FET Gate Driver w/ BJTs

[dknguyen]

Hi. I am trying to put together a PNP/NPN totem pole FET gate driver since I added an isolator and theh bootstrap IC doesn't work anymore (or is more than needed).

THe problem I have is that the gate current pulse needed (that the bootstrap IC could provide) is 3A. Now looking at BJTs, beta is drops like a rock at the collector currents I need (for any BJT of "reasonable size"). THe datasheet charts don't even make up to the currents I need.

Am I missing something here? Or are the BJTs inside the boostrap IC far larger than the ones found in discrete packages? Because I kind of doubt that.

 

[Roff]

Do you really need 3 amps?

 

[dknguyen]

Yeah. 252nC gate charge and even if I switch at 50ns which is what 3A gets me, I still get 3W of switching losses. The other 4W comes from I^2R.

I found some really high current BJTs in small packages from Zetex though meant for the purpose. I'm still slightly suspicous though.

 

[Roff]

You must be switching at a fairly high frequency.

 

[dknguyen]

You must be switching at a fairly high frequency.

THe switching frequency is 30kHz due to the sensorless BLDC scheme being used. I'm currently trying to change it to use floating supplies so the sampling time is independent of the switching frequency. SHould help to reduce losses, but it won't change the frequency by much.

 

[Roff]

You must be switching at a fairly high frequency.

 

[Hero999]

Roff, you've just said that. You must have pressed the reply button twice.

 

[Styx]

Roff, you've just said that. You must have pressed the reply button twice.

no its just dknguyen maths doesn't add up

so... " You must be switching at a fairly high frequency."


252nC and switched in 50ns will require a gate-resistor of X-ohms ok BUT since you are switching a FET it only draws current during turn-ON and turn-OFF for ... 50ns

To calcuate the true wattage of the gate resistor needed you need to calculate the RMS current AND that is dependent on the switching freq, in this case 30Khz

How did you calc the gate charge for this FET? For a 1200Volt, 1000Amp IGBT's (and even then the gate-capacitance is that high cause there are 8 in parallel in the package) I get a gate-charge 5x that, so something not right there

EVEN switching such devices at 20KHz with a peak gate current of 15A, using +/-15V to switch the gate as well (and devices hitting Millar in 500ns and gate-current negligiable in 1us) I only need 2W worth of gate resistance

you seem to have gone astray in yr maths
please provide the FET you are using as well as the drive circuit (ie what voltage you are pulling the gate-source between)

I do these calculations weekly so I can give it a go for you
I have a nice excel spreadsheet that chucks out the numbers for me so give me some numbers



BUT to answer your question BJT's will be fine, put something like 6uF worth of capacitance near the BJT's to provide a source for the current-spike at turn-ON and turn-OFF. I use 44uF for my cct's to give you an idea

 

[Roff]

Styx, I think he means FET power dissipated due to the finite rise time. As I'm sure you know, slower rise and fall times translate to higher transistor dissipation.

 

[dknguyen]

no its just dknguyen maths doesn't add up

so... " You must be switching at a fairly high frequency."


252nC and switched in 50ns will require a gate-resistor of X-ohms ok BUT since you are switching a FET it only draws current during turn-ON and turn-OFF for ... 50ns

To calcuate the true wattage of the gate resistor needed you need to calculate the RMS current AND that is dependent on the switching freq, in this case 30Khz

How did you calc the gate charge for this FET? For a 1200Volt, 1000Amp IGBT's (and even then the gate-capacitance is that high cause there are 8 in parallel in the package) I get a gate-charge 5x that, so something not right there

EVEN switching such devices at 20KHz with a peak gate current of 15A, using +/-15V to switch the gate as well (and devices hitting Millar in 500ns and gate-current negligiable in 1us) I only need 2W worth of gate resistance

you seem to have gone astray in yr maths
please provide the FET you are using as well as the drive circuit (ie what voltage you are pulling the gate-source between)

I do these calculations weekly so I can give it a go for you
I have a nice excel spreadsheet that chucks out the numbers for me so give me some numbers



BUT to answer your question BJT's will be fine, put something like 6uF worth of capacitance near the BJT's to provide a source for the current-spike at turn-ON and turn-OFF. I use 44uF for my cct's to give you an idea

I know the power on the gate drive BJT is low, but I was having problems finding one that could actually had a large enough beta value at higher collector currents. THe signal ones I was looking at had betas that dropped to about 1 at 1A (but I found some that don't now from Zetex. It's like oversized BJTs in an undersized packaged for high current, low powwer apps)

I read the gate charge off of the datasheet and used Q=IT where T is the transition time. I'm using 15V gate drive switching at 30kHz with a rise time of 50-100ns. Right now I have two 4.7uF decoupling caps for each driver. Each gate driver is being powered by a 15V@66mA floating power supply.

https://www.electro-tech-online.com/custompdfs/2008/02/irf2804s-7p.pdf
I am using the Figure 2 drive circuit:
https://www.electro-tech-online.com/custompdfs/2008/02/62207.pdf
except there is a Q1 emitter resistor and Q2 emitter resistor that are used to control the gate resistance and shoot-through, but they may be zero. I haven't gone through gate resistance calculations yet.

About that Figure 2 circuit:
I also finally figured out why the NPN is on the high-side and the PNP Is on the low-side- to keep the thing in quasi-saturation for faster switching. I remember some time ago I was telling a new member asking about his gate driver and he had them this way and I told him he needed to flip them around. THat was MOSFET half-bridge thinking and although it would have worked, the BJTs would have been in saturation so the efficiency would have been higher but the switching would have been slower, and the pulsed nature of the driver makes the efficiency not as important and would have reduced performance. Sorry!

ALSO:
Is there some confusion going about whether W means watts or ohms? I mean watts when I use W, and I use R or "ohms" for resistance. I've seen my power prof use W as ohms though and Styx seemed to just do that to. So when I say 3W and 4W, I mean the power dissipated due to the transition time is 3W at 50ns and the I^2R losses are 4W at full load.

 

[Nigel Goodwin]

Is there some confusion going about whether W means watts or ohms? I mean watts when I use W, and I use R or "ohms" for resistance. I've seen my power prof use W as ohms though and Styx seemed to just do that to. So when I say 3W and 4W, I mean the power dissipated due to the transition time is 3W at 50ns and the I^2R losses are 4W at full load.

No, W means watts - to be kind, perhaps your professor is using the lower case omega, which looks like a 'babys bum' as we were told at college!

See https://en.wikipedia.org/wiki/Omega

 

[Styx]

I know the power on the gate drive BJT is low, but I was having problems finding one that could actually had a large enough beta value at higher collector currents. THe signal ones I was looking at had betas that dropped to about 1 at 1A (but I found some that don't now from Zetex. It's like oversized BJTs in an undersized packaged for high current, low powwer apps)

Yer Zetex have some fantastic discrete components. That or use a good FET drive chip, I quite like the IR4426 (and its non-inv equiv) - although in this case it would be too low a power

I read the gate charge off of the datasheet and used Q=IT where T is the transition time. I'm using 15V gate drive switching at 30kHz with a rise time of 50-100ns. Right now I have two 4.7uF decoupling caps for each driver. Each gate driver is being powered by a 15V@66mA floating power supply.

https://www.irf.com/product-info/datasheets/data/irf2804s-7p.pdf
I am using the Figure 2 drive circuit:
https://www.farnell.com/datasheets/62207.pdf
except there is a Q1 emitter resistor and Q2 emitter resistor that are used to control the gate resistance and shoot-through, but they may be zero. I haven't gone through gate resistance calculations yet.

mmm I tend to work with Q=CV and use both that figure (and the gate-drive voltage) as well as looking at the charge curve for the voltage & charge for the end of the millar plateux.

With a capacitance AND a switching time needed I can work out the peak current that needs to be injected into the gate, and from that the gate resistors can be set

About that Figure 2 circuit:
I also finally figured out why the NPN is on the high-side and the PNP Is on the low-side- to keep the thing in quasi-saturation for faster switching. I remember some time ago I was telling a new member asking about his gate driver and he had them this way and I told him he needed to flip them around. THat was MOSFET half-bridge thinking and although it would have worked, the BJTs would have been in saturation so the efficiency would have been higher but the switching would have been slower, and the pulsed nature of the driver makes the efficiency not as important and would have reduced performance. Sorry!

The thing is this part has a rise time of 150ns, if you are trying todo it in 50ns you are more then likely going to setup a pearce oscilator so be warned. NOTE the operating area of the FET as well!!!

ALSO:
Is there some confusion going about whether W means watts or ohms? I mean watts when I use W, and I use R or "ohms" for resistance. I've seen my power prof use W as ohms though and Styx seemed to just do that to. So when I say 3W and 4W, I mean the power dissipated due to the transition time is 3W at 50ns and the I^2R losses are 4W at full load.

na (I use "R" for resistance, not "W" as a std-keyboard lazy omega), the confusion (asumption on my behalf sorry) was unknown FET and for 3W worth of switching and using:
W = 1/6*V*I*Ts*Fs lead to some more real voltage and some more real amperage and well kinda didn't think that was going to happen
Seeing the FET in question however

 

[dknguyen]

No, W means watts - to be kind, perhaps your professor is using the lower case omega, which looks like a 'babys bum' as we were told at college!

See https://en.wikipedia.org/wiki/Omega

Nope, that's what he did for the homework. It took me a while to figure out that the load was 10R and not 10W (even though he typed it out as 10W) because the problem was impossible to solve otherwise.

 

[dknguyen]

mmm I tend to work with Q=CV and use both that figure (and the gate-drive voltage) as well as looking at the charge curve for the voltage & charge for the end of the millar plateux.

I read some stuff about how the capacitance can be misleading so I try and just stick to the using Q rather than C. Something about C changing or whatever. I think I also read that if capacitance is given and Q wasn't then it's best to take C, calculate Q from it and then use that to recalculate C at your operating voltage.

Yer Zetex have some fantastic discrete components. That or use a good FET drive chip, I quite like the IR4426 (and its non-inv equiv) - although in this case it would be too low a power

The thing is this part has a rise time of 150ns, if you are trying todo it in 50ns you are more then likely going to setup a pearce oscilator so be warned. NOTE the operating area of the FET as well!!!

THanks. I'll check that out. The continuous load is around 20-30A. I just assumed 50A for maximum. I also assumed max voltage for the circuit iself. In practicality the battery voltage is probably going to be half that so the losses are quite a bit less.

na (I use "R" for resistance, not "W" as a std-keyboard lazy omega), the confusion (asumption on my behalf sorry) was unknown FET and for 3W worth of switching and using:
W = 1/6*V*I*Ts*Fs lead to some more real voltage and some more real amperage and well kinda didn't think that was going to happen
Seeing the FET in question however

Why do you divide by 6? THe switching losses formula I use is:
P = (1/2)(V*I)f(ton+toff), the average of the on/off state power averaged out over time and frequency.

I had originally gone with the Direct Back EMF sensing scheme, but the power losses (since the sampling speed is related to the PWM frequency which is related to motor speed) was increasing too much. Lately I've been trying to change that scheme to the regular virtual ground formed by 3 resistors, except I am making a sensing circuit that floats at that neutral point rather than regular ground so I don't lose sensitivity and deal with the common-mode noise of the changing netural point a bit better. Then I can use whatever switching speed I want while grabbing BEMF data whenever I want. Requires more floating supplies though, but whatever.

IT would have been dead easy if there was a DC transformer to differentiall re-reference the neutral-to-phase BEMF reading for the electronics so no attenuation, filtering would be needed, or worrying about voltage ratings or accuracies of differential amplifiers. But that'd be too good to be true.

 

[Nigel Goodwin]

Nope, that's what he did for the homework. It took me a while to figure out that the load was 10R and not 10W (even though he typed it out as 10W) because the problem was impossible to solve otherwise.

OK, so he's a total idiot then!

 

[Styx]

I read some stuff about how the capacitance can be misleading so I try and just stick to the using Q rather than C. Something about C changing or whatever. I think I also read that if capacitance is given and Q wasn't then it's best to take C, calculate Q from it and then use that to recalculate C at your operating voltage.

Thats why you don't use the stated capacitance BUT use the gate-charge curves for your operating voltage

Qg Total Gate Charge ––– 170 260

This is from the 2nd page and is ONLY valid for: Id = 160A, Vds=32V, Vgs=10V

A better curve to use is: Fig6
**broken link removed**

From this you can see how much charge needs to be transfered for a given Gate-source voltage & gate-drain voltage. My main focus is getting out of the millar plateux as fast as possible (during this period the device is in its linear stage AND it this part of the curve where the gate capacitance looks infinite). Its also the period where the FET/IGBT can go unstable.

Why do you divide by 6? THe switching losses formula I use is:
P = (1/2)(V*I)f(ton+toff), the average of the on/off state power averaged out over time and frequency.

Sorry equation wrong (looked at the wrong cell in my excel sheet). I use

**broken link removed**

P = (1/3)*(V*I)*ts*F
Where P == power dissipated in that one switching event
V = peak voltage switching to/from
I = peak voltage switching into/out of
ts = switching time (in this case 50ns)
F = switching freq (in Hz)

its 1/3 due to the waveform of the voltage and the current changing at the same time (see assumptions)

v(t) = (-V/ts)*t
i(t) = (I/ts)*t

so integrate V*I by t over ts:


int( (-VI)/(ts^2)t^2,t,0,ts)
gives E = (1/3)*I*V*ts
so multipying by F will give you the power for all turn-ON (if that ts was the turn-ON time)

The assumptions made in this are
1) that the voltage falls/rises linearly (in practice it doesn't)
2) that the current rises/falls linearly (in practive it doesn't)
3) it neglects reverse-recovery of any freewheel diodes at turn-on
4) neglects any voltage overshoot due to stray inductance at turn off
4) neglects the delay in voltage falling w.r.t. current rising at turn-on & the delay with current falling w.r.t. voltage rising at turn-off

These assumptions are valid for low amp/voltage operation (you should be ok). Even at higher voltages and currents these assumptions can lead to a switching loss that, while not spot-on, isn't too far out.

I had originally gone with the Direct Back EMF sensing scheme, but the power losses (since the sampling speed is related to the PWM frequency which is related to motor speed) was increasing too much. Lately I've been trying to change that scheme to the regular virtual ground formed by 3 resistors, except I am making a sensing circuit that floats at that neutral point rather than regular ground so I don't lose sensitivity and deal with the common-mode noise of the changing netural point a bit better. Then I can use whatever switching speed I want while grabbing BEMF data whenever I want. Requires more floating supplies though, but whatever.

IT would have been dead easy if there was a DC transformer to differentiall re-reference the neutral-to-phase BEMF reading for the electronics so no attenuation, filtering would be needed, or worrying about voltage ratings or accuracies of differential amplifiers. But that'd be too good to be true.

mmm getting interesting, capturing the backEMF can be tricky and must be sync with your PWM and modulation depth (so you sample during periods of no switching). Still there is only two instances when you would go to the effort of trying to measure the BackEMF
1) space-vector modulation of a BLAC machine
2) a means to sense speed

#1 is bad, there are better ways to derive it indirectly
#2 alot of hassle, might as well strap an encoder disk and count pulses

 

[Roff]

I don't understand why gate driver power dissipation is not simply F*Q*V when the charge is given, or F*C*V^2, where only total equivalent capacitance is given. F is the switching frequency, Q is the gate charge, and V is the p-p swing of the gate voltage. If there is a discrete series resistance between the driver and the gate, the dissipation would be distributed proportionally between that resistance and the internal resistance of the driver.
Can someone comment on that?

 

[dknguyen]

I don't understand why gate driver power dissipation is not simply F*Q*V when the charge is given, or F*C*V^2, where only total equivalent capacitance is given. F is the switching frequency, Q is the gate charge, and V is the p-p swing of the gate voltage. If there is a discrete series resistance between the driver and the gate, the dissipation would be distributed proportionally between that resistance and the internal resistance of the driver.
Can someone comment on that?

THe gate driver power dissipation is something like that isn't it? WHat I have been talking about is the switching losses due to the MOSFET passing through the intermediate on/off region in a finite amount of time.

mmm getting interesting, capturing the backEMF can be tricky and must be sync with your PWM and modulation depth (so you sample during periods of no switching). Still there is only two instances when you would go to the effort of trying to measure the BackEMF
1) space-vector modulation of a BLAC machine
2) a means to sense speed

#1 is bad, there are better ways to derive it indirectly
#2 alot of hassle, might as well strap an encoder disk and count pulses

I am not measuring back EMF, technically. I am just detecting when it becomes zero for commutation purposes. I'm getting speed by just averaging the time between commutations, not from measuring back EMF strength.

THe original sensing scheme I used was this one:
https://www.electro-tech-online.com/custompdfs/2008/02/T.pdf
(Page 26-33, for the most concise explanation)

WHere you detect it referenced to ground so there is no common-mode switching noise or attenutation so you can get very high sensitivity. But the sample rate is dependent on PWM frequency. You need at least 3 off-times per commutation period to accurately detect the zero crossing so the PWM frequency has to be at least 3x higher than the highest electrical frequency of the motor. 9800Hz - >~30kHz. That's considerably faster than what I need mechanically and it makes for quite a bit of switching .losses

Since then, I have changed my circuit to use the tranditional "forming a "virtual ground" method using 3 Y-connected resistors in parallel with the BLDC's delta-connected windings" (page 13). But instead of using a diff-amp to measure the voltage between the neutral and each phase (where I would need attenutation since the diff-amp is referenced from ground and filtering due to common mode switching noise of the neutral) I used a floating supply to float all the sensing circuitry at that neutral point so then I can clamp the voltage rather than needing to attenuate it in order to detect the zero crossing.

I'm not sure if it will cause problems for the floating supply though or if any switching noise will still be experienced since the neutral is still jumping up and down during switching, just the entire circuit is floating at that neutral point now.

 

[Roff]

WHat I have been talking about is the switching losses due to the MOSFET passing through the intermediate on/off region in a finite amount of time.

That's what I was originally talking about, but styx went off on a tangent about gate resistor dissipation:

To calcuate the true wattage of the gate resistor needed you need to calculate the RMS current AND that is dependent on the switching freq, in this case 30Khz

Then he posted an equation,

W = 1/6*V*I*Ts*Fs

but as far as I can tell, he never said what it applied to. Maybe I missed that part. I thought the equation was for gate resistor dissipation, since that was the topic of his previous post.

 

[dknguyen]

W = 1/6*V*I*Ts*Fs

is for the power MOSFET dissipation due to finite switching time.

He misread his datasheet though and it is actually W = 1/3*V*I*Ts*Fs