gearheadgene said:
Not sure what you think is cheap, but 1pc price is around 4 bucksI'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
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).
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).
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😉classd4sure said:
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 🙂
[OFF]
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]
Do you want to product something without building a prototype? 😱
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.
Then first set high dt, and decrease it slowly while checking drain current!
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]
Do you want to product something without building a prototype? 😱
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.
Then first set high dt, and decrease it slowly while checking drain current!
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?
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?
gearheadgene said:
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?
gearheadgene said:
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.
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.
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.
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.
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 🙂
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 🙂
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.
It would be quicker, if you could reach higher current, and if current generator itself could be fast enough.
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?
Pafi said:
Good point. If this method could work, the current source/sink has to turn on and off. Maybe that's slower.
No, q is fixed value at the initial time. Vgate = Vdrive[1-exp(-t/(rc))]
tf -ti : t(final) - t(initial), in other works delta T.
the turn-off can be achieved automatically with a current source that maxes out at a desirable Vgs (for example 10 or 12v).
I try to figure out what could you think about:
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?
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.
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.
Maybe, for a very ideal current source. It would certainly have a place in high EMI applications.
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