High side driver not functional - Page 1
Forum section: electronics / beginners

author: CranberryFire
author_rank: Contributor
timestamp: June 13, 2020, 08:10:53 pm
content: I am trying to use an IRS2127(https://www.infineon.com/cms/en/product/power/gate-driver-ics/irs2127/data sheet here) as a high side driver, and it does not seem to be functioning at all. Here is what I have hooked up:V_hi is currently 34v, but will be driven high in actual application. The load is a high power led. I have tried various values for the boot strap cap C2, with 0.1uf being the lowest. I have also tried attaching CS to Vs as shown in an appnote, but I do not need current sense for my needs.At all times, regardless of what the GPIO signal is, HO always reads 10v with respect to ground.Any clues as to what mistake I could be making?

author: wraper
author_rank: Unknown
timestamp: June 13, 2020, 08:20:41 pm
content: Well I do not see in the datasheet that you are allowed to simply leave CS pin unconnected. If you don't need it, use different driver.

author: CranberryFire
author_rank: Contributor
timestamp: June 13, 2020, 10:30:44 pm
content: Quote from: wraper on June 13, 2020, 08:20:41 pmIve tried with CS connected to Vs(as shown in IR's buck converter example) and it does not function.

author: wraper
author_rank: Unknown
timestamp: June 13, 2020, 10:47:08 pm
content: Where did you buy the IC and IGBT, are they from reputable place?

author: CranberryFire
author_rank: Contributor
timestamp: June 13, 2020, 11:13:21 pm
content: Arrow. Packed in ESD tubes, inside of ESD bags. The IGBT is tested and functional. I have tried two different chips for the IRS2127, seems unlikely they would both be kaput.

author: Ian.M
author_rank: Super Contributor
timestamp: June 13, 2020, 11:41:53 pm
content: What's the frequency and duty cycle of the GPIO pin you are driving it from?

author: wraper
author_rank: Unknown
timestamp: June 13, 2020, 11:55:43 pm
content: Just in case, aren't you trying to operate it at DC (always stays ON) or a very low frequency? The thing is the bootstrap will not work if you do so, there must be a pulsed operation. Otherwise circuit won't be able to rise Vb/gate voltage high enough. Also because there is no low side switch, with LEDs there is an issue that they will pass barely any current below some forward voltage and Vs voltage may not drop to the potential of GND, again preventing proper bootstrap operation. In such case try connecting resistor in parallel with LEDs and see what happens.BTW, do you really need a high side switch for LED? It's better to use low side switch unless you have a good reason to do otherwise. Also there is no current regulation for LED, so this circuit seems dubious.

author: Dmeads
author_rank: Regular Contributor
timestamp: June 14, 2020, 12:06:04 am
content: https://www.infineon.com/dgdl/Infineon-Using_Monolithic_Voltage_Gate_Drivers-AN-v01_00-EN.pdf?fileId=5546d462584d1d4a01585242c11947b1Have a look at this note for Calculating the boostrap cap value on pg. 2. Your should be able to get all the values from the IRS2127S datasheet as well as the IGBT data sheet.Also keep CS tied to Vs like it says in the datasheet (which you already did).Also is the ground of your microcontroller(or whatver youre using to generate the signal) connected to the ground on the driver? ( I have made this mistake before lol)

author: CranberryFire
author_rank: Contributor
timestamp: June 14, 2020, 12:41:49 am
content: All grounds are tied together. I have tried frequencies up to 10khz at 50% duty cycle.Yes, a high side switch is needed, this is not the final functionality. The high side supply is current limited. Resistor in parallel did not change anything.

author: jmelson
author_rank: Super Contributor
timestamp: June 14, 2020, 02:33:59 am
content: Quote from: CranberryFire on June 13, 2020, 08:10:53 pmWhen HO is 10 V, what voltages are VS and VB? There should normally be about 12 V from VS to VB.If HO is 10 V, and the transistor is "off", then VS should be at 0 V, and the transistor should be turned "on".If the transistor is turned "on", then VS should be about 32 V, so HO at 10 V would not turn it on.So, I'm confused. Ase you sure you have the IGBT connected correctly?Oh, do you have a load connected? The bootstrap circuit REQUIRES a load to be connected. And, if you have a multi-chip (series) LED hooked up, then the circuit will NOT work! For the bootstrap capacitor to charge from a 12 V supply, the emitter of the IGBT (VS) must be held close to ground when the transistor is off. With nothing other than the LED string, a little charge will leak across the gate - emitter barrier and charge up VS-Emitter-LEDs, and then the bootstrap cap will not get charged.For the bootstrap to work, VS must be pulled ALL THE WAY to ground during each cycle. This circuit would work fine if the load was a real resistor, but not for a diode string.Jon

author: CranberryFire
author_rank: Contributor
timestamp: June 14, 2020, 04:14:38 am
content: Quote from: jmelson on June 14, 2020, 02:33:59 amThis appears to be the reason. I disconnected the LED and put in a measly 2/3w resistor, and it went straight to resistor heaven. I added the led back with a resistor to ground and limited the high side current alot and it appears to 'function'(the resistor wants to draw alot of current). Is there a way to fix this circuit with out requiring a crazy high wattage resistor?Thanks for the answer.

author: CranberryFire
author_rank: Contributor
timestamp: June 29, 2020, 01:08:00 pm
content: Quote from: jmelson on June 14, 2020, 02:33:59 amI have spent more time trying to look into this, but can not find any info.Does any one know how to run a high side driver with an led (high wattage COB type) as the load? Or are there resources on this topic somebody knows of that address this issue?

author: Zero999
author_rank: Super Contributor
timestamp: June 29, 2020, 05:49:27 pm
content: Quote from: CranberryFire on June 14, 2020, 04:14:38 amWhat was the value of the resistor? Try a higher value which will give a lower power dissipation. P = V2/R

author: CranberryFire
author_rank: Contributor
timestamp: June 30, 2020, 05:07:55 am
content: Higher value resistors from emitter to ground do not work so well. They cause the the circuit, and led to draw less power than it should, and also make the pwm frequency much lower. For instance if I am driving the input at 10khz, and have a 10k resistor pulling VS to ground, the LED on the output will be getting driven at like 400hz for what ever reason, and the total power draw will be significantly lower. High value resistors exaggerate the problem. And of course the problem with a really low value resistor, is the power draw gets split between the resistor and the LED. So adding a resistor from VS to ground seems like a losing proposition.

author: Ian.M
author_rank: Super Contributor
timestamp: June 30, 2020, 06:48:57 am
content: The LED string has significant junction capacitance holding the voltage across it up at turn-off. There is no way a bootstrapped driver for a high side N-MOSFET is going to work well unless you either add enough series inductance to the string to 'ring' the MOSFET source right down to ground (and clamp it there with a catch diode), or use a half H-bridge, so the source is actively pulled hard down by another N-MOSFET, driven in antiphase to the high side one (probably with some small anti-shootthrough delays to the leading edges of the complimentary drive signals). As the lower MOSFET only has to handle discharging the junction capacitance and charging the bootstrap cap, and doesn't carry the LED string operating current, it could be a lot smaller than the high side one.OTOH personally, I'd avoid the whole problem and simply use a high-side P-MOSFET, with a pullup resistor on its gate and a switched BJT current sink as a level-shifting driver for it.

author: Zero999
author_rank: Super Contributor
timestamp: June 30, 2020, 07:45:19 am
content: Quote from: Ian.M on June 30, 2020, 06:48:57 amYou're right about the capacitance, but it might not be a big deal, if the LED stops illuminating when the voltage falls below its forward voltage. The voltage across the LED isn't always a good indication of whether it's actually producing any light or not. I suspect it could be the forward voltage, preventing the bootstrap capacitor charging.Quote from: CranberryFire on June 30, 2020, 05:07:55 amPlease post some oscilloscope shots.Try a constant current sink, say 10mA.

author: CranberryFire
author_rank: Contributor
timestamp: July 06, 2020, 01:13:28 pm
content: Quote from: Ian.M on June 30, 2020, 06:48:57 amI considered using pmos, but for my end goal they dont seem ideal due to the power requirements.Do you have any resources of what using a half bridge would look like for an application like this?Is there an example of what a circuit with added inductance and a clamp diode would look like? I can not find anything searching.