Need help with troubleshooting a gate driver circuit - Page 1
Forum section: electronics / projects

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 02:56:35 pm
content: I designed an H bridge using the DGD0506A gate drivers.https://www.diodes.com/assets/Datasheets/DGD0506A.pdfI used the FDMC7696 for both high side and low side mosfets.I went with the suggested circuit as provided in the gate driver datasheetI just used a transistor to invert the IN signal to the second chip, so the opposite mosfets turn on (e.g. HO on the U1 and LO on the U2)and this is the arrangement of H bridge itselfafter ordering the parts and the PCB, I started testing it, but I had no luck and it wouldn't work.I checked everything and found out I'm missing a connection here:I just botched a few wires to make that connection but it still wouldn't work.I have tried it with a DC motor and a 5k resistor connected to the load, no chance with either of them. I checked a few volrages and the Vs is sitting at around 9.5v with respect to ground (drops quickly to 7.8 volts when measured by a multimeter or oscilloscope) but the voltage across both tantalum capacitors is 0. I've created this exact same circuit 3 times and I'm running out of them but I've had no luck finding what the problem is.(both EN1 and DIR1 are 3.3v signals,DIR1 is digital High/Low, but EN1 is a PWM @25KHz. VCC is 12v and VDD is not being utilized here but it would be 3.3)I was wondering if you guys can help me troubleshoot this issue and perhaps get this board working. I have enough parts to make 1 new board but that's pretty much it.Thank you very much.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 03:04:40 pm
content: update: I looked at the data sheet again and it says EN and IN signals should be 0 to 5v but I missed that and I'm connecting 12v to the IN of U2, all the other signals are proper 3.3v though. Could this be causing a mishap in the circuit because the second half of the bridge is getting bricked? I honestly have no idea.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 05, 2018, 04:09:41 pm
content: Looks like there's a solder short on, whatever the hell it is, the chip left of R1?The soldering looks sloppy, not sure if it actually is or it's just the lighting. Might find it useful to pick up some paste flux? It's just about indispensable working with SMT.Also with miswiring or such, the chip might be blown, you'll have to figure out which of a few faults may be at play here.On a separate note, the clamp diodes look to be very far from the supply bypass capacitors (if that's what they are), and I have no idea where the transistors are (bottom side I guess?) or what the routing looks like. It's also hard to see if there is any ground via stitching. The diodes aren't necessary anyway (the transistors have body diodes internally, and I guess you're driving them full wave so the channel resistance should dominate), but being close to the capacitors is necessary, and having a solid ground (with stitching) on both sides is important.Tim

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 04:12:29 pm
content: update: I looked at the data sheet again and it says EN and IN signals should be 0 to 5v but I missed that and I'm connecting 12v to the IN of U25V recommended, but IN and EN can be up to Vcc on the data sheet.I'm quite confused by the rest of it, if you're stopping one haif of the bridge toggling, are you sure that the floating supply is still produced.

author: Wolfram
author_rank: Frequent Contributor
timestamp: September 05, 2018, 04:29:43 pm
content: Quote from: StillTrying on September 05, 2018, 04:12:29 pmThis is an important point, this driver uses a bootstrap derived power supply for the upper MOSFET gate driver. This means that the circuit requires the bottom MOSFET to be switched on periodically to recharge the supply capacitor for the upper MOSFET gate driver. With a fixed high input signal, the top MOSFET will only be on for as long as C1/C2 can supply the upper gate driver, typically a few milliseconds per microfarad.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 04:54:51 pm
content: Quote from: Wolfram on September 05, 2018, 04:29:43 pmI'm inverting the IN signal to U2 which turns on the bottom mosfet on it if U1 is switching it's high side fet and vice versa, is there something I'm missing?If yes, can you please let me know how will be able to change the circuit so that I can have a full bridge instead of a half bridge?The mosfets are placed on the bottom side of the board indeed and I made sure there's no short anywhere on the board, even if it looks like it in the picture.

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 06:38:08 pm
content: I think the problem/solution is here, - somewhere.Edit: Bottom left.During PWM/EN switching, the lower MOSFET never turns on to recharge the Vb boost cap.Without the 12V+10V boost voltage, the upper MOSFET will be limited to ~12V at its gate and ~8V at its source.Probably.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 07:00:47 pm
content: Quote from: StillTrying on September 05, 2018, 06:38:08 pmIt would be really nice to know where thoI seriously looked at everything that could possibly be an issue. how can I test each half bridge sperately? if anyone has any ideas, I'd love to hear it.

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 07:51:42 pm
content: If you've got a 2 ch scope, I'd use ~1k as a load, to view both ends of the load at the same time.At the times the 2 waveforms are at different voltages there's voltage across the load.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 08:02:27 pm
content: Quote from: StillTrying on September 05, 2018, 07:51:42 pmThat's not really helpful because they both are at Vs (floating supply return) which is 9v, which means there is 0v across the load.

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 08:10:57 pm
content: "which means there is 0v across the load."Well that's the problem then.Have you looked at one IC's LO and HO together, to check that they're switching on alternately.I don't thinkyou can just keep switching just one of them, because the float/boost voltage won't be produced.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 08:21:26 pm
content: Quote from: StillTrying on September 05, 2018, 08:10:57 pmthey would be pointless if they must be switched alternatively, they are designed for H bridge applications (just read the data sheet). and I know that's the issue, but I have no idea why the chip is producing that output! the example circuit provided by the datasheet is exactly the same. I don't know what I'm missing here

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 08:48:29 pm
content: "they are designed for H bridge applications"The DS sayshigh-frequency half-bridgeand you seem to be using it as a full bridge at DC, I think that's different. Unless point "Vs" is switching, the floating voltage for the top FET's gate won't be produced.The DGD0506A is a high-frequency half-bridge gate driver capable of driving N-channel MOSFETs in a half-bridge configuration.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 05, 2018, 09:12:30 pm
content: Quote from: StillTrying on September 05, 2018, 08:48:29 pmif you go down a few lines, the clearly state: motor control in one of their applications ... they meant an H bridge because you can't control a motor with a half bridge.

author: StillTrying
author_rank: Super Contributor
timestamp: September 05, 2018, 11:10:55 pm
content: "they clearly state: motor control in one of their applications"Yes but every FET H bridge that google has found so far is continuously switching between high and low.If you're going to switch the top FET on for a bit, you have to first start with the lower FET on so that the boost cap is charged to ~12V, when Vs then goes high the top of the cap goes above 12V to supply the >12V for the top FETs gate through HO.This isn't very good, but it says:"The bootstrapping method for achieving the high mosfet gate voltages avoids the need for an additional DC to DC power supply, but for the bootstrapping to work, there needs to be time to recharge the bootstrapping capacitor. That means that this H-bridge cannot be run at 0% or 100% duty cycle. At 20kHz, the maximum forward and reverse speeds occur at about 10% and 90% duty cycle."http://hades.mech.northwestern.edu/index.php/Driving_a_high_current_DC_Motor_using_an_H-bridgeIf you haven't got continuous high/low switching for the H-bridge High side I think you might have to arrange it.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 06, 2018, 01:18:20 am
content: Quote from: StillTrying on September 05, 2018, 11:10:55 pmHaving a limited range PWM is different from switching High side and Low side, don't get carried away. it just means the ON time vs OFF time of each mosfet. this cannot be the possible cause of the gate driver not functioning correctly. I'm still nowhere closer to having an answer as to why it won't work with the schematics provided above

author: Phoenix
author_rank: Frequent Contributor
timestamp: September 06, 2018, 01:34:42 am
content: Hi OM220Can you please show on an oscilloscope the signals DIR and EN. I believe the other posters understand your problem, but are not describing it well.Consider just a half bridge of your circuit, let's take Q1 and Q2 half bridge and U1 gate driver. Your gate drive IC is has a bootstrap supply for the high side gate drive of Q2. In order for this to operate correctly Q1 must switch on to charge up C1 -> only then can you switch over to turning on Q2. When Q2 is on C1 will be discharging, before C1 discharges completely Q2 needs to be turned off and Q1 turned on to recharge it. This is summarized in the table:EN IN Q1 Q2 C11 0 ON OFF Charging1 1 OFF ON DischargingTherefore during normal operation Q1 must be on some of the time to allow Q2 to operate. Hence the limitation of approximately 10% to 90% duty cycle. Using the driver for simple on/off and direction control does not meet this requirement - the DGD0506A must be used for PWM.The other half-bridge is independent and identical in it's bootstrap supply operation. If you can please provide DIR and EN signals from an oscilloscope this can confirm our suspicions on what's happening.

author: StillTrying
author_rank: Super Contributor
timestamp: September 06, 2018, 10:40:44 am
content: "I'm still nowhere closer to having an answer as to why it won't work with the schematics provided above"LOL.I've drawn what about 3 of us think the problem is on the diagram inReply #6.I'm quite sure that's the problem myself, I don't know an easy solution using the EN input as the PWM input.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 06, 2018, 12:47:50 pm
content: Quote from: Phoenix on September 06, 2018, 01:34:42 amThis is the EN signal. The IN signal is just a static 3.3 or 0v. I'm guessing you mean I should be using PWM to control IN signal as well if I'm not mistaken, then the half bridge charges C1 through itself (not the other half bridge (which is what I read happens in a bootstrap circuit)). If that's the case, wouldn't I be losing efficiency at higher frequencies due to deadtime at each cycle?

author: Phoenix
author_rank: Frequent Contributor
timestamp: September 06, 2018, 01:08:59 pm
content: Quote from: OM222O on September 06, 2018, 12:47:50 pmEach half-bridge independently charges its own bootstrap capacitor. Therefore each half-bridge needs to turn on it's lower switch at least 10% of the time.QuoteWhat do you mean by this?

author: OM222O
author_rank: Frequent Contributor
timestamp: September 06, 2018, 01:46:00 pm
content: Quote from: Phoenix on September 06, 2018, 01:08:59 pmthe deadtime is a fixed value (about 70 to 100ns), which the entire circuit is off when you switch from high side to the low side fet. at higher frequencies this will be a higher percentage of the cycle time,therefore reducing the effective PWM duty cycle.If you go high enough it might even not allow the circuit to turn on at all. I know it's really small compared to the 40us (25KHz) but at higher frequencies this might be an issue? for example a DC-DC converter at a few mega hertz.by the way, this means that the two PWM signals should be synced up and IN should be switching when EN is high. How can I sync the two signals on a micro controller? are they synced automatically?

author: Phoenix
author_rank: Frequent Contributor
timestamp: September 06, 2018, 11:31:22 pm
content: For your chip deadtime can be adjusted using the resistor on the DT pin. Deadtime is usually tweaked based on observing the switching speeds of the transistors. You have correct identified a limitation of deadtime - just something to live with; but in general if you want to switch at MHz you will have much faster switches and less deadtime, there might also be minimum and maximum duty cycles that the circuit can achieve because of it.In terms of your circuitEN - The enable signal, must always be high for the gate driver to operateIN - The signal you need to PWMFor simple control the IN on each chip should be the same but inverted. Your micro should be able to produce this complementary pair. I don't want to go into details of 2 level and 3 level PWM patterns, but you can google the terms if you're interested.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 06, 2018, 11:46:01 pm
content: Quote from: Phoenix on September 06, 2018, 11:31:22 pmIf I understand that correctly, it measn 50% duty cycle means motor is off, below that (down to 10%) means for example clockwise rotation and above that (up to 90%) means counter clockwise rotation?

author: Phoenix
author_rank: Frequent Contributor
timestamp: September 07, 2018, 03:28:19 am
content: Quote from: OM222O on September 06, 2018, 11:46:01 pmYupNow the next question is - how big is the motor you are controlling? Do you have any current monitoring?

author: StillTrying
author_rank: Super Contributor
timestamp: September 07, 2018, 11:58:02 am
content: One option could be to use a 3rd self-oscillating DGD0506A to produce the Vb for the other 2.DIR/IN and PWM/EN would stay exactly as they are, and the H would work down to DC or 100% PWM because the Vb for the upper mosfet is always available. I guesstimatethe averageVb current needed for one full H @ 20kHz PWM to be only about 30mA.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 07, 2018, 12:16:07 pm
content: Quote from: Phoenix on September 07, 2018, 03:28:19 amThe specific ones for these project are fairly small (1200mA / 4.5A stall) but I'm planning on using the board as a general H bridge in the future, so I chose mosfets that can handle 20A. I don't think I'll have an issue with any motors in the near future. I am using INA213s for current monitoring (it's a dual H bridge if you couldn't tell from the PCB) but they're really annoying to solder. I was considering just getting an op amp and some resistor to create my own current monitoring circuit. If you know any chips that are not fine pitch ICs please let me know.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 07, 2018, 12:18:33 pm
content: Quote from: StillTrying on September 07, 2018, 11:58:02 amI'm guessing that the second capacitor (Vvb 20v) should be connected to Vs on the negative side not ground? if no, then what should be connected to Vs? GND?

author: StillTrying
author_rank: Super Contributor
timestamp: September 07, 2018, 01:18:42 pm
content: "I'm guessing that the second capacitor (Vvb 20v) should be connected to Vs on the negative side not ground?"No, just GND/COM. The final cap already has the full 20V Vb on it, it doesn't have to do the 12V+10V Vs H-switching thing.LO is a 0V to 12V square wave, Vb would be a 11V to 22V squarish wave, 20V Vb stays charged at the ~20V Vb voltage."what should be connected to Vs"Nothing, it only uses the 2 internal mosfets on the LO side.It's just an idea to create a simple voltage doubler for an independent Vb.

author: OM222O
author_rank: Frequent Contributor
timestamp: September 07, 2018, 01:24:36 pm
content: Quote from: StillTrying on September 07, 2018, 01:18:42 pmThe same result could be achived with a simple charge pump tho, you just need a discrete transistor to turn the 3.3v PWM to 12V PWM, then use a double diode and a capacitor. that's gonna be cheaper and uses less components, plus it's easier to solder by hand. Isn't that true?

author: StillTrying
author_rank: Super Contributor
timestamp: September 07, 2018, 01:44:54 pm
content: "The same result could be achived with a simple charge pump tho"That's what it is, the advantage in using a mosfet driver would be the higher +/- output drive current."3.3v PWM to 12V PWM, then use a double diode and a capacitor."You could use the PWM/EN signal, but you'd have to keep the duty within ~10% - ~ 90%.I did think of mangling the PWM/EN and DIR/IN signals to keep the Vbs charged, or using 4 signals, a separate IN and EN for each half of the bridge, but I think producing an independent Vb would be better because it keeps the simple DIR/IN and PWM/EN arrangement you already have, and allows the full PWM 0% to 100%, but it's up to you which method you want to try!