EEVblog #260 - Tracking Pre-Regulator Simulation in LTspice - Page 1
Forum section: electronics / general

author: BravoV
author_rank: Super Contributor
timestamp: March 19, 2012, 05:26:23 pm
content: Very interesting series, especially turning the pre-regulator control loop from digital to analog, yay.Tinkered with ltspice after watching and made this on the control loop for 2 different output voltages.The shaded area is my addition on the control loop using dirt cheap discrete and single jelly bean bi-polar transistor, the rest are identical with Dave's.This gives me a steady drop out approx 1.3 volt across the linear regulator, I "guess" with this 1.3 volt should be ok right ? I mean its not too high sacrificing the power efficiency, while not too low for better load transient response/recovery.12 volt linear output result, top signal is the difference voltage across switcher output vs linear output (this can be made by pressing Ctrl button while clicking at two measurement points), and the bottom chart is two output voltages from switcher and linear. The signal's label is self explanatory.5 volt linear result, with R4 changed to 500K.Comments are welcome, btw, this is an entry level enthusiast grade mod, not an EE, so please be gentle, especially from BAW, I know you're lurking somewhere around here on this particular topic .PS : Attached below also the zipped LTSpice file from above circuit.

author: Rufus
author_rank: Super Contributor
timestamp: March 19, 2012, 06:24:02 pm
content: The simulation doesn't work with this MOSFET or that MOSFET so blame the simulator or models - oops.If the MOSFET threshold is less than the LT3080 dropout voltage the circuit will regulate to whatever output voltage/current it takes to make the voltage across the LT3080 the same as the MOSFET threshold.

author: amspire
author_rank: Super Contributor
timestamp: March 19, 2012, 10:28:50 pm
content: BravoB, although the ides of replacing the MOSFET with a transistor circuit is that the transistor behave a bit differently to the MOSFET. The MOSFET acts as a voltage controlled resistance and the transistor with the 10K resistor input acts as a voltage controlled current source. Without getting into the specifics, it is enough of a difference to mess up the control circuit of the switching regulator. You can see this in your plot - the ripple on the switching regulator has dropped from over 1MHz to something like 200kHz. This means the regulator circuit is overcompensating causing it to boost the inductor current too much for several cycles, then turn off completely for a few cycles.The result may look OK, but it can be causing big problems like pushing the inductor towards saturation. This is where you have to remember that LTSpice has ideal inductors that never saturate, so the real result may be a lot more ugly.Richard.

author: amspire
author_rank: Super Contributor
timestamp: March 19, 2012, 11:59:32 pm
content: Arguments for the original circuit over the tracking regulator.For a start, variations in gate turn-on voltages is just to wide, but I think the digital control of the switching regulator is superior anyway.There are a several benefits of allowing the micro independent control over both the switching regulator voltage and the linear regulator voltage.One benefit is that you have the option under remote software voltage control of allowing the switching regulator to rise to its new voltage first before you change the linear regulator's voltage. This way the rise will be at the full speed of the linear regulator, and the regulator will be in control though the whole voltage rise.If you use the the tracking regulator option, then during the change the LT3080 will be hard on so basically the load will be connected to the switching regulator until the new voltage is reached. During the transition the LT3080 will have no ability to reject any switching ripple on the switching regulator output.It just seems you have more options for the micro if it can control the switching regulator. Without the remote software control, it probably wouldn't be an issue - how fast can you turn the voltage control knob anyway? But there is remote control.The second point is protecting the load. With a micro controlled switching regulator, if the LT3080 shorts, the voltage on the load will rise by 4V before the micro shuts everything down. With the tracking pre-regulator, if the LT3080 shorts, the tracking regulator's divider MOSFET will switch off, and so the regulator will jump to its absolute maximum voltage. If the regulator needs to be able to go to 24V, the absolute maximum may be something like 30V or more.Basically once you have code written working code for the micro to control the switching regulator voltage, then the tracking-regulator option offers no real advantages, and has several disadvantages.Richard.

author: IanB
author_rank: Super Contributor
timestamp: March 20, 2012, 01:27:37 am
content: I think this is probably covered by the arguments here and elsewhere, but it is also perhaps worth mentioning that every design should consider not only when the power supply is operating in voltage limited mode, but also in current limited mode. The pre-regulation comes into its own when the output is current limited and the terminal voltage is close to zero. This is where the linear regulator could potentially be dissipating the maximum power and the pre-regulator saves it. There may be a tendency for the naive designer to think only of voltage regulation and forget current regulation.I am sure the experienced people have it covered, but it seems worth a mention anyway.

author: EEVblog
author_rank: Unknown
timestamp: March 20, 2012, 05:30:46 am
content: Quote from: amspire on March 19, 2012, 11:59:32 pmI don't follow that.If the load shorts, the pre-regulator drops to the offset voltage (say 3V).Whats' the problem?Dave.

author: enclis
author_rank: Contributor
timestamp: March 20, 2012, 03:59:50 pm
content: Maybe it sounds stupid but why dont you use LTC3600 for pre DC-DC conversation?

author: Bored@Work
author_rank: Super Contributor
timestamp: March 20, 2012, 04:18:15 pm
content: Quote from: enclis on March 20, 2012, 03:59:50 pmWhy did you pick a random buck converter to replace a boost converter?

author: armandas
author_rank: Frequent Contributor
timestamp: March 20, 2012, 04:21:23 pm
content: Quote from: enclis on March 20, 2012, 03:59:50 pmProbably because LT parts, while nice, tend to be much more expensive.Quote from: BoredAtWork on March 20, 2012, 04:18:15 pmThat too

author: enclis
author_rank: Contributor
timestamp: March 20, 2012, 07:15:17 pm
content: Quote from: BoredAtWork on March 20, 2012, 04:18:15 pmLTC3600 is not just a random converter, its output voltage is down to 0V, actually there is a schem in datasheet that uses both LT3080 and LTC3600.

author: Bored@Work
author_rank: Super Contributor
timestamp: March 20, 2012, 07:37:04 pm
content: Quote from: enclis on March 20, 2012, 07:15:17 pmThen let me rephrase the question. Why did you pick a non-random buck converter to replace a boost converter?

author: enclis
author_rank: Contributor
timestamp: March 20, 2012, 08:36:11 pm
content: Quote from: BoredAtWork on March 20, 2012, 07:37:04 pmOk, ok - you right

author: amspire
author_rank: Super Contributor
timestamp: March 20, 2012, 08:41:59 pm
content: Quote from: EEVblog on March 20, 2012, 05:30:46 amI meant if the LT3080 fails by shorting. With the micro control, the switching stays at a fixed voltage. The tracking regulator goes to maximum voltage.Sorry for the confusing words.Richard

author: Pentium100
author_rank: Frequent Contributor
timestamp: March 20, 2012, 09:03:43 pm
content: Quote from: amspire on March 20, 2012, 08:41:59 pmIf the LT3080 shorts (input to output) with the software control the output voltage will be a few volts higher than the set voltage. With the analog regulator, the output voltage will be 3V (or whatever the threshold voltage of the transistor is).

author: EEVblog
author_rank: Unknown
timestamp: March 20, 2012, 09:26:12 pm
content: Quote from: Pentium100 on March 20, 2012, 09:03:43 pmYes, so either way I don't see the problem.Dave.

author: Rufus
author_rank: Super Contributor
timestamp: March 20, 2012, 10:02:39 pm
content: Quote from: EEVblog on March 20, 2012, 09:26:12 pmNo, if the LT3080 fails short the switcher and so the output will go as high as it can. That said, I don't think possible single component failure mechanisms should have much influence on the design.

author: Pentium100
author_rank: Frequent Contributor
timestamp: March 21, 2012, 01:41:46 am
content: Quote from: Rufus on March 20, 2012, 10:02:39 pmOK, here's the full list of what can happen to the LT3080:1. Short input to ground - the DC-DC converter gets shorted, 0V at output.2. Short output to ground - the DC-DC converter goes to maximum voltage, 0V at output.3. Short input to output - the output voltage becomes equal to the gate threshold voltage of the transistor.4. Open - the DC-DC converter goes to maximum voltage, 0V voltage at output.The maximum voltage may be a problem to other components connected to the DC-DC converter, that is why you should set the resistor so that the voltage without the transistor is not too high.

author: Rufus
author_rank: Super Contributor
timestamp: March 21, 2012, 02:17:05 am
content: Quote from: Pentium100 on March 21, 2012, 01:41:46 amNope, that's what you think will happen and you got half of them wrong.Why don't you try simulating it and when it doesn't do what you think it should you can blame the simulator like Dave?

author: Short Circuit
author_rank: Frequent Contributor
timestamp: March 21, 2012, 02:38:08 am
content: Quote from: Pentium100 on March 21, 2012, 01:41:46 amGetting late here, so my logic might be flawed, but my failure modes are a little bit different;-2- The gate of the FET is at ground, so the FET is fully open (as in 0 ohms), which couples the output directly into feedback. Hence Vout = Vfb. But Vfb is probably lower than Vgs(th), so FET resistance increases and output regulates to Vgs threshold (pretty much same scenario as in the Blog simulation of the problematic FETs)-3- Means Vgs is 0V, FET fully closed and output voltage is determined by the resistive divider-4- Truly open is very unlikely, leakage currents will pull Vg to 0V, see -2-No real problems, except that I would choose to set the resistors for a safe value.This is also (and more practical) useful if a higher input voltage is applied to the outputs. That will couplethrough the back diode anyway, but there is not much use in overdoing it by allowing the booster to raise it even further

author: Frenchie
author_rank: Regular Contributor
timestamp: March 21, 2012, 10:29:07 am
content: I was looking into doing something similar a while back, but the thing that was doing my head in was inductor selection. What would be the best way to size it, just pick an average output voltage and target that or is it a case of throw a bunch of different values at the problem and see what sticks?

author: bfritz
author_rank: Regular Contributor
timestamp: March 21, 2012, 10:21:58 pm
content: Quote from: Frenchie on March 21, 2012, 10:29:07 amMost of the manufacturers try to make this stuff pretty easy for you. Often the data sheet will give you a set of equations, and some manufacturers give you an excel spreadsheet that allows you to choose parameters, and it calculates the component values for you. Going through the calculations by hand, and deriving the equations can give you better insight into why it works. For choosing the inductor, you need to know the output current, input and output voltage, and switching frequency. Typically you choose an Inductor Current Ratio (LIR), and that sets the ripple current in the inductor as a ratio of the output current. The following article I found by googling, and discusses various LIR, and derives the appropriate equations. If you can go through it, and work through how they derive the equations, this will no longer "do your head in".http://my.ece.ucsb.edu/Bobsclass/194/References/NonIsolated/Buck/Buck%20Converter%20Design%20Demystified%20606PET25.pdf

author: mldelibero
author_rank: Newbie
timestamp: March 22, 2012, 12:45:42 am
content: If anyone needs another tutorial to get started with LTspice, I gave a talk here:for a student chapter of IEEE.

author: amspire
author_rank: Super Contributor
timestamp: March 22, 2012, 12:51:09 am
content: If you are designing a switching circuit for a fixed voltage load and current, it is not hard making a good choice of inductor, filter cap, etc.If you are designing a switching circuit that is going to have a wide ranging input voltage, a wide ranging output voltage and supplies a wide ranging current, you are going to be in for a fun time. You have to work through every combination of the extremes of the specs and try and find a combination that will work. As long as you start knowing that there will not be some kind of magical optimal solution, then it is just a case of finding a choice that can work at all the extremes, and you live with the results.It may mean that at maximum input voltage, minimum output voltage and current, the switching regulator is only turning on one cycle in ten, but at least it can still output the correct voltage that way. This could be a problem if you are powering audio gear as it can bring the switching frequency down into audible range.Don't look for perfection - just look for a solution.In choosing inductors, the bigger the inductor, the more constant the current flowing through the inductor. This is good for reducing the ESR demands for the filter caps, the resulting ripple voltage, and the current rating of the inductor. However the negative is the bigger the inductor, the slower it can respond to changes of output voltage and current. In particular, if the load goes from 0% to 100%, the output voltage will drop until the supply has had time to adjust. You can add a bigger output filter cap, but then a big cap will make it slow to change voltage. At every step, you will find conflicting choices.Richard.

author: Frenchie
author_rank: Regular Contributor
timestamp: March 24, 2012, 03:38:21 am
content: Thanks Guys,Should've been a bit clearer, but ampspire hit the nail on the head anyway. No issues selecting inductors for fixed output voltages, it's when things get variable that I was struggling a bit. Anyway, basically what I've been doing, so I'll stick with it.Cheers.

author: amspire
author_rank: Super Contributor
timestamp: March 24, 2012, 05:07:03 am
content: As long as you take care of the extremes, the rest should be fine. Other then knowing switching frequency, you can do the calculations without looking at the data sheet for the switching devices at the start. You have to eventually to make sure your parameters are within the specs for the regulator IC.He is an example, you are designing an inductor-based boost converter that takes from 2.8V to 4.5V input and you want 5V to 25V output at 2A.I would first go to one of the most important extremes - the highest voltage at the highest current with the lowest input voltage. That's a 9:1 output:input voltage ratio, so with ideal parts, the inductor will be charging 90% of the cycle, and discharging 10%. The average current out is 2A, so the average current during the 10% discharge time is 20A. Now how much are you going to allow the current in the inductor to discharge during that 10% ? Pick two extremes. It the first, it only discharges by 5%. In the second, it discharges by 100%. In the first, the peak inductor current will be 2 x 5%/2 = 20.5A. In the second case the peak current will be 40A. We know we will have inefficiencies so lets boost those numbers by 20%. We have to allow for component tolerances so add another 20% on top. That gives a peak inductor current between 30A and 58A. Now you look at the spec sheets for the inductors, the regulator, the MOSFET's, can you work at either end of those current specs.At switch-on you probably want some extra power available to get up to working voltage, so can the components be pushed an extra 20% for a one off start?If the numbers just do not work, then a single inductor solution will not work and you will have to go to a transformer. With a transformer, you could reduce the peak current to more like 4A to 8A for these specs.Then you look at what happens again at the minimum voltage in and maximum power output conditions - probably will be at 25V out. With the two inductors chosen, if the load changes from 0% to 100%, how many cycles will it take for the current in the inductor to build up to the required level? The bigger inductor will take something like 10 times longer then the small inductor. During this time the output will droop by an amount. The size of the filter caps on the output will decide the amount of the droop as they have to sustain the current until the current in the inductor can build up.You might want to look at how quickly the output voltage can change, if that is of importance.Notice I have been talking voltages and currents, but I haven't touched the device spec sheet much. I now at least have a feel for the numbers, and an idea if it is possible. If it looks possible, pick a real inductor, real mosfets, etc and repeat the calculations in more detail allowing, regulators speca such as minimum ON time for resistance of the inductor windings, Diode voltage drops if it uses diodes, resistance of the mosfet, rise and fall times of the mosfets, and regulator specs such as minimum ON time/OFF times.If it works comfortably, you have your solution. If not, you probably either can guess what values you do need, or you will know that the specs need rethinking.Richard.

author: mightyzen
author_rank: Contributor
timestamp: March 24, 2012, 10:23:24 am
content: @amspire thanks for your example!Quote from: amspire on March 24, 2012, 05:07:03 amCould you elaborate a bit on the transformer? That was the only part that was not clear to me.

author: ejeffrey
author_rank: Super Contributor
timestamp: March 24, 2012, 11:41:14 am
content: Quote from: mightyzen on March 24, 2012, 10:23:24 amA transformer coupled SMPS lets you take advantage of the turns ratio so that you can operate more near 50/50 duty cycle. This keeps your peak currents to a more manageable level. You can do this as a flyback converter which uses the transformer as the energy storage (as a direct replacement for an inductor) or a forward converter where the transformer is used in the usual fashion, and there is a separate inductor for energy storage.

author: amspire
author_rank: Super Contributor
timestamp: March 24, 2012, 11:52:30 am
content: A boost converter with a transformer instead of an inductor is basically a flyback converter. Since you can choose your ratios, you could pick a 2.8:25 ratio. The transformer primary is connected exactly the same way as the inductor was, but the discharge cycle now happens from the secondary winding.This would mean that at maximum voltage, the duty cycle is 50%. So if you have a 50% duty cycle on the secondary, then for a 2A average output, you would need 4 A peak in the secondary or more.What I forgot to explain properly is that with the transformer ratio, it would mean 36A peak in the transformer primary or more. It was 20A or more when the primary had a 90% charging duty cycle. It makes sense that it goes up to about 36A with a 50% charging duty cycle.So the transformer can be designed to give you much less ripple current in the secondary circuit at the expense of more ripple and a higher peak current in the primary. Sometimes it is worth the effort if the quality of the output, or the ESR peak current rating of the secondary components is the prime concern. Transformers are always more expensive then inductors and tend to be much harder to design.Richard.

author: mightyzen
author_rank: Contributor
timestamp: March 24, 2012, 04:58:41 pm
content: Clear, thanks, the diagram helped.