Questions using the high-side of mosfet driver IR2130 - Page 1
Forum section: electronics / projects

author: htm
author_rank: Regular Contributor
timestamp: March 22, 2025, 12:13:41 am
content: Hii'm trying to use the 3 phase mosfet driver IR2130 (datasheet link).By following the application note (link) i got a single phase working-ish.i realized that the chip requires to toggle the input-signal periodically, so that the voltage at the high-side mosfet's gate gets recharged.But i'd like not strobe the mosfet. i'd like to have it continually open as long as the input is active.So i thought to make a voltage doubler with a 555 timer and feed the chip's Vb1 with this high-enough voltage.(i looked through both documents, but i don't see a text explaining how to supply a constant voltage. Maybe it's explained too technical why i didn't pick it up)I'm reasonably newbe which could explain why i don't get why it isn't working.I'd like to know what i'm doing wrong.The chips Vcc is connected to 12V.When i feed the 19V to VB1 and i measure the voltage at HO1.Then i see 12V at HO1 when the inputs Hin1 and Lin1 are not active(i.e. they're unconnected. At Hin1 and Lin1 i then reads 5v with multimeter),and i see 0V(?!) at HO1 when the inputs Hin1 and Lin1 are active(i.e. connected Hin1 and Lin1 to ground).i've attached a picture of the schematic of my setup,i also added picture from the application note, for which i have some additional questions:- the yellow marked part; What are those circled-arrow symbols?- the green marked part; the diode&capacitor are the charge pump. i think i don't need it if i supply a high enough steady voltage(19V) to VB1,VB2,VB3 ?Any extra insight is much appreciated

author: PCB.Wiz
author_rank: Super Contributor
timestamp: March 22, 2025, 03:38:45 am
content: Quote from: htm on March 22, 2025, 12:13:41 amThose are current sources.Keep in mind that this type of level shift is edge based, and does a SET/RESET for each rising/falling edge.That means you cannot run it fully static, as it will stay in the POR state.Of course, they are ok with that as the boost cap design cannot do 100% anyway.Your added charge pump can get closer to 100% duty, but you would need periodic refresh pulses to know the output state.Addit: You are not driving a high voltage here, so an alternative is newer gate drivers like 1EDN7550B. or opto or isolated gate drivers.That's in a SOT26 package & has differential drive, so you can level shift using resistors, and still use your charge pump.This will not lose state with power interruptions on the secondary, and the small package means you can place next to each MOSFET.Low side drivers can use something simpler.

author: htm
author_rank: Regular Contributor
timestamp: March 22, 2025, 08:32:57 am
content: ThanksAh, current sources.. (i guess that text "Vdd/Vbs level translator" is then related to the top 2 yellow components?)In that picture the VB signal is split up to have two paths with 2 current sources.And the bottom two current sources are influenced by the 'pulse generator' (i guess to let current flow in one of two paths)So i guess, that the 'pulse detector' can detect a pulse from the 'pulse generator/Hin/Lin logic' once the availability of the VB1 voltage.Quotein my application i would be able to pulse it once after power-on.But i want to get around having to to pulse Hin/Lin indefinitely (and with that the power mosfet will switch off/on)QuoteI don't get the following;When i removed the green marked parts, and instead supplied constant 19V (generated by voltage doubler) to VB1.Then the HO1 signal goes to 0V(?) whenever i activate HIN1(connect to Gnd) even though there's still 19V on VB1? i.e. in neither Hin1 state i see the 19V on the HO1 pin?Can you explain what could be happening there?

author: PCB.Wiz
author_rank: Super Contributor
timestamp: March 22, 2025, 09:20:45 am
content: Keep in mind the high side also has UVLO, I think on this part they say 10-20V, so 19V is only 7v above 12V, so not enough. Try it with > 22v

author: Benta
author_rank: Super Contributor
timestamp: March 22, 2025, 11:25:03 am
content: How fast do you need to turn on/off the upper MOSFET?If it's not very fast, you could look at using a photovoltaic optocoupler instead, eg, Vishay VO1263.

author: mtwieg
author_rank: Frequent Contributor
timestamp: March 22, 2025, 12:20:43 pm
content: Quote from: htm on March 22, 2025, 12:13:41 amYou mean you want the high side MOSFETs "continually open", or all MOSFETs "continually open"? Because these are both doable without any modifications to the typical circuit. The only requirement is that before turning the high side MOSFETs on, you turn on the low side MOSFETs briefly to charge the bootstrap caps.QuoteShort story: This isn't the right way to work around the issue.Long story: You might have overlooked that the bootstrap supply has its own undervoltage lockout threshold, referred to as VBSUV. In order for the high side driver to operate, the bootstrap supply has to first rise above VBSUV+, which could be as high as 9.2V. If the bridge supply is 12V, and the bootstrap diode drops 0.5V, this means you'd need a Vb of 12+0.5+9.2 = 21.7V to guarantee the UVLO is satisfied when the low side FETs are off. But this also means that when the low side FETs are on, VBSwill be over 20V, which may easily damage the driver or the FETs.The more "normal" way to ensure VBSis always charged is to use a charge pump. See figure 16 in this white paper:https://web.mit.edu/6.131/www/document/float_drive.pdf.

author: htm
author_rank: Regular Contributor
timestamp: March 22, 2025, 06:43:16 pm
content: Thanks for the detailsi didn't notice the undervoltage protection but also not that VB123min=VS+10 VB123max=VS+20 is something to watch out for.The ranges 0v+(10..20) and 12v+(10..20) don't have any overlap, so providing a fixed voltage of 19V won't work indeed.i will go with figure 16 description.if i read that scheme correctly,i should keep the default diode and bootstrap capacitor, andmake my 555 timer be fed by the VS voltage, so that the doubling follows VS level, and feeds into the bootstrap capacitor.

author: langwadt
author_rank: Super Contributor
timestamp: March 22, 2025, 07:33:52 pm
content: Quote from: htm on March 22, 2025, 06:43:16 pma ~12V zener in parallel with each bootstrap capacitor feed via a resistor from > (12V + ~10V) could work

author: htm
author_rank: Regular Contributor
timestamp: March 25, 2025, 11:25:33 pm
content: Thanks everyone.i have it working(i.e. continuous drive) by using Application Note Fig 16 / chapter "8. HOW TO PROVIDE A CONTINUOUS GATE DRIVE".i still have it on a breadboard though.Seeing this charge pump's "+voltage" is connected to the high-side mosfet'source, i guess there's no way around having to make 3 of these 555 parts.i increased the cap to store the gate voltage to 330uF (instead of 0.1uF / 100nF as in the application note). Maybe i should try as low as possible? i'm not sure.But i'm using the LM555 (i don't have the CMOS type 555 as shown in application note). i had to reduce the ground-resistor from 100K down to 2K (10K also didn't work)(Comments on how to improve the schematic are welcome.. #learning)

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: March 26, 2025, 08:21:05 am
content: 330uF is definitely too high. I would suggest taking the total capacitance of all the FETs (using the gate voltage and total gate charge), 10x that number and use that as the bootstrap capBoostrap cap is usually in the 10nF to 100nF range, I have only once seen even a 1uF one. Do the numbers just to be sure but beyond that, its not terribly important

author: htm
author_rank: Regular Contributor
timestamp: March 26, 2025, 07:06:01 pm
content: @BadeBhaiyai tried replacing C17 with0.47 uF(64V) - dropout within a second6.8 uF(250V)22 uF(450V) - dropout after about 2 secondsThe 555 won't maintain its function.The VB1 voltage slowly drops below a voltage where the 555 circuit would operate.Should i make the charge pump charge larger?I checked 3 versions of datasheets for the 2SK1018.i see: Ciss is 1800-2700 pFI asked ChatGPT to help out;--Assuming:Vgs (Gate Drive Voltage) = 10V (a typical value for MOSFETs)Ciss (Average) = 2250 pF (midpoint of 1800–2700 pF range)Qg ≈ Ciss × Vgs = 22.5 nC​For a fast switching time of 100 ns:𝐼𝑔 = ( 22.5 × 10−9) / 100 × 10−9= 225 mA​if i use 10V what should be the capacitor size if it needs to deliver 225mA instantaneously?C = ( I * dt ) / dV = 45 nFTo support 225 mA instantaneously for 100 ns with only a 0.5V drop, you need a capacitor of at least 45 nF (0.045 µF).If you need even better performance, use a 100 nF (0.1 µF) capacitor for extra stability.---Is actually spot on to what BadeBhaiya suggested!But i don't know why it drops off with smaller caps.i did notice that depending on the load, there's a different voltage drop across the 555 circuit.When i connect a motor with too little resistance, i don't get a voltage drop across the 555 circuit.i guess that is the reason why in theApplication Notethey used a CMOS 555 version..Current Draw TTL NE555: Higher (~10mA - 15mA)Current Draw CMOS LMC555, TLC555, and ICM7555 :Lower (~50µA or less)

author: langwadt
author_rank: Super Contributor
timestamp: March 26, 2025, 10:26:59 pm
content: who not just a resistor from a ~24V supply and a zener on VB/VS

author: htm
author_rank: Regular Contributor
timestamp: March 26, 2025, 11:38:06 pm
content: Quote from: langwadt on March 26, 2025, 10:26:59 pmThanks.i'm trying to understand the function of the zener12v parallel to the capacitor;if H1 on, L1 on, then vSource is 12V, and zener12v doesn't need to clamp because the capacitor is 12V on top of vSource. But chops of excess >12V.if H1 off, L1 off, then vSource is Open, and zener12v doesn't need to clamp.if H1 off, L1 on, then vSource is Gnd, and zener12v is passing the excess 12V..24V(current is limited by resistor from doubler, and also limited by the 10 Ohm near Gnd) to Gndif H1 on, L1 off, then vSource is Open, and zener12v doesn't need to clamp.?