Half-bridge problems - Page 1
Forum section: electronics / projects

author: Aleksandar B.
author_rank: Contributor
timestamp: August 18, 2014, 11:43:04 am
content: Hi guys,First a little about my project.. I have to design a BLDC driver as a part of my master thesis. Since we had a few TI DRV8332 BLDC drivers lying around that were never used, I was asked to try and use them. After having them fail horribly, and having no control over them nor a way to localize the error, I decided to make a driver using gate drivers and discrete MOSFETs.The gate driver I've chosen is the IR2104. It has two inputs. One is the input signal for the PWM, and the other one is the shutdown pin.A short overview of how it works is basically the high side output is the same as the input signal (with a larger amplitude), and the low side output is inverted. When the shutdown pin is driven low (active low) both the MOSFETs should turn off, which they do, but I still get 10V on the output. I have connected the components like the "typical connection" (Page 1 in the IR2104 datasheet) shows. The output is connected to the drain of the low side FET, source of the high side FET, and Vs which is the floating supply for driving the high side FET. This voltage is not coming from the Vs pin (I think), since it disappears when I turn of the power supply which powers the load. Another thing I noticed is that the 10V output does not change with the power supply voltage. To me it looks like the body diode of the high side FET is directly polarized and drops the voltage a little, and the body diode of the low side FET is inversely polarized, thus regulating and outputting 10V.Now the question is, am I right, or is this caused by something else? Is this a normal thing? For me it feels wrong to have 10V on a coil that should be floating..Like mentioned before, the circuit is connected as shown in the IR2104 datasheet, with the addition of two flyback diodes.Components used:Driver: IR2104,FETs: IRFZ44NPBF,100Ohm resistors in series with the FET gates,Flyback diodes: 1N5822,Bootstrap diodes: 1N4148,Bootstrap caps: 22uF (calculated based on the driver itself, and the MOSFETs its driving) electrolytic in parallel with a 100nF ceramic,1uF and 100nF in parallel on Vcc, and 470uF between the high side drain and low side source.Everything else is working as I expect it to, its just this thing that's bothering me.Also sorry if something like this was asked before, I searched the forums and couldn't find anything similar.Regards,Aleksandar

author: Richard Head
author_rank: Frequent Contributor
timestamp: August 18, 2014, 01:30:05 pm
content: It sounds like you've done everything correctly.I wonder how much load to gnd you have on the centre point of the MOSFET's? It could be that the leakage current through the high side MOSFET is raising the centre point potential. If you load it a bit it may well drop to gnd.

author: Aleksandar B.
author_rank: Contributor
timestamp: August 19, 2014, 06:49:48 am
content: Yesterday I only had a 100k resistor in my reach which I knew would be allot but I didn't think it would have that high of an effect.Today I bought a 10Ohm resistor and it works like a treat.Thanks.

author: Richard Head
author_rank: Frequent Contributor
timestamp: August 19, 2014, 09:05:07 am
content: Great.

author: bktemp
author_rank: Super Contributor
timestamp: August 19, 2014, 10:07:51 am
content: The 10V comes from the supply voltage of the IR2104: The voltage is flowing over the bootstrap diode into Vb and through the highside driver into Vs: And Vs is connected to the output.This explains why this voltage is basically independent from the output supply voltage, it is a bit less than the driver supply voltage which I assume is 12V.

author: Aleksandar B.
author_rank: Contributor
timestamp: August 19, 2014, 10:40:02 am
content: It's independent of the value of the output supply voltage, as in it doesn't matter if its 24V, 15V or 30V. When I turn off the output supply voltage, it slowly fades away (because of the 470uF cap). When the IR2104 is powered, without the output power being on, there is no problem.