once more depending on the characteristics of the SCR and IF the voltages actually permit it to. Effectively this is a strange way to cathode switch the SCR, attempting to force it to switch at the same rate as the buck switch - which is complete nonsense. The results of any attempt to simulate such a circuit will depend greatly on the characteristics of the SCR chosen for the model. The control loop of the buck will be tormented - it will go to maximum duty cycle as long as the SCRs are off or the instantaneous voltage out of the SCR bridge is less than the voltage on the filter caps at the output of the buck. It will spend most of its time trying to cope with the large signal problem. This is a useless circuit. It becomes useful by adding capacitive or LC filtering between the SCR bridge and the buck regulator. The shape of the current waveforms in the SCR bridge, and hence the RMS value, vary dramatically with the values of the components in the filter. The average current from the bridge is easy to evaluate.

Assuming the questions have been translated properly, it is my contention that whoever ask the questions does not understand what he/she is asking.

Using four SCRs instead of 2 with 2 diodes isn't wrong, it just makes the model and simulation unnecessarily complex.

Hi,

Yes there are problems with this and that's why i brought up the question in the first place.
Also, i have never seen a buck without an input cap of decent size.
The problem was stated as it was however, so he has to fly with that.

If we digress, we can investigate a real circuit. Most of the circuits i have worked on would have a buck ahead of a linear, not a SCR bridge ahead of a buck, but for some reason they felt like posing this problem.

I might also guess that they have an inductor between the SCR bridge and the input to the buck. That helps keep the peak currents in the SCR's down and also adds to the filtering for the input DC of the buck. The waveforms shown in some of the posts suggest this because it looks like an inductor input filter load on a diode bridge.

 

Morning all...

I know it's not easy for experienced guys to work with this kind of stuff, disregarding a bunch of technical details that are needed for this to work, even in a minimal environment. But the more details you plug in, the harder this becomes.

I made some simple calcs in my post #21 that were "contested" by @Jony130 and then fixed by me but I got no confirmation yet about if that fix is now more reasonable or not for the conditions given! Can anyone confirm it? I've been in this question in the last few days and I have exam in one week and I would like to post here more problems to try to solve them with your help!

Thanks
Psy

 

@Jony130 would you consider calcs from post #23 correct?

Also I was thinking about this now and if current is 1.25A, then P = R*I² <=> R = 64Ω and not 18Ω.

I'm confused right now!

 

The math looks good. But I "don't like the idea" of using this equation in this practical case. Also, I do not like the idea of connecting the SCR bridge directly to the buck converter input.

You need to ask your teacher for extra details about this problem.

 

The math looks good. But I "don't like the idea" of using this equation in this practical case. Also, I do not like the idea of connecting the SCR bridge directly to the buck converter input.

You need to ask your teacher for extra details about this problem.

Well, after thinking better about this, I think this was meant for the Buck to simply become an inductive load, as RL at the output of the rectifier!





The plots are repeated, but intentionally to compare Vs and Is, Vout and Iout and Is and Iout.

But doing things like this, Ro = 64Ω and not the 18Ω from Buck problem to be able to get the 80V problem says. And this is from calcs, because I can't confirm these values in LTSpice!

However I don't know if it is possible to have Vo = 80V, Ro = 18Ω with Pin = Po = 100W

 

For the resistive loads to get average voltage across the resistor equal to 80V you need to set alpha to:

arcos [ (Π*Vo - Vm)/Vm ] = arcos [ (Π*80V - 230V)/230] = 84.68°

But in Europe the mains voltage is 230V RMS. So, the Vm = √2*230V and the firing angle is 103.1°

But still this voltage is far away from DC. And this is the main concern right now. Because for constant duty cycle ( for the buck converter in CCM ) the output voltage is Vout = D*Vin.

 

For the resistive loads to get average voltage across the resistor equal to 80V you need to set alpha to:

arcos [ (Π*Vo - Vm)/Vm ] = arcos [ (Π*80V - 230V)/230] = 84.68°

But in Europe the mains voltage is 230V RMS. So, the Vm = √2*230V and the firing angle is 103.1°

But still this voltage is far away from DC. And this is the main concern right now. Because for constant duty cycle ( for the buck converter in CCM ) the output voltage is Vout = D*Vin.

And where did you derived that formula from?

 

And where did you derived that formula from?

Well I used this site to do the math for me.
http://www.wolframalpha.com/input/?i=1/pi+Definite+integral+from+a+to+pi++V*sin(t)++dt
Or this one
https://www.symbolab.com/solver/definite-integral-calculator/\frac{1}{\pi}\int_{a}^{\pi} V sin\left(t\right)dt

Vo = Vm/Π * (1 + Cos(a))

Where Vo is a average voltage across the resistor

And LTspice confirms the results.

 

Well I used this site to do the math for me.
http://www.wolframalpha.com/input/?i=1/pi+Definite+integral+from+a+to+pi++V*sin(t)++dt
Or this one
https://www.symbolab.com/solver/definite-integral-calculator/\frac{1}{\pi}\int_{a}^{\pi} V sin\left(t\right)dt

Vo = Vm/Π * (1 + Cos(a))

Where Vo is a average voltage across the resistor

And LTspice confirms the results.

Yes, I've seen LTSpice matches the math which is always good! I also use Wolfram alpha many times. I dind't know the other site!