Blog for students taking ECEN5807 Modeling and Control of Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2013
Thursday, April 28, 2011
18.14 Vcontrol
I'm not sure how to go about finding Vcontrol. Using equation 18.50, and assuming we are in equilibrium so Pav=Pload, I get Vcontrol=500V. This is obviously wrong. So I'm missing something here. Any hints?
18.14 ripple
Is it correct to use equation 18.90 to find the capacitor voltage at pi/2 and 3pi/2? Then the difference should be the peak to peak ripple. Correct?
Wednesday, April 27, 2011
HW12 Spice simulation not converging
HW12 Spice simulation takes forever and is not converging. Any tips would be appreciated.
Regards,
Nitish
Regards,
Nitish
18.10b.ii
In response to Jares, I believe you're correct. I think we use the equations found in table 18.3 for rms and peak inductor currents to answer 18.10b.ii. Does anyone disagree?
Equation 18.55 to 18.56
I'm not able to post comments to postings due to our IT security so I'm creating a post as a response. In the text it states that equation 18.56 is the averaged control to output transfer function. For that reason, vg(t) is set to zero in equation 18.55 before applying Laplace to get equation 18.56.
Chapter 18 Question (Equation 18.55)
I asked the professor about this, but it still did not make sense to me.
I do not see how you go from equation 18.55 to equation 18.56.
when I take the Laplace of equation 18.55 I get:
L*s*ig(s)= vg(s)-d'(s)*V
now I thought V is a constant. How do I get rid of the vg(s) term, and put the equation in the form of ig(s)/d(s) = V/(s*L) ????????????
thanks
Mark
I do not see how you go from equation 18.55 to equation 18.56.
when I take the Laplace of equation 18.55 I get:
L*s*ig(s)= vg(s)-d'(s)*V
now I thought V is a constant. How do I get rid of the vg(s) term, and put the equation in the form of ig(s)/d(s) = V/(s*L) ????????????
thanks
Mark
Tuesday, April 26, 2011
18.14 output voltage dynamics
I am a little confused about the dynamics in problem 18.14.
When we came up with the linearized model for the feedforward boost rectifier in class, we neglected any changes in v_hat and assumed the output voltage to be basically DC. This led to a simple G_id(s) which I used for the current loop in the problem. However, this is not sufficient for the voltage loop of the problem since it is trying to regulate variations in output voltage. Does it make sense to be using a different small signal model to solve the different loops in the circuit? I was thinking about using the same average equations as in 18.3 but assuming that the input current is a perfectly regulated DC signal and deriving a new small signal model from there. Does this approach seem reasonable?
When we came up with the linearized model for the feedforward boost rectifier in class, we neglected any changes in v_hat and assumed the output voltage to be basically DC. This led to a simple G_id(s) which I used for the current loop in the problem. However, this is not sufficient for the voltage loop of the problem since it is trying to regulate variations in output voltage. Does it make sense to be using a different small signal model to solve the different loops in the circuit? I was thinking about using the same average equations as in 18.3 but assuming that the input current is a perfectly regulated DC signal and deriving a new small signal model from there. Does this approach seem reasonable?
Monday, April 25, 2011
18.10a
Another question for clarity...in part 'a' of problem 18.10 we are to derive an expression for the duty cycle of the universal-input boost rectifier. It states that there are no losses and we can neglect converter dynamics. That being the case, I assume that this will just end up being the ideal version of the duty cycle as was discussed in lecture 39 for section 18.6.1. Is this an accurate assumption?
HW13 question a
It seems like the duty cycle in part should be a function of Ppv in addition to the other variables listed, because d controls the power flow in DCM, right? Anybody agree/disagree?
Sunday, April 24, 2011
HW12 prob. 8.14 (shouldn't ramp slope of PWM be negative?)
In HW12 prob. 8.14, shouldn't the ramp slope of PWM be negative, so that when vg(t) is high, the duty cycle is low? But then if the ramp slope is negative, will it not make the loop gain negative...but don't we need the loop gain to be positive for negative feedback (because va=ig(t) is already being subtracted from the vref1)?
Any comments would be appreciated.
Nitish
Any comments would be appreciated.
Nitish
Chapter 18 Question
In figure 18.29b on page 669 of the text it shows a model of the recifier output port with the constant and time-varying power souces separated. The time-varying source is negative and shown with a squared cosine function. Equation 18.93 is a calculation of the instantaneous output power which is a sum(in this case it is a difference due to the negative term) of the two power sources. In equation 18.93 the cosine function is not squared but in the model it is squared. What am I missing?
Saturday, April 23, 2011
HW12 Prob1
I have a somewhat general question about problem 1 of homework 12. There is a statement in the problem description that says, "There is no compensator in the inner wide-bandwidth average current control feedback loop." Does this mean that we are to assume Gc(s)=1 for our bode plots of Ti(s)?
Thursday, April 21, 2011
HW 11, Prob 2
I know this is coming in last minute, but I'm stuck on finding the normalized equation describing the DCM input characteristic for the buck-boost. In following the buck example in the book, I looked at the DCM buck-boost averaged large-signal circuit from Fig 11.11. However I'm not sure if I'm supposed to develop equations from this model or simplify it into a DC equivalent model and go from there. Also, am I supposed to try to solve for input current "ig" like in the buck example? Any pointers would be appreciated.
HW#10 prob.1 solution
Hello everyone,
I am aving trouble understanding the solution posted for HW10 prob. 1:
The solution states: Tc(fci) = (Rf*Gio*Ki/VM)*fo^2/(fz*fci) = 1. I do not understand how this expression was arrived at. It appears as though (Rf*Gio*Ki/VM) was assumed to be equal to the flat gain, say, Tco, of the compensator. I don't see how that is true. Even if it were true, I don't see how the rest of the expression makes sense.
I can see that (fci/fi)*(fi/fo)^2 = Tco, the flat gain of the compensated loop gain. However, one cannot know Tco without knowing the fi (the zero of thecompensator)….so there are three unknowns here (fi, fci and Tco).
My thinking would be: Choose fi first (say, for instance, fi = fci/10)….and assume fci = fs/5 = 200/5 = 40kHz. Then fi = 4kHz. Once we know fi, onlythen we can get Tco and then determine fci.
I would really appreciate if someone can shed light on this (or perhaps the grader or the student who did the solution).
Regards,
Nitish
I am aving trouble understanding the solution posted for HW10 prob. 1:
The solution states: Tc(fci) = (Rf*Gio*Ki/VM)*fo^2/(fz*fci) = 1. I do not understand how this expression was arrived at. It appears as though (Rf*Gio*Ki/VM) was assumed to be equal to the flat gain, say, Tco, of the compensator. I don't see how that is true. Even if it were true, I don't see how the rest of the expression makes sense.
I can see that (fci/fi)*(fi/fo)^2 = Tco, the flat gain of the compensated loop gain. However, one cannot know Tco without knowing the fi (the zero of thecompensator)….so there are three unknowns here (fi, fci and Tco).
My thinking would be: Choose fi first (say, for instance, fi = fci/10)….and assume fci = fs/5 = 200/5 = 40kHz. Then fi = 4kHz. Once we know fi, onlythen we can get Tco and then determine fci.
I would really appreciate if someone can shed light on this (or perhaps the grader or the student who did the solution).
Regards,
Nitish
Tuesday, April 19, 2011
HW11 Problem 1
This seems like a straightforward problem to me, but I am having trouble getting results for the CCM/DCM boundaries that make sense.
The fundamental difference for the buck-boost compared to the boost is that the inductor current is not equal to i_g, but rather only equal to i_g when the MOSFET is on and equal to the output current otherwise.
I used the normal buck-boost conversion ration to determine my d(t) equation, and I replaced d(t) in the inductor current equation with this. Then I calculated the CCM boundary by < delta i_L(t). This is a little more complicated then the boost since in the boost =.
Basically, I am getting a value for R_e,crit that is a function of Ts, L, and R. It is not dependent on V or v_g so it creates a hard boundary for CCM/DCM. This doesn't seem right. Does anyone see where my approach might be wrong?
The fundamental difference for the buck-boost compared to the boost is that the inductor current is not equal to i_g, but rather only equal to i_g when the MOSFET is on and equal to the output current otherwise.
I used the normal buck-boost conversion ration to determine my d(t) equation, and I replaced d(t) in the inductor current equation with this. Then I calculated the CCM boundary by < delta i_L(t). This is a little more complicated then the boost since in the boost =
Basically, I am getting a value for R_e,crit that is a function of Ts, L, and R. It is not dependent on V or v_g so it creates a hard boundary for CCM/DCM. This doesn't seem right. Does anyone see where my approach might be wrong?
Thursday, April 14, 2011
HW10 Problem 1 Spice simulation
Hello, In verifying the solution for the HW10 problem 1 (ACMC), when I simulate the uncompensated loop gain in LTSpice, it matches the hand calculation (here the duty cycle was input using a voltage source). However, as soon as I put the PI compensator in the loop and connect the output to the duty cycle input of the CCM-DCM1 model, all currents and voltages of the converter go out of range (the inductor current becomes in kilo amps and the output of the compensator is in megaVolts)....I am not sure what is going on. I would expect the output voltage of the compensator to be twice the steady state value of the duty cycle (when multiplied by 1/VM would give the duty cycle). Any hints would be greatly appreciated. Regards, nitish
Tuesday, April 12, 2011
HW10 Prob 1
I'm working on finding the range for Vc in problem one. I found the range of duty cycle using V/Vg=1/D' and used it as a reference for finding Vc. Looking at the mid-band gain I come up with Vm(t)=(R2/R1)Vrf, with Vrf=Rf*I. At turn-off Vm(t) should be at maximum. I used R2=(R1/Vrf)*Vm(t) to find R2. I get R2=350K. This results in a range of .286 to .5 for Vc, using Vc=(10R1/R2)*Vm(t), but it ends up being a lot of mid-band gain for the compensator, ~30dB. I'm not confident in this approach. I'm using Vrf=Rf*I=.05V which is true at steady state but it doesn't account for the inductor current ripple. I haven't had a chance to simulate it yet. Is my lack of confidence in this approach warranted?
Saturday, April 9, 2011
New Edtion of text book?
the information presented in this course, that is not in the book is quite valuable. It was mentioned that some of this material would be in the new edition of the book.
Is there any estimate of when the new edition would be made available?
thanks
Mark
Is there any estimate of when the new edition would be made available?
thanks
Mark
Friday, April 8, 2011
Thursday, April 7, 2011
any way to re-create a schematic using .cir files from HW7 solution ?
Hi, For HW7 flyback solution, we were provided .cir Spice files. Is there any way for Spice to create a schematic (.asc file) out of it? Regards, Nitish
Wednesday, April 6, 2011
SPICE Boost
Hey everybody,
I was having trouble getting a fast enough op-amp for my comparator to handle the sawtooth at 200kHz. I found an op-amp that has a good enough gbw, but the slew rate kills everything and I cant turn the MOSFET on.
Any thoughts?
danny
I was having trouble getting a fast enough op-amp for my comparator to handle the sawtooth at 200kHz. I found an op-amp that has a good enough gbw, but the slew rate kills everything and I cant turn the MOSFET on.
Any thoughts?
danny
Tuesday, April 5, 2011
HW9 P1 - Voltage Loop Gain
For HW9 P1a, I was using figure 12.25a of the text as a guide, and as such I was able to get an equation for Gvc based on eqn 12.73. However, the voltage loop gain using this model would incorporate Fv as a multiplier (Tv = Gc(s) * Fv * Gvd(s)), and since Fv = 0 in model 4, I am inclined to think maybe I need to change the model in 12.25a to compensate for model 4 in some way. Could anyone give me a hint here?
Monday, April 4, 2011
HW 9 part(c)...do we use model(4) results that match 2nd order Pade approx, sampled data model results?
In HW9, problem 1(a), it appears that to evaluate the given compensator design, we need to use averaged model (4) results that match the first order Pade approximation results of the sample data model. Then in 1(c), we are asked to use the "more accurate" model to design the compensator. Are we expected to use the modified averaged model (4) whose results match the 2nd order Pade approx. of the sampled data model (or do we still use the first order results)?
Saturday, April 2, 2011
HW10 Boost ACM (a)
I was just wondering if anyone out there had looked at part (a) for the Boost ACM problem from HW10. I'm a little confused on exactly what parameters we can set for the sawtooth since the switching frequency and the amplitude are given.
I'm thinking it has something to do with the minimum voltage of the sawtooth. I've read the papers on ACM posted online, and Dixon makes it seem like the slopes need to be the focus, though we have no control over these.
The problem does say to just sketch the "waveshape," so perhaps numbers are not important?
Anyways, can anyone point me in the right direction?
Thanks,
Danny
I'm thinking it has something to do with the minimum voltage of the sawtooth. I've read the papers on ACM posted online, and Dixon makes it seem like the slopes need to be the focus, though we have no control over these.
The problem does say to just sketch the "waveshape," so perhaps numbers are not important?
Anyways, can anyone point me in the right direction?
Thanks,
Danny
Friday, April 1, 2011
IEEE 1547 standard used in USA for connecting distributed sources to the power grid
In reference to the standards discussion in lecture 31, I would like to state that all distributed power sources in USA (anyone producing power other than the utilities) must meet the IEEE1547 standard, in order to legally connect to the power grid.
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