Haha that is quite a tongue twister there...so you do not get trapped in the traps of bootstraps?

Wow thanks man

You you are right, during the delay the MOSFET is just waiting, so no switching losses.

Thank you! I know the Slayer Exciter, and it is not a metal band from the 80's. I have no experience but I expect you can use the low side MOSFET driver push pull part to make your slayer exiter more efficient. I did see that for Tesla coils it is key to drive it pulsed, e.g. with 5-10Hz PWM with low on cycle (5% or so) to get nicer sparks and prevent burning your MOSFET. You can use the arduino for that (or a 555....) So yes I am interested and if I find the time I will build one myself 🙂

Did you check this video, seems like a robust design to me. kzfaq.info/get/bejne/rq-doNp9mrO2YKM.htmlsi=BMNipvCvJpJh1Tgh

Thanks Rick_er2481! More will come!

Thank you Wliterow!

haha cool, yes these component shortages...you could still buy old stuff like transistors and NE555's.

Thank you, that is a great appnote, found it: educypedia.karadimov.info/library/pwm%20vegfok%20philips%20UM10155_1.pdf

The "UM1055" app note circuit provides no provision against C5+C6 overdischarge so insufficient gate drive of the top mosfet, e.g. when the top transistor is requested to be ON for way too long (in case of the amplifier, when the input signal overdrives it towards clipping on the top rail - so pretty much normal operation unless the input has properly set and even supply tracking limiter). It is even worse as the Q8 draws current from the C5+C6, so is actively discharging them, so the critical maximum ON time is not that long (barely covering the normal switching frequency).

@@annaplojharova1400 So , we need simple forward one-transistor isolated converter to supply high side - if you want to turn on high side during long time.Or we can use a usual half-bridge gate driver IC and supply high side (without bootstrap diode) with mentioned one-transistor forward self-oscillating isolated DC/DC converter.

@@andrewandrosow4797 You can use the simple bootstrap, but the predriver has to provide protections that ensure the bootstrap capacitor gets recharged (e.g. by forcing an off pulse in the class D amplifier or the buck converter examples) and/or the transistor gets shut down properly if it can not handle the current with the available gate drive (in fact this could be done by a simple 2-transistor, few resistors and a capacitor addon circuit monitoring the ON state Vds, if we want to stay with "simple" discretes). Or use one of the many available gate driver ICs, which do have at least the bootstrapped supply monitoring build in (that is enough, when either the current is monitored elsewhere, and a direct short circuit is not an issue, like e.g. in an active speaker box amplifier). Here I see the gate driver IC just as the simplest practical solution, offering at least the minimum protection (essentially against input undervoltage for the buck converter, or the signal overdrive for the amplifier). There are also ICs that provide the overload protection (either the large Vdson, or current sense, or both).

@@annaplojharova1400 I had made in past the class D amplifier using this application note.There also was simplest protection against short-circuit. The amplifier works well. I am totally agree - an IC gate driver much reliable and robust that discrette driver.Mosfets much resistant against avalanche breakdown when voltage becomes high.

thank you!

Interesting, do you mean a circuit as mentioned in this discussion (2nd schematic, scroll down). Never tried that! electronics.stackexchange.com/questions/296879/logic-level-converter-using-transistors

@@smartpowerelectronics8779 Pretty much yes, but as I’ve said, I tend to use a low-power MOSFET for this, just for the convenience (no need for a base current limiting resistor). Gate is connected directly to Vdd (+5V or +3.3V, whatever the chip is powered by), and source to the output of the chip. The typical Vgs ON of these FETs is abt. 1.5-2V, so it needs to be either a rail-to-rail output or open drain. Might not work with the old TTL logic. HIGH level at the output of the chip turns the power MOSFET ON, so there’s no inversion like with the common-emitter level shifter.

My pleasure Carlosmf6954!

Optocouplers are a solution, I do wonder what the propagation delay of the opto is, maybe you will need some special fast types.

I use Simetrix, the free version.

@@smartpowerelectronics8779 Thank you!🙌

Great to hear you learned something william4749. For practical use I do recommend an IC, much easier to build and it also has some protection against low voltage etc. ....but circuit2 works OK

Great to hear that, the discrete cheap components can really show how this circuit works, success with studying transistor operation !

There is no such thing as a stupid question, every question may make you smarter. Let me try to explain: 1 - the 1 kΩ resistor connected to the gate "Pulls" the gate to +12V of the capacitor 2 - this way the MOSFET G0S has ~ 12V volt and the MOSFET is ON 3 - If you switch on the BC547 a current will flow from the 12B via the 1 k, and the BC547 this current "pulls the gate down" if (I visualize a schematic like elastic bands connecting the parts ) 4 - If the current is too small (say 0.0001mA) nothing will happen, if the current is big (say 1A) the gate will be pulled down very fast. In this case the current is 20mA, 5 how to know if 20mA is enough? well: 12V/1kΩ is 12mA, it must be more than that for it to "win" the pulling, so 13mA would be just ok. To be sure we set it at 20mA. (also no too much to keep your transistor from breaking )

They are different indeed. delay will not cause any switching losses because the MOSFET is just waiting. The speed is very important because during the slopes/transitions you have current and voltage at the same time = power loss...

The IC version is for sure the easiest and very robust. However understanding the basic principles is key to make a good design with integrated circuits. By the a different version of circuit 1 has been produced in the millions for a 20W, 300V half bridge running in ZVS at 50kHz and proved robust. When I saw the circuit the 1st time I was actually shocked just like you... but it was cheap, did the job and reduced the electromagnetic noise so much that more savings were found in reducing EMI filtering.

Thank you Steven! Here is the link: drive.google.com/file/d/1_vD16bBGvdzPY8TEF4GOULbfYH5T9_QT/view?usp=sharing I also added in the video comments. Please let me know it it works!

Thank you - yes the simulations work and will save me much time @@smartpowerelectronics8779

At these currents, it’s best to use some kind of resonant topology (normally ZVS). Rather than battling the FET parasitic capacitance by brute-forcing the shortest rise & fall times possible, you can take advantage of it. :) It’s only applicable if you’re driving an inductive load of course, but at these frequencies any load becomes somewhat inductive. Re: the bootstrap push-pull circuit, replacing the pull-up resistor with a current source and using a common base or common gate transistor instead of common emitter/source would be a big improvement.

If you search the datasheet of the IRFZ44 you can find the Thermal resistance is "60 K/W". You can then calculate the temperature rise: T= Tamb + Powerloss*Rthermal , in this case 25C + 0.85*60 = 76 C. A TO220 without a heatsink is 60K/W a small BC547 in TO92 is 200 K/W 🙂

Yes you can, for the simulationin the Simetrix tools, efficiency calculator. I did not measure the efficiency, however it can be done with some multi-meters. Regarding the RMS value...I checked a bit more online and you are right I should use mean...... never too old to learn thank you for pointing that out!

You can use it for an inverter. You can not use it for a DC Motor: all circuits cannot handle DC output in thje HIGH state (LO high) because the bootstrap capacitor will slowly lose its charge. So you have to switch "low" once i a while to top up it's charge. Solutions for motor drive: 1) stay at 99.5% (1kHz PWM) or 0.5% 2) a charge pump, this adds extra complexity - would not recommend 3) Motor drive IC like L293

Isolated methods are much more expensive than non-isolated.

You are correct, transformer drives have been still popular though, especially for bipolar transistor half bridges

sure hoe so! Thanks mumbaiverve2307!

www.simetrix.co.uk/ the demo version

Thank you infoolyugun! A level shifter can drive a MOSFET with a "floating" source. So the IR2104 is a level shifter. It drivers the MOSFET AND handles that the MOSFET source is flying at some high voltage. A MOSFET driver only drives the MOSFET at ground level, see this video: kzfaq.info/get/bejne/rdxkqKSrnay6Yqs.html The IR2104 has a MOSFET driver at the LO (we did not use it in the video) and a Levelshifter for the HO.

undustund thank You 🤟@@smartpowerelectronics8779

Thank you p0lyglot, you must be a professional in this field. The IR2104 has built-in logic and non-overlap time for that. For the discrete solution you will need to build in the dead time and logic in your logic signals. Half bridges are tricky for beginners, especially prevention of capacitive mode and hard switching are complex to manage.

The false triggering thing isn't a dead-time issue, but a matter of physical design, parasitics and low-side turn-on speed. It's not a shortcoming of this high-side video but if you do a half-bridge or full-bridge video, it'd be worth including some additional considerations for high-side driver design. The problem is that when the low side comes on, the high-side source voltage goes down very fast. Any stray C between high side gate and a power supply rail mean that the high-side driver has trouble "following" the FET (keeping a fixed low Vgs) so suddenly you have a bump in Vgs, the high side turns on and all the smoke gets out.

Yes the 2N2222 is faster and has more current capability than the BC547!

it's in the corner. openhantek6022

Appears to be some version of SIMetrix/SIMPLIS Elements.

No that is my oscillosope! The simulation software is Simetrix, free version

yes you are correct, I added the simulation models in the comments today

Yes, you are right, and it will be very fast. For my demonstration I try to use the most common parts possible so the circuit is easy to build.

That would be about the minimum, expect it works ok

That is out of the scope for this video...

@@annaplojharova1400 Even a small mention like "You can replace this BJT with an opto if you want isolation" and show a simple circuit would make this video better without any drawback.

@@XiaZ These "simple basic circuits" are already quite bad from safety perspective (microcontroller fault -> output ON; microcontroller supply failure -> output ON; in both cases even not completely, so mosfet meltdown ensured). With an optocoupler attempt the same issue becomes way more complex (more power supplies to handle, parasitic capacitive coupling between LED and the phototransistor within the optocoupler, dealing with the typical optocoupler parameter instability in general,...). By the way optocouplers are normally not used with fast switching, it goes usually via transformers or balanced capacitive coupling (when the insulation barrier is within ICs).

@@annaplojharova1400 That's why I said you need to address it. You just proved my original point.

Good discussion guys! I did consider the opto-solution but that would make the video even longer because it comes with many caveats (speed being one of them....) I do agree that could have mentioned it.

Correct, free version. I now added the simulation files download link in the comments

Good point, that is also why I recommend an IC.

Hi Zeeebrenn, yes that part is tricky, but the key point of a bootstrap supply. If you look at circuit #1 (0:07 in the video, make a screenshot) *If the MOSFET is off, the 100n cap is charged from 12V via the top 4148 and the load to about 12V right? *OK, so we have 12V on the capacitor, that will remain there if nothing discharges it... *Now if we let go of the gate (BC547 off) the 100n has 12V and will charge the G-S of the MOSFET, the G-S is only 2-3nF so the 100nF keeps at 12V! *If the MOSFET is on...D-S will be like a short circuit, so the bottom of the 100n is now connected to +24V...therefore the top of the 100n will be at almost 36 Volt! Please draw or print the schematic and use a pen to draw the current s and voltages, hope you understand it! 🙂

Yes thanks, I understood it by realising the mosfet D-S just acts as a short when it turns on, resulting in 24V on the source, which is then added to the cap's 12v. I've seen these bootstrapping used in amplifiers and didn't understand until now. I don't however understand your explanation of the gate diode. You mentioned that there might be cases where the gate voltage can be -24V, but I don't see how. Could this be simulated with a capactive load or something? @@smartpowerelectronics8779

Well that used to be my opinion too, for instance if you simulate a astable multivibrato r with 1 npn's, it does not even start because the parts are 100% balanced mathematically..(you add a 100G resistor anywhere and it works ) ..But if you know what you do and accept the limitations of simulation it can be helpful. And I fully agree with you, build & testing is gold and more fun to do! :-)