| gearheadgene |
What is the best way to weed through all the MOSFET selections available? Seems to me that turn-on/off time is important because if it's short then the dead-time can be programmed short giving decent distortion (hey, is that a run-on sentence?) Anyway, if true, then how short is good?
I'm sure that Rds is important too, with low being better for minimal losses and heat generation. But, with Rds in the range of sub 100milliOhms or so, how low is really important?
Seems that gate charge is important too. I think it affects the turn on time, but am not certain how to use this.
Can someone put me on the right path?
thanks |
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| BWRX |
Here are some of the important characteristics to look for: on state resistance, gate threshold voltage, total gate charge, turn on and rise time, turn off and fall time, reverse recovery charge, and reverse recovery time. Of course don't forget about the drain source voltage and drain current ratings. Pay careful attention to conditions used to generate some specs. Not all manufacturers use the same conditions and thus some characteristics between mosfets can't always be compared directly.
A good mosfet to check out is the STP14N10FP. It's the one used in the UCD180s and has some very good specs. While there are others that excel in certain areas, it's not terribly easy to find other mosfets that have such good overall numbers. |
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| classd4sure |
I dont' know if I'd recommend that as being a great mosfet, maybe a generation or two ago.
RDSon is pretty high!
Best you can do is start hitting app notes and figuring it out the hard way, it's not an easy thing to answer.
What BRWX mentioned isn't all encompassing.
Start digging for some power mosfet application notes I say, buck converter info is most relevant here. |
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| BWRX |
| quote: | Originally posted by classd4sure
I dont' know if I'd recommend that as being a great mosfet, maybe a generation or two ago.
RDSon is pretty high! |
I never said it was great, just that it had some very nice specs. Rdson is a little high, but that's all part of the trade off :) Post filter feedback enables you to pick mosfets with higher Rdson and better switching characteristics. |
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| fokker |
| quote: | Originally posted by BWRX
Here are some of the important characteristics to look for: on state resistance, gate threshold voltage, total gate charge, turn on and rise time, turn off and fall time, reverse recovery charge, and reverse recovery time. . |
I have no doubt that those attributes are important. However, are they important enough to make a real life difference?
Alternatively, are the run-off-the-mill type mosfets good enough for switching applications in audio class D amps? |
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| gearheadgene |
| I'm sure a book could be written on this, but how do you use this information? I will start looking for app notes, but maybe someone can point me at some good ones? |
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| BWRX |
There's no simple answer to that question. As with anything, you have to look at the whole picture when selecting components. What you look for ultimately depends on your design goals and budget (unfortunately, this is true for most of us).
Post filter feedback can correct for some things like higher Rdson but it certainly wouldn't hurt if the STP14N10FP had lower Rdson!
If you want an amp with a lot of power you'll probably want to use mosfets with very low Rdson. The downside is that they usually have a higher gate threshold voltage and much more gate charge which means they need a more capabale gate drive circuit to switch them fast.
These compromises are a result of the way the devices are fabricated. A little research into semiconductor physics will shed some light onto that topic. |
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| gearheadgene |
| quote: | Originally posted by BWRX
A little research into semiconductor physics will shed some light onto that topic. |
ugh, not semi physics
:bigeyes:
I had a hard time with that one in college!!!! |
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| classd4sure |
I have a whole folder reserved for such app notes....
AN-936
AN-937
AN-9010
Some reading on "Cdv/dt induced turn on in buck converter" would be most useful. Search google.
Also research the reverse recovery effects in buck converters
AN7019 is very interesting
AN1005 deals with avalanche breakdown as does
AN10273_1
Then you'll be familiar with alot of the problems in design with them for such an application.
Other papers like perhaps the class d basics tutorial from IR might be a quick list of a few factors that need to be optimal, and its' a good read anyway.
Minimal output capacitance is important, Cdv/dt induced turn on immunity ..... by evaluating the gate charge ratio's around Vth, could prove worthwhile, if you research that you'll find differing views on it though, some are more accurate than others. Fully avalanche rated... its a topic worthy of a mean headache but when it's all said and done take your best educated guess order up some free samples and give it a shot. :)
Just going with a good recommendation can get you startequickly, but if something goes wrong.. it pays to be aware of the above.
Sorry I can't just give you the links, but like I said I save them all to a folder so.. :confused: |
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| darkfenriz |
I second fokker's question, or rather paraphrase:
How bad is it to use in class d an industry standard inexpensive mosfet? Like, say IRF540N ? |
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| classd4sure |
"does it make a real world difference"
Hmmmmmmm.... why do you suppose they used to say class d was only good for subwoofer amp use?
Yes it makes a world of difference. You can still use old school industrial type mosfets for low power, phase 1 testing. After you're done blowing those up, put some good ones in and go for some real efficiency/reliability. |
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| gearheadgene |
This app note has all the info I was looking for, thanks. Turns out I've read it already, but completely forgot about it :D One thing sticks out though - it's not clear in the doc whether they mean rms or peak. For example, they calculate the BV recommendation based on output power and load - but no mention if it's rms or peak. Peak makes more sense to me. |
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| fokker |
| quote: | Originally posted by classd4sure
You can still use old school industrial type mosfets for low power, phase 1 testing. After you're done blowing those up, put some good ones in and go for some real efficiency/reliability. |
I am sorry how "low power" was your low-power amp? 1mw? 1w? 1kw?
and what is a typical amp that we build? 1mw? 1w? 1kw?
maybe if we establish a common baseline we can have a real discussion, not a conceptual one where we talk about the theoretical merits of better / best specs. |
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| classd4sure |
| quote: | Originally posted by fokker
I am sorry how "low power" was your low-power amp? 1mw? 1w? 1kw?
and what is a typical amp that we build? 1mw? 1w? 1kw?
maybe if we establish a common baseline we can have a real discussion, not a conceptual one where we talk about the theoretical merits of better / best specs. |
I don't know why you insist on having the obvious spelled out for you, but I can't hold your hand. |
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| FastEddy |
" ... AN-9010 ... Some reading on "Cdv/dt induced turn on in buck converter" would be most useful. ..."
Actually reading the entire application notes from Fairchild on the -9010 or the others will clarify most of the definations and procedures in selection & design using MOSFETs (Metal Oxide Semiconductors / Field Effect Transistors) ... the physics is sometimes overwhelming, but Fairchild's clarity of information and diagrams of what is going on here will carry any one through the deep stuff.
A great read: http://www.fairchildsemi.com/an/AN/AN-9010.pdf
Example:
"1) Off State
(1) BVDSS: This is the maximum drain-to-source voltage where the MOSFET can endure
without the avalanche breakdown ..."
:smash: |
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| gearheadgene |
OK, so Qg is important and needs to be low. How come there's no mention of Tr, Tf, and Ton and Toff? Everything I've read says that to minimize distortion, you have to keep the dead-time low, but without allowing shoot-through. Doesn't the Ton and Toff switching info dictate the dead-time?
If I am right, it takes the Ton + Trise for a fet to be fully on. Same for the on-to-off state, just Toff+Tfall. Right so far? To prevent shoot-through, the gates need to be staged properly (i.e. dead-time).
So, if (Ton + Tr) = (Toff + Tf) = x, then I need x dead-time. When x = 0, no dead-time (ideal case). But what happens when x is non-zero? Is it proper to wait the worst-case (Ton + Tr), or (Toff + Tf), which could be very long time. Or, is it ok to allow some cross-conduction between the high and low side FETs? Say, 1/2 way for example, into the (Ton + Toff). Seems it would heat up the fets. |
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| fredos |
Take a look at the old STW34NB20, that's cheap, lo Qc, lo RDSon, easy to drive and a VERY good rise time and fall time for that price, the only drawback is the slow body diode...But I'm sure that lot of people know how to overpass this...That's what I use in the 4000HVI...2Kw per side with 4 of these in full bridge configuration.
Fredos
www.d-amp.com |
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| gearheadgene |
| quote: | Originally posted by fredos
Take a look at the old STW34NB20, that's cheap, lo Qc, lo RDSon, easy to drive and a VERY good rise time and fall time for that price, the only drawback is the slow body diode...But I'm sure that lot of people know how to overpass this...That's what I use in the 4000HVI...2Kw per side with 4 of these in full bridge configuration.
Fredos
www.d-amp.com |
Not sure what you think is cheap, but 1pc price is around 4 bucks:bawling: I've seen very good specs on other devices for at least 1/2 that price.
Having said that, I'm still at a loss on how to figure out the dead-time needed based on my prior post. Any ideas?
gene |
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| classd4sure |
| It's not done by math, but rather adjustment |
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| fredos |
| quote: | Originally posted by gearheadgene
Not sure what you think is cheap, but 1pc price is around 4 bucks:bawling: I've seen very good specs on other devices for at least 1/2 that price.
Having said that, I'm still at a loss on how to figure out the dead-time needed based on my prior post. Any ideas?
gene |
4$ ??? What your source? I pay about 1.50$ for a strip of 30...No need for dead time adjustement with this part...No dead time in my 4000HVI with this output! Only ''natural'' delay of parts!
Fredos
www.d-amp.com |
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| fokker |
my sense on this is that there is always a desire to have better or the best parts. and quite frankly, even a monkey can decide to use the best part.
What would have been much more interesting, and far more challenging, is to find out a "good enough" part to use in a given application.
One of the companies I have affiliation to makes dc converters for military and commercial applications. They tend to be a similar designed but sourced differently for their respective market.
It is far more difficult to source for the (lower grade) commmercial application as it requires the use of "good enough" parts, whereby in the military application they just source the best part for the nature of the contract.
and it is not uncommon to see people advocating the use of the best or better parts in the DIY world. and if you ask one or two more questions, you will find people using them simply because they don't know what is good enough for that particular application and they just default to the dumb approach (of using the best or better parts). |
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| gearheadgene |
| quote: | Originally posted by classd4sure
It's not done by math, but rather adjustment |
Really? That may be ok for a one of a kind circuit, or when prototyping, but not really good for production. Definitely seems ok for tinkering, though. But I'd really like to know with some accuracy if the thing is going to work, or be a smoke signal generator;)
Hey Fredos, I got the price from Digikey. Checked some other places too and got similar numbers. Arrow had them cheaper, but still $2. Maybe you could scalp those you get, and make some extra dough :) |
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| Pafi |
[OFF]| quote: | | 1pc price is around 4 bucks |
I don't really understand why only people from USA, Sweden, etc. always crying because of high prices! In opposition: people from Russia, China, India, etc. are ready to pay the price despite the fact that salaries are 1/10th of salaries in the above mentioned. How much does a USA man have to work for this kind of parts? 15 minutes? If this is too much, then what about the time spent with designing?
[ON]
| quote: | | not really good for production. |
Do you want to product something without building a prototype? :eek:
You can't set dead time properly without building a prototype. Times are not known exactly, especially since they are specified on datasheets with resistive load. Gate driver affects this very much, and temperature too a little.
As soon as protoype works, you don't have to compute dead time any more, you just have to copy values from working prototype! But devices are not exactly the same, so you have to add a little safety margin to avoid cross conduction, or have to do some (+/-5...10 ns) post production adjustment.
| quote: | | I'd really like to know with some accuracy if the thing is going to work, or be a smoke signal generator |
Then first set high dt, and decrease it slowly while checking drain current! |
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| gearheadgene |
Has anyone experimented with a current source/sink for driving the output mosfets? Seems like it could be quicker way to charge or discharge the gates. If q=integral[i(t)dt] , and i(t) is a constant source, the the equation simplies to:
q= I * integral(dt)
q = I * (tf - ti)
If so, then the gate charge time and discharge time are completely set by the current source.
On the other hand, if the gate driver is a voltage source, then the time to remove the charge is an exponential function and is dependant on the gate resistor value.
Anyone ever consider this before? |
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| fokker |
| quote: | Originally posted by gearheadgene
Anyone ever consider this before? |
it is an interesting idea. However, how do you decide when to stop it? presummably, you can time it: it requires a high precision timer that can account for device variations. a better way might be to monitor Vgs. it has the added advantage of automatically adapting to differences in gate charges from device to device.
If so, how much benefit do you get from a voltage driver? |
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| BWRX |
| quote: | Originally posted by gearheadgene
Has anyone experimented with a current source/sink for driving the output mosfets? |
BJTs are used as current sourcing and sinking components to drive mosfets. I don't know if anyone uses them to provide constant current charging and discharging of the gate, but you want the mosfets to turn on and off as quickly as possible for any class d topology. So it would make more sense to have drivers with a lot of surge current capability as opposed to constant current capability. The basic UcD schematic has a PNP device connected between the gate and source of the mosfets with a diode between the PNPs emitter and base and a resistor between the base and the source. This allows for a very fast turn off by directly discharging the gate capacitance through the local PNP device.
The mosfet gates are also charged by PNP devices and are current controlled as well. |
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| gearheadgene |
good question. Let's see. If it's removing charge, then it's easy. The current source is referenced to the source voltage, and can only suck out gate charge until it's all gone. No need to time anything.
When charging, that's a little bit of a thinker. Don't they spec the mosfets with some Vgs-on, say around 10V? If so, and my current source is referenced to 10V highere than the source, again I don't need to time it. |
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| BWRX |
| To turn on the mosfet you need the gate to source voltage, Vgs, to be at least the threshold voltage, Vth, above the source. To fully turn on the mosfet and reduce the on state resistance, Rdson, Vgs must be charged to a potential higher than Vth. |
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| gearheadgene |
While this is true, I believe that to turn on the transistor fully, the gate must be charged up to Qg. Qg is made up from the Qgs + Qgd. Not sure how Qgd plays into it, but Qgs is formed by the capacitor from gate to source. Given that, charging up the cap, Cgd, and therefore Qgd, brings the gate voltage up above the source - and is presumably above the threshold voltage. I think all this stuff is interrelated.
q=c*v
Vg(t) = 1/c * integra[(i(t)dt]
Anyway, I've read at least on definition of Qg to say it's the charge required to fully turn on the transistor. Also, the data sheets I looked at have a max value for Qg. Yet, Ciss only shows a typical value. Hmmm. That's what got me thinking about using a current source instead of voltage. Now the turn on time is easy to control.
stuff like that :) |
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| BWRX |
| The gate of a mosfet has a number of capacitances that need to charge. The result of these capacitances is what is known as the Miller plateau (part of the turn on charactersitic of a mosfet). The gate voltage will rise until it reaches the Miller plateau, which is the threshold voltage. When it reaches the threshold voltage the mosfet will turn on. The gate voltage then continues to increase. Qg is useful in that it can help you calculate gate drive requirements, but is not as helpful as an actual gate charge diagram. |
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| Pafi |
| quote: | | Has anyone experimented with a current source/sink for driving the output mosfets? Seems like it could be quicker way to charge or discharge the gates. |
It would be quicker, if you could reach higher current, and if current generator itself could be fast enough.
| quote: | | On the other hand, if the gate driver is a voltage source, then the time to remove the charge is an exponential function and is dependant on the gate resistor value. |
You mean q is exponential, and time is proportional to R, don't you? What is the big difference? With current source time is proportional to 1/I.
What is "ti", and why do you subtract it from tf? |
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| gearheadgene |
| quote: | Originally posted by Pafi
It would be quicker, if you could reach higher current, and if current generator itself could be fast enough.
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Good point. If this method could work, the current source/sink has to turn on and off. Maybe that's slower.
| quote: |
You mean q is exponential, and time is proportional to R, don't you? What is the big difference? With current source time is proportional to 1/I.
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No, q is fixed value at the initial time. Vgate = Vdrive[1-exp(-t/(rc))]
| quote: |
What is "ti", and why do you subtract it from tf? |
tf -ti : t(final) - t(initial), in other works delta T. |
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| fokker |
| the turn-off can be achieved automatically with a current source that maxes out at a desirable Vgs (for example 10 or 12v). |
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| classd4sure |
| quote: | Originally posted by fokker
the turn-off can be achieved automatically with a current source that maxes out at a desirable Vgs (for example 10 or 12v). |
really.. |
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| gearheadgene |
| quote: | Originally posted by fokker
the turn-off can be achieved automatically with a current source that maxes out at a desirable Vgs (for example 10 or 12v). |
Yep, that's how I see it, too :D |
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| Pafi |
| quote: | | No, q is fixed value at the initial time. Vgate = Vdrive[1-exp(-t/(rc))] |
I try to figure out what could you think about :
| quote: |
q= I * integral(dt)
q = I * (tf - ti)
|
This means that you think of q as Qfinal-Qinitial (=deltaQ). So wich of the two meaning is right?
BTW: in my country small letters refers to time dependant variables, not constants.
I think q should be the instantaneous charge of gate. If gate capacitor were linear, then gate charge during sourcing would be q(t)=vgate(t)/Cg=Vdrive[1-exp(-t/(R*Cg))]/Cg*heaviside(t).
tf usually means "fall time" in relation of MOSFETs. Please don't over-specify values without mention!
Question is still on: why do you think constant I drive is better then R charging? |
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| fokker |
let me give it a shot:
q=i*delta(t) - current is the rate at which charge is added or removed.
the amount of electronic charge removed / charged is proportional to the (constant) current and the time of discharge / charge.
the advantage, as far as i can tell, is rapid discharge or charge of the gate. In the traditional voltage driver, the charge is proportional to the voltage differential between the driver output and the Vgs. the driver output is usually 10v, and as the Vgs is brought up to 10v, the charging or discharging slows down.
With the current driver, you no long have that issue and the turn-on and turn-off of the mosfet is harder. as such, this idea may have a place in high power or high frequency applications. |
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| classd4sure |
| quote: | | With the current driver, you no long have that issue and the turn-on and turn-off of the mosfet is harder. as such, this idea may have a place in high power or high frequency applications. |
Maybe, for a very ideal current source. It would certainly have a place in high EMI applications. |
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| Pafi |
delta(q)=I*delta(t). (Initial charge should be taken account.)
| quote: | | the charge is proportional to the voltage differential between the driver output and the Vgs. |
The charge change rate is what proportional to this.
| quote: | | as the Vgs is brought up to 10v, the charging or discharging slows down |
- Why do you think this is the most important part of charging process? Above Vgs=6V MOSFETs are almost fully turned on. At normal drain currents the only effect of this low gate voltage is a little more Rds, but since this state lasts for ~100 ns, there is no significant drawback. End of discharging is a little more important: at a high chip temperature treshold voltage can be as low as 3,5V. Around this voltage a high Vds and significant Id occure coincidentally, so low voltage sinking capability is important feature of gate driver in order to avoid the long egzistence of this state. At a typical professional gate driver you can find that sinking capability at low voltage difference is higher then sourcing. (This is not specified most of the times, but you can check it by measurements.) But in both regions effective Cg is much lower then capacitance in miller plateau region, this because this is the most important part of switching. Some gate drivers specified especially in this region.
- Do you think the faster the FET switches the better the amp works? Some drawbacks are there.
- If you found some device wich can supply high current fast at low voltage difference, then why would you limit its current capability at high voltage difference?
You may say that constant I current is better then I(1-exp(-t/RC)). But is I better then 2...3*I(1-exp(-t/RC))? Because this is the technically realisable choice! |
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| classd4sure |
| Good stuff Pafi! Kill.. Kill! :smash: |
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| poobah |
There are simplifying assumptions that may have been overlooked here (except BWRX). During the Miller step the gate voltage remains nearly constant... about 4 Volts depending on your FET.
Now, because the gate voltage is constant, during the Miller step, the gate current is constant as well when fed by a standard totem pole driver and series resistor.
The Miller step really tells all about FET performance. Gate waveforms are always the first thing thing I look at.
:) |
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| classd4sure |
| quote: | Originally posted by poobah
There are simplifying assumptions that may have been overlooked here (except BWRX). During the Miller step the gate voltage remains nearly constant... about 4 Volts depending on your FET.
Now, because the gate voltage is constant, during the Miller step, the gate current is constant as well when fed by a standard totem pole driver and series resistor.
The Miller step really tells all about FET performance. Gate waveforms are always the first thing thing I look at.
:) |
Thing is, we're enduring theoretical drivers by people who haven't bothered to study the gate waveforms. So all the math and such counts for nothing more than alot of verbal masturbation, as Pafi pointed out in a slightly more wordy way. |
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| fokker |
| quote: | Originally posted by Pafi
delta(q)=I*delta(t). (Initial charge should be taken account.)
The charge change rate is what proportional to this. |
that's the same thing as we charge the gate from either fully discharged state (to a fully charged state) or from a fully charged state (to a fully discharged state). so the starting point doesn't matter in this particular discussion.
| quote: | Originally posted by Pafi
- Why do you think this is the most important part of charging process? |
why do you think I think this is the most important part of charging process?
| quote: | Originally posted by Pafi
- Do you think the faster the FET switches the better the amp works? Some drawbacks are there. |
absolutely. Anything has its positives and negatives.
| quote: | Originally posted by Pafi
- If you found some device wich can supply high current fast at low voltage difference, then why would you limit its current capability at high voltage difference? |
you wouldn't, short of for emission concerns. The point I was trying to make is that the traditional voltage drive's ability to charge or discharge a gate diminishes as Vgs approaches the output voltage of the drive.
That is not an issue with the current drive design, and that is an advantage for the current drive design and in some areas people may find it desirable.
another way to say it is that current drive design has its draw backs and may not be as desirable in other applications. |
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| fokker |
| quote: | Originally posted by classd4sure
Thing is, we're enduring theoretical drivers by people who haven't bothered to study the gate waveforms. So all the math and such counts for nothing more than alot of verbal masturbation, as Pafi pointed out in a slightly more wordy way. |
it is nice to see that people realize their own deficiencies every once in a while. |
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| Pafi |
| quote: | | that's the same thing |
I'm not sure what does "that" refer to. You quoted 2 different sentences.
| quote: | | so the starting point doesn't matter in this particular discussion. |
I don't see wich way did you define those quantities, so you must be right, but I can't find out what do you think. With usual signs the definiton of current is i=dq/dt, so at constant current delta_q=I*delta_t. You can't just forget a delta and keep the other!
| quote: | | why do you think I think this is the most important part of charging process? |
Because if not, then it's not an answer to my question. (However you were not the one I asked, so you may answered to a different question.)
| quote: | | The point I was trying to make is that the traditional voltage drive's ability to charge or discharge a gate diminishes as Vgs approaches the output voltage of the drive. |
Yes, this simple fact comes from Ohm's law, but I tryed to point that this has virtually no influence on performance of amplifier.
| quote: | | in some areas people may find it desirable. |
Yes, there is a possibility... |
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| gearheadgene |
funny stuff, you guys :cool: (classD, you cracked me up)
I was just thinking out loud here - seems to have stirred up a bees nest.
i = dq/dt comes from the integral form of current: q = integral[i(t) * dt]
If you take the derivative of that equation, you get:
dq = i(t) * dt, or
i(t) = dq / dt
I like the math for the integral form. As long as i(t) is constant (a constant current source), the equation is:
q = I * integral[dt]
or
q = I * (tfinal - tinitial)
(btw, yes I left out the initial condition, or q(0). It should be there, you are right. I was assuming inital charge of 0,)
For a voltage source, the math is different because i(t) is not constant. In that case, q = Cg*Vg, and Vg = Vs * [1-exp(-t/RgCg)]
where
Cg = gate capacitance
Rg = series resistance at the gate
Vg = gate voltage
Vs = gate driver voltage.
Plug that in and have some fun . . . that's why I asked the question about using constant current source. The math is sooo much easier. Hey, I am not an expert with mosfets, and yes, I am likely simplifying out of ignorance. Just posed a for instance, and seems to have stuck a nerve :devilr: |
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| fokker |
| quote: | Originally posted by Pafi
I don't see wich way did you define those quantities, so you must be right, but I can't find out what do you think. With usual signs the definiton of current is i=dq/dt, so at constant current delta_q=I*delta_t. You can't just forget a delta and keep the other! |
the most rigirous form is given by gearheadgene:
i(t)=d(q)/d(t).
as you integrate over t=[0...T], you have
Q(T)-Q(0)=I*[T-0], assuming constant current.
As I pointed out earlier, we usually start with the mosfet fully charged to fully discharged (Q(T)=0, Q(0)=q), or the reverse (Q(T)=q, Q(0)=0), the form degenerates to
q=I*T, which gearheadgene presented quite a while back.
it has nothing to do with exponential q, proportional time to R, or ohm's law. it is simply how current (I) is defined: the rate at which charges are added or removed.
BTW, the same math applies to a voltage drive. except there you cannot pop the current out of the integration as it is time-variant. |
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| fokker |
| quote: | Originally posted by gearheadgene
yes, I am likely simplifying out of ignorance. |
and that is science: you simply the non-important factors and focus on the ones you want to understand or to explain. so it is wrong to be complete, to be 100% correct and to insist on modeling the reality perfectly. A lot of times we just wanted to be directional correct and let the experiment verify the rest.
and it wouldn't surprise me if this current drive idea is used somewhere. |
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| gearheadgene |
| quote: | Originally posted by TOINO
... |
man, that is hillarious!!!!! |
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| fokker |
the first article is quite interesting. It essentially confirmed the issue that gearheadgene was trying to address, and the solution is remarkably simple and similar to gearheadgene's proposal: the inductor is effectively the constant current source gearheadgene was talking about. when the "gate driver" turns off, the current going throught he inductor wouldn't stop right away and it went to charge / discharge the mosfet.
the benefit: a hardened turn-on or turn-off of the mosfet reduces the conduction loss in the switcher, a major benefit accounting to the author.
Those guys at VT are a major player in the power electronics mkt. |
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| Pafi |
fokker!
Read it again with more attention! This article is not about conduction loss in power MOSFET, but conduction loss in gate driver! Our original discussion has no deal with gate drive loss.
| quote: | | the most rigirous form is given by gearheadgene: |
I talked about Your equation. Please don't confuse everything!
Your original equation: q=i*delta(t)
Your new equation: Q(T)-Q(0)=I*[T-0]
1.:Q(T)=0, Q(0)=q
2.:Q(T)=q, Q(0)=0
And this is what I said! In the left side of equation there is Q(T)-Q(0) wich is deltaq, not q. In first case deltaq=-q. Big difference!
Lets substitute Q(T) and Q(0) in equation:
1.:-q=I*[T-0]
2.:q=I*[T-0]
You can see that q=I*T is correct only in case of zero initial charge as gearheadgene admitted already.
| quote: | | it has nothing to do with exponential q, proportional time to R, or ohm's law |
Please don't deny something wich is not stated by anybody! I mention Ohms law in connection with a completely different statement.
gearheadgene!
Thank you your explanation, now it's correct and clear. | quote: | | The math is sooo much easier. |
I think theoretical simplyness of a method in a simplified modell is not enough reason to change real life gate charge method when realisation is already difficult, and desired method is even more difficult to achieve. But of course you can play with it. |
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| gearheadgene |
| maybe I should change my nickname to "lunkhead":smash: |
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| fokker |
| quote: | Originally posted by Pafi
You can see that q=I*T is correct only in case of zero initial charge as gearheadgene admitted already. |
and you can see that is precisely what I said earlier on, and it is precisely the case we are concerned about, unless you intend to charge up your mosfet to 25% or discharge it down to 45%. |
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| fokker |
| quote: | Originally posted by gearheadgene
maybe I should change my nickname to "lunkhead":smash: |
or luckhead, :).
Conceptually, there is a lot of advantages to current-driven circuitry: they are far less suspect to noise / interference, and they tend to be very fast. Just look at the popularity of current-mode pwm controllers or current-feedback amps.
so it wouldn't surprise me if the future is really in some sort of current drivers for high-end class d amps - it is happening in the analog world now. |
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| poobah |
This is getting really weird... stupid really
Current is not any faster than voltage... EMF is what causes current in the first place. One circuit may be faster than the other... that is a question of topology.
:rolleyes: |
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| anatech |
Inductance.
That's all I'm going to say.
Nah, I lied. For voltage the issue is capacitance. Therefore a happy middle ground. Ground. Hmmmm. That'll be an issue with current systems. It's impedance is now on the same order. Not good.
-Chris |
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| Pafi |
| quote: | | quote: | | in case of zero initial charge |
and you can see that is precisely what I said earlier on, and it is precisely the case we are concerned about, |
Have you written this?
Isn't Q(0) initial charge? I think so. Is q=0? Doubtfully.
Let's finish this!
gearheadgene!
Your nick is OK! You are learning, trying. No problem. I'm doing it too. |
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| fokker |
| quote: | Originally posted by Pafi
Have you written this?
Isn't Q(0) initial charge? I think so. Is q=0? Doubtfully. |
I am at a loss how to explain things this simple. But we are talking about charging and discharging.
And in this particular instance, we are talking about discharging. So the mosfet at t=0 is fully charged and Q(t=0)=q. The end state is to have it fully discharged, Q(t=T)=0. so in the whole discussion, q is never zero. It reprsents the amount of charge that turns a mosfet fully on, and consequently it is the amount of charge the driver needs to remove from the gate to turn the mosfet (from fully on) to fully off.
Let me know what else you have an issue with. |
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| poobah |
That link is much better than the links you posted earlier... The second link in post 53 has very poorly drawn waveforms... very misleading in fact.
The duration of the Miller step matches the duration of the transistion in drain voltage. The author is correct is in stating that it is not in step with the current waveform... I can't agree with his windbag mathematics.
;) |
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| TOINO |
Right!
I will go to their site again and in the bottom left corner “Rate This Article” = 1 (Worse)
:D |
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| classd4sure |
Yup, that recent link has a few neat ideas.
Don't forget though those "active" drivers (never seen a passive one) are all for motor control... much slower frequencies.
I'd imagine the ideas carry over very well, but I think the real issue is with dead time control. |
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| TOINO |
Take a loock at page 56.
I like that one!:up:
hei classD... you must found an avatar for yourself ! |
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| fokker |
| quote: | Originally posted by classd4sure
Don't forget though those "active" drivers (never seen a passive one) are all for motor control... much slower frequencies.
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looks like you haven't tried transformer based gate driver. and a optical coupler or magnetic coupler may belong in that category as well.
yeah, the whole PWM thing came out of motor control, and it hasn't prevented its use in lowly audio as well. |
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| classd4sure |
| quote: | Originally posted by fokker
looks like you haven't tried transformer based gate driver. and a optical coupler or magnetic coupler may belong in that category as well.
yeah, the whole PWM thing came out of motor control, and it hasn't prevented its use in lowly audio as well. |
If you'd consider those passive, I dare say you haven't looked at them close enough. |
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| fokker |
| quote: | Originally posted by classd4sure
If you'd consider those passive, I dare say you haven't looked at them close enough. |
maybe you should define "passive" for us. hopefully it is different from your definition of efficiency. |
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| fokker |
| quote: | Originally posted by TOINO
Take a loock at page 56.
I like that one!:up: |
that in essence is gearheadgene's current-drive idea: using a current source to speed up the turn-on / turn-off of the switcher.
On further reflection, it is probably true that most drivers are indeed current-driven. Taken the discrete driver in the philipse ucd for example. The lower pnp transistor is activitated by the voltage signal (off the resistor on its b-e junction), and drains the charge off the gate. it may not be a constant current but the removal of the charge is certainly less dependant than the voltage differential between the voltage input signal (which activates the upper pnp) and Vgs. |
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| classd4sure |
| quote: | Originally posted by fokker
maybe you should define "passive" for us. hopefully it is different from your definition of efficiency. |
Who is us? Are you performing for an audience, or do you always speak of yourself in the plural form. I'd find either very believable.
Where do you come up with "efficiency". I believe the term you were struggling to understand was application specific. Get your facts in order.
Figured out your gate driver IC yet or have you just been too busy blowing wind. |
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| fokker |
| quote: | Originally posted by classd4sure
Who is us? Are you performing for an audience, or do you always speak of yourself in the plural form. I'd find either very believable.
Where do you come up with "efficiency". I believe the term you were struggling to understand was application specific. Get your facts in order.
Figured out your gate driver IC yet or have you just been too busy blowing wind. |
thanks for your non-answer. I believe the question for you is which part within a transformer would you consider active? the copper wires? the laminated steel core? the epoxy packaging? the shield / shoe or the screws?
alternatively if you have defined "activie" differently, as you defined "efficiency", maybe you can lay it out for us in the open so that we can follow you.
As to "efficiency", I will let you memory come to you first. |
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| classd4sure |
Ok, I'll pretend for a moment that it's a sincere question.
The transformer simply provides galvanic isolation. It's not a driver in itself. That's why you might say it's a transformer "based" driver.
The active aspect, is obviously what drives the transformer. Further to which, if you actually do a little reading on such a driver, you'll see there's some active circuitry in the secondary /gate side of it as well, which can be anything from a simple turn off pnp transistor, to a pass transistor+turn off.
Now, if I have to explain the same for an opto coupler you're just going to look silly, so I won't bother.
There's no such thing as a passive driver, "driver" very much implies active devices.
Good 'nuff? |
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| fokker |
| quote: | Originally posted by classd4sure
The transformer simply provides galvanic isolation. It's not a driver in itself. That's why you might say it's a transformer "based" driver.
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Wow, that's nice. so how's that active carbon film resistor or active film capacitor doing in your class D amp? they must be pretty active too, :) |
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| classd4sure |
| quote: | Originally posted by fokker
Wow, that's nice. so how's that active carbon film resistor or active film capacitor doing in your class D amp? they must be pretty active too, :) |
Now you've just leaped well beyond the boundary of being completely absurd, and aren't worth the time. |
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| fokker |
| quote: | Originally posted by classd4sure
Now you've just leaped well beyond the boundary of being completely absurd, and aren't worth the time. |
we are all in the spirit of your active transformers. |
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| Netlist |
Guys, refrain from picking on each other. It's a plain waste of time and IIRC this is not the first warning.
/Hugo :cop: |
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| Pafi |
| quote: | | the mosfet at t=0 is fully charged |
So initial charge is NOT zero. Previously you said that it IS zero.
There is nothing to talk about. You made a mistake. You should admit it, and step over.
| quote: | | that in essence is gearheadgene's current-drive idea: using a current source to speed up the turn-on / turn-off of the switcher. |
No, it's not. It is not simply a current driver, but a complex controlled driver consist of two independent controlled current source, a miller plateau sensor, plus a regular gate driver, and the goal is not to speed up switching, but to reduce EMI and overshoot by reducing di/dt. This is a very useful and interesting article, but very different from gearheadgene's idea. As you can see on next page this gate charging method can be achieved without current generator, with conventional, but variable driver stage. Current generator is only an intermedial theoretical step. |
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| fokker |
| quote: | Originally posted by Pafi
So initial charge is NOT zero. Previously you said that it IS zero.
There is nothing to talk about. You made a mistake. You should admit it, and step over. |
I thought I had said from very beginning that the case we are concerned about is to get the mosfet fully charged up or fully discharged off. so the initial state is either Q(0)=q (fully charged up), or Q(0)=0 (fully discharged off).
so depending on which case you wanted to talk about, the initial charge is one of the two.
Let me know what else you have an issue with. I am really at a loss as to how one can have an issue with that. |
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