delta(q)=I*delta(t). (Initial charge should be taken account.)
The charge change rate is what proportional to this.
- 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!
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.
🙂
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.
🙂
poobah said:
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.
Pafi said:
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.
Pafi said:
why do you think I think this is the most important part of charging process?
Pafi said:
absolutely. Anything has its positives and negatives.
Pafi said:
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.
classd4sure said:
it is nice to see that people realize their own deficiencies every once in a while.
I'm not sure what does "that" refer to. You quoted 2 different sentences.
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!
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.)
Yes, this simple fact comes from Ohm's law, but I tryed to point that this has virtually no influence on performance of amplifier.
Yes, there is a possibility...
funny stuff, you guys 😎 (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
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
Pafi said:
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.
gearheadgene said:
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.
TOINO said:
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.
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.
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.
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.
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.
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.
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.
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.
Pafi said:
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%.
gearheadgene said:maybe I should change my nickname to "lunkhead"
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.
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.
🙄
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.
🙄
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