Odd Voltage Behavior under Odd Conditions for H-Bridge Driver - Page 1
Forum section: electronics / projects

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 19, 2022, 04:16:40 pm
content: To start off, I built my own H-Bridge Driver! It was part of a request to turn a DC-voltage input to an AC-voltage output. I got some help with that, and I truly appreciate all the help you guys have given me for this circuit. After testing it, it works....but with a few caveats. I wanted to make a post here to ask for some help in troubleshooting the thing.In a nutshell, my circuit has two voltage inputs: a +15 V source and a DC source that is converted to an AC-square wave voltage. The 15-volts power a +5V regulator which powers an ICM7555 clock and a CD4013B D-flip flop to generate two clock signals out of phase by half a cycle. These signals go into two separate LT1158 H-Bridge drivers to drive MOSFETS on and off, and being out of phase allows the LT1158s to drive the MOSFETs in such a way as to generate an AC output. Simulation shows it works well in theory. I posted images of my schematic below showcasing the schematic as a whole as well as just focusing on the H-Bridge portion. I tested this using a dual-power supply (while they are independent sources, they are connected to the same ground plane on my board) and an oscilloscope to monitor the output AC voltage. While it works as intended, there was an issue with the output voltage. I'll explain the problem and describe it in detail (sorry, I'll do my best to be as clear as I can).To better explain, with my dual supply, the two outputs are independent of each other. I kept one output at +15 volts to power the circuit while using the other input as my DC voltage to be converted to an AC square wave. There was no load at my AC output, but the output did go to an oscilloscope. Varying the DC voltage did showcase the correct AC square wave output. The first problem was noted when I tried to turn the DC voltage to 0 (turning voltage to 0 while in constant voltage mode, limiting output current to something very small). It did not go to 0 volts, but rather it went as low as +1/-1 volt. This was seen on my oscilloscope and my DC supply. I even turned the current control knob to 0 (setting my DC output to constant current mode and turning the output voltage knob to 0), and it still showed 1 volt (with an output current reading of 10 mA, quite odd). This was odd, so as a test, I used an old, single output power source that's independent of my dual supply, just to eliminate any potential issues when drawing from the same source. Turning on the single source while keeping the knobs at 0, it still showcased 1 volt. Apparently, when there's no voltage or current being outputted at the DC voltage terminals, it acts as a voltmeter of sorts.I've been thinking about why this occurs. At first, I thought it might have been reading a voltage difference between the two different SENSE+ pins of the LT1158s, but now I'm not so sure. For the first case, by having the voltage source be 0 volts, you are (ideally) shorting the DC voltage input to ground. According to the LT1158 datasheet, since short circuit protection wasn't required, I connected SENSE+ and SENSE- to TSOURCE on each LT1158 and left FAULT open. If there's no voltage, then there would be no current, but somehow there is voltage and current, if only a little.It's important to keep in mind that the MOSFETs are still being turned on and off whether or not there's a voltage at the DC input, so there has to be a difference between gate/source voltages to turn them on. Datasheet for my MOSFET (the STD95N2LH5) gives me a gate-source threshold voltage of 1 volt. Could the voltage I'm seeing on my DC supply be the voltage needed to turn on the MOSFET? Is it possible this problem hinged on my choise of MOSFET? The only connections the DC input has to the LT1158 is via TSOURCE through the MOSFETs, so I don't think it could be anything else.(On a somewhat related note, as a test, I kept the DC input voltage terminals open and measured the AC output. The output voltage was a square wave that had a max of +15 and -15 volts. According to the LT1158 datasheet, the TSOURCE and SENSE+/- pins have an absolute max of VCC+5 volts, so what I'm seeing is probably the max voltages being reached by the pins accordingly.)EDIT: As a third test, rather than have the DC voltage be 0 via the power supply, I used a wire to short the terminals together. The AC output voltage showed 0 volts. So, I suppose from this test, I may not be seeing the MOSFET gate-source voltage as I thought before. Perhaps there's something with the way the bench power supply is wired inside that is causing me to see a 1-volt output.

author: jwet
author_rank: Frequent Contributor
timestamp: October 19, 2022, 08:36:59 pm
content: A DMM has an input impedance of 10M generally. The FET's couple some of their gate voltage through to the source and gate. This is more than enough to light up a 10M load. There is also DC leakage on the FET's that can light up a 10M load. Add a 1k or so load across AC out and the circuit will behave more like you expect it to.Also, with a perfect square wave from a flip flop driving the bridge, you go from driving one fet directly to drving the other. The gate drivers are designed to do this very fast, so fast that both fet's will be on for a short time directly across the supply. This called shoot through. Some FET drivers account for it, you can control it somewhat by adding resistors in the gate legs. Often you can use a diode and a 2 R's here to have control off the turn on time and turn off time independently. What you'd probably want to do is to have a fast turn off and a slower turn on unless the LT1158 does it for you. I looked at 1158- this is likely a non issue but be aware.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 19, 2022, 10:16:39 pm
content: Thank you for replying. How could a FET's gate voltage be coupled through the source? I always thought that the gate voltage stays at the gate (following the principles behind an ideal MOSFET). Is this due to inherent parasitic components of the MOSFET?Thanks a lot for your suggestion on adding a 1k-load. It kind of solved the problem by dropping the voltage down from 1 volt to 0.6 volts. Not exactly where I want to be, but it's a step in the right direction. I would have never thought that adding a load would help things. Let me see if I understand the reasoning behind it. With the power supply, assuming it's outputting 0 volts, taking into account the input impedance, it's as if you're seeing a 1 Meg-ohm load between the DC input and ground. With the gate voltage bleeding through to the source and the LT1158s actively driving the MOSFET regardless of the drain voltage, the "source voltage" has a path back to my zeroed-DC input going through the input impedance. Looking at which MOSFETs are turned on in this H-driver, adding a resistor in between the AC loads would bridge the active MOSFETs together allowing for a path for the "source voltage" to go to ground, thereby "lowering" the voltage being seen at the zeroed-DC input source? It seems that, more or less, I'm just seeing leakage voltage from the MOSFET source terminals. My design has space for a resistor to go between the gate and source in order to help discharge the gate (as shown in the LT1158 datasheet in the section of low current shutdown), but I thought that it was causing my issues, so I removed it.Thanks as well for your warning about shoot-through. Yeah, the datasheet says it protects against shoot-through currents, but it doesn't seem to go into detail...especially since I disabled the short circuit protection. The DC voltage source will be current controlled (very slowly, but up to 2 amps at 12 volts), and I picked my MOSFET specifically to handle higher current than what was rated. Even in the worst case scenario, I was hoping that the MOSFET could handle it (famous last words). With what I mentioned above, would a large resistor between gate and source help mitigate the turn on/off times?(The goal of this is to convert a current-controlled DC voltage source to a current-controlled AC source, and that will be passed to an object with low resistance. Assuming that there will be a load at the AC output terminals, is it necessary to always have a fixed load between the AC sources? With your suggestion, I could solder a resistor in between the AC loads. All that would do is draw more current from the DC source, but it would allow a way for the "source voltage" to discharge when there is no load or DC voltage).

author: jwet
author_rank: Frequent Contributor
timestamp: October 20, 2022, 02:45:12 am
content: Big MOSFET's have tremendous gate capacitance. Banging on the gate with a square wave drives a lot of current into this capacitance, it has to go somewhere- Cdv/dt with fast rise times and maybe 1000 nF or C. 1K ohm might not be low enough really. In analog swtiches that operate at low levels- signal type level, the effect is called charge injection. Imagine that you're making a sample and hold by closing a switch at precisely the right time to sample some waveform. Each time, you close or open the switch, you add or subtract 100 pico colombs (charge = I*t). Your sample and hold has these little hold steps. What you're doing in similar but at a larger scale.You can't really get rid of charge injection by slowing things down- its an integral type thing, it will happen slower, but the same amount of charge will eventually get to the load.Thank me on line- I'm trying to up my "thanks". Good luck.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2022, 02:16:03 pm
content: Thanks a lot for your explanation. That helps me understand things better. Yeah, clock transition time going from low to high is really small at ~100 ns. As a test, using a 100-ohm resistor at my AC-output managed to get down to a low enough voltage...until it got really hot. That's on me for using a typical 1/4 watt resistor. I'll have to run more tests then and see if all parties understand what's going on. If they're okay with it, then great. If not, then...back to the drawing board. At least we know this circuit works...fundamentally.Thinking about it....with what you said....I am seeing some strangely large voltage spikes at the transition points. Sometimes, they're 1-volt above/below the magnitude of my DC source, and they happen only for ~100 ns, the same as the clock transmission time. I thought that adding an RC-filter (8.2k ohms, 100 nF) would filter that out, but unfortunately not. These voltage spikes might be related to what you said. If that's the case, then adding a fixed value resistor might help. Interesting how this goes. If the charge injection will end up going to my "load", then I'll have to make sure that it doesn't reach my actual load and go through the fixed value resistor instead.It sounds like the charge injection can be lessened by a MOSFET with a lower gate capacitance. I picked my MOSFET due to the max end-application voltage/current requirements, so I thought that having a MOSFET that can handle parameters greater than the max values would be better for the circuit. Also, to pass as much voltage as I could to the actual load, I went with a smaller R_DS_ON parameter as small as I could so that the voltage drop would be insignificant. Charge injection never occurred to me.EDIT: Observing the behavior with a 300-ohm 2-watt resistor at the load, it didn't help with the voltage spikes at my transition points. Great. Really hope I don't have to redo my PCB.

author: Ian.M
author_rank: Super Contributor
timestamp: October 20, 2022, 02:56:38 pm
content: A RC filter isn't practical with your very low load resistance. Look at symmetrical LC filters, possibly damped by resistors in parallel with their inductors.Beware: many higher power low ohm resistors are wirewound and typically inductive, so don't load the output significantly during brief glitches.Even if you are operating at 100Hz and getting two 1V (above/below the input DC level) 100ns glitches per cycle, they are insignificant as their contribution to the total RMS output is reduced by their effective duty cycle of 100ns in 5ms (a factor of 50000). Also, unless you are using a well-shielded differential probe with excellent CMRR, they could well be an artifact of your scope probing setup!

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 20, 2022, 03:22:29 pm
content: Thanks for replying. Yeah, I see what you mean by an RC filter not being practical. That's my go-to when I think of LPFs; in terms of application, I can see how using inductors would be better. It may not even be necessary, but I suppose that will depend on the demo/end application. If my set up will have the AC output be connected to a load consistently, then maybe that might be okay. It was only when there was no load that I found that there was an issue.Yeah, with such a small glitch period given the frequency of the waveform, it does sound insignificant. I can only hope that it remains insignificant in practice (i.e., hoping that a small, very brief glitch doesn't destroy anything). My setup is definitely not well shielded. I merely used a banana-to-BNC adapter and connected it to my oscilloscope. Not exactly the most robust, but it was the most convenient.

author: jwet
author_rank: Frequent Contributor
timestamp: October 21, 2022, 01:32:15 pm
content: On the topic of smaller FET's, and lower gate charge FET's, this is a holy grail problem. Its one of the things that really limits efficiency in high power high frequency SMPS and why you don't see many converters running much over a Mhz with most in the 300-600 Khz range. Don't use a FET that is too much larger than required- you're wasting efficiency for no real gain. New devices using compound semiconductors like Gallium Nitride and Silicon Carbide (most useful at high voltages) were developed to combat this and let you make faster switchers. GaN devices have much lower gate charge at a given Ron and are available today. This is a figure of merit in use "milliohms per nanocoulomb"- you'll see it in the literature. Silicon FET's have improved in this respect over the years but GaN is a game changer.BTW- Maxim, my old employer, worked on recycling the charge used for FET turn on and turn off. I believe we have a patent on something in this area. Instead of just discharging the gate charge and wasting it, you would store it somehow and use it on the next cycle. If you could recover and reuse even 50%, this would be like having a 50% lower gate charge. The circuits are complicated though and didn't see much light of day. These techniques are similar to resonant techniques where you maintain this high circulating energy- it can be very efficient in some applications like old CRT Flyback circuits.Electronics is pretty darn cool isn't it!

author: jwet
author_rank: Frequent Contributor
timestamp: October 21, 2022, 01:45:32 pm
content: Also forgot to mention, AC couple the resistive load so it only snubs the AC components and doesn't have to see the full class A DC when you operate the H bridge. This AC stuff won't be require a lot of power.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: October 21, 2022, 04:17:00 pm
content: Thanks again for your replies. Indeed, it's quite amazing to think about, especially when things don't work quite as you expect. I never delved into different material types and their pros/cons before, so it's great to learn about new stuff for future applications. Thanks a lot for this.Forgive me for asking such a rudimentary question, but how would you AC couple the output? Typically, at least from what I learn in small-signal amp design, an AC source is represented as a voltage source with one end being the AC-output and one end connected to ground. You AC-couple the input by putting a capacitor in series with your AC source before it goes into the small-signal amplifier. Would you do it the same way here, putting a capacitor in series with your load, or would you do it another way? I ran a demo with the end user, explaining the outcome of the design, and they seemed okay with it. Although, I did add a capacitor in between my AC-output terminals (in essence, putting the cap in parallel with the resistive load). I thought putting it there would help filter the signal (or at least help mitigate the transitions), but I may have been fundamentally wrong about that.

author: jwet
author_rank: Frequent Contributor
timestamp: October 21, 2022, 05:19:07 pm
content: Sorry- I didn't explain this very well. It looks likes you havle an AC problem and a DC problem. Put the low value resistor in series with cap (ac couple) across the load. This is for AC. Almost any cap will look like a short with to these fast edges- something like 10 nF might be a place to start with a 100 ohm resistor. The duty cycle of these edges is low enough (% of rise time to dwell time) that the power should be low- 1/4W probably. This is similar in function to a Zobel network commonly used in audio output circuits. You can tame the DC portion with a DC connected R (no cap) across the output but probably a higher value maybe a 1K or so, this will see the full swing so do your V2/R calc for power- this will likely be a pretty small current just FET leakage, etc but the voltage swing is high which will drive power dissipation. If its 15v into 1k, this is only a quarter watt- use a 1/2W or 1W- you don't it getting hot. Resistor ratings are often for a 50 deg C rise- makes for low rel circuits.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 21, 2022, 06:46:43 pm
content: LT1158 is $10, looks very much a legacy part. Is it really worth it for this? Especially if you're not using its protective features?What is "VCC"?You don't need anywhere so many capacitors. Parallel staggered values are almost never helpful, either. In particular, 7555 and CD4013 draw very little shoot-through current (at most, 10s mA), at little rate (dI/dt). One 0.1uF, or whatever the regulator requires really, will suffice for the whole 5V net, even with fairly long routes to them (10cm?).Tim