Diode Driver Control - Page 1
Forum section: electronics / projects

author: apraxophobia
author_rank: Contributor
timestamp: June 20, 2024, 10:54:34 pm
content: I am making a (laser) diode driver current source that needs to supply (max) 13A with a 6V diode drop.I have a 7V source lined up, and the driver circuit attached. It's a P channel FET controlled by a precision op amp (with appropriate high-power components).I'll probably wind up using a gate driver since I have to modulate this at 70 kHz.In this scheme, the current output becomes I = (Vreg - Vset) / Rs, highly dependent on the sense resistor. I want Rs to be small to avoid power loss/heat, but the smaller it becomes the more difficult it is to make a DAC to control this, because the range of Vset from 0 to full current gets smaller and smaller. Currently I am planning to make a 6.35-7V DAC based on a 0-5V DAC (Rs = .05, DAC made with all voltages, including 0V, referenced to the 7V for stability), using some scaling after the DAC. This is insanely messy, though.Is there no better way to configure the current controller so I don't need to make such a finicky DAC controller? Some kind of current mirror trick? And if not, is this DAC even possible/a good idea? I have very little experience with those.

author: moffy
author_rank: Super Contributor
timestamp: June 21, 2024, 03:16:56 am
content: An easier way is to use an N channel MOSFET with the source connected to Rs and the other side of Rs to ground. The Anode of the diode goes to Vreg and the Cathode to the drain of the N channel MOSFET, that way all of your control signals can be ground referenced which usually makes life easier.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: June 21, 2024, 03:38:47 am
content: Quote from: apraxophobia on June 20, 2024, 10:54:34 pmThat's many watts, does this need a grounded cathode/case for cooling ?Quote from: apraxophobia on June 20, 2024, 10:54:34 pmHow much overshoot can the laser diode tolerate ?It's common to see a RC element added OpAmp out to INV pin, to give you some control over the stability/overshoot.Quote from: apraxophobia on June 20, 2024, 10:54:34 pmCurrent mirrors are not difficult, you have effectively already made one here.If you cannot use a NFET as suggested in #1, you can level shift a reference voltage.Trickier will be to modulate the diode at 70kHz with fast edges and no overshoot.A shunt circuit would be fastest, the current stays fixed and you steer where it goes, but that will mean ~100W continual system power.

author: apraxophobia
author_rank: Contributor
timestamp: June 21, 2024, 11:42:49 am
content: Quote from: PCB.Wiz on June 21, 2024, 03:38:47 amInteresting idea, although I'll need this to do both the 70 kHz modulation and a steady-state, stable DC (at low-ish power), so I doubt I can arrange that. Re level shifting: my thought was to define several different voltages, all off of the 7V Vref, in order to make the DAC. This gets very messy very quickly though (precision reference, dividers, buffers), unless I am missing a simpler way.

author: apraxophobia
author_rank: Contributor
timestamp: June 21, 2024, 11:44:28 am
content: Quote from: moffy on June 21, 2024, 03:16:56 amThanks a lot! Do I have this right in the attached image?I am working off of this roughly (https://arxiv.org/pdf/0805.0015).If I do it with the low side, N channel switch, how do I make the VREF of my control DAC? In the high-side P Channel scheme, the benefit was that you can make DAC VREF and DAC GND off of one VREF, so changes in the setpoint will mirror those of VREF. With low side, DAC GND can be board ground, but won't DAC VREF swing with whatever other voltage I use to make it (e.g. 7V VREF)? Or am I missing something?

author: moffy
author_rank: Super Contributor
timestamp: June 21, 2024, 11:51:39 am
content: Quote from: apraxophobia on June 21, 2024, 11:36:12 amThe drawing looks good, but I can't quite understand your problem with the DAC Vref. I would assume, maybe incorrectly, that Vset is the output of the DAC which is something like Vset = k*Vdac = k*Vref*(0x111/0xFFF), where 'k' is a gain, can be less than 1 and 0x111 is the current digital DAC value and 0xFFF is the DAC max value, unipolar.

author: apraxophobia
author_rank: Contributor
timestamp: June 21, 2024, 02:13:02 pm
content: The drawing looks good, but I can't quite understand your problem with the DAC Vref. I would assume, maybe incorrectly, that Vset is the output of the DAC which is something like Vset = k*Vdac = k*Vref*(0x111/0xFFF), where 'k' is a gain, can be less than 1 and 0x111 is the current digital DAC value and 0xFFF is the DAC max value, unipolar.I think you are right and I was misunderstanding- thanks again! Makes me wonder why it was done the other way in the paper.As for making a 0-.65V DAC, I assume this isn't too tricky given I can just scale the output of, e.g., a 0-5V DAC. With this factor of ~-ten gain, I assume I lose some resolution, right?

author: ajb
author_rank: Super Contributor
timestamp: June 21, 2024, 03:09:46 pm
content: Quote from: apraxophobia on June 21, 2024, 02:13:02 pmSome laser diodes have one side of the diode or the other bonded to the case (depends in part on the semiconductor technology used), which would limit which driver structures are practical, or you might want to take certain measurements at the diode that are easier with the cathode grounded. Or since the paper says it's based on a previous design, they may simply be inheriting the older design's limitations/priorities.QuoteIf you do the scaling digitally, then you will lose a lot of resolution: using only 0-0.65V out of a 0-5V range means you're only using about 1/8 of the range, which at 8 bits means you only have 33 possible levels instead of 256. But if you do the scaling in the analog domain, either by using a 0.65V reference for the DAC or by dividing its output down to that level, you will not lose any resolution. However, you will be dealing with relatively small voltages (especially if you have a high resolution DAC and want the low end of the modulation to be useful), so you will need to be cognizant of that depending on your requirements. But since we're talking about a 13A laser diode, you probably don't care about anything below, what, 500mA? and it'll probably be fine. The other option is to amplify the current feedback, which could let you use a really small shunt value while maintaining larger voltages from the DAC through the main analog signal path, but this necessarily introduces lag into the feedback loop and makes high frequency operation more challenging. Lots of possible tradeoffs here depending on your priorities.Quote from: apraxophobia on June 20, 2024, 10:54:34 pmYou can't really use an off-the-shelf gate driver for the FET here, if that's what you were thinking. Those are designed for switching operation, to take the FET from fully on to fully off and back again as quickly as possible -- they are not meant for analog operation like this. An op amp will be fine, but the part choice bears some consideration depending on the FET chosen. Speaking of the FET, note that you will need to choose that carefully as well. There are a ton of parts that will appear to meet your requirements based on the first page of the datasheet, but these too are largely designed for switching operation, and may not survive for very long in a partially-on state. You will need to look at the safe operating area chart and ensure that your current/voltage operating point is in the DC region of the chart.

author: apraxophobia
author_rank: Contributor
timestamp: June 21, 2024, 03:40:08 pm
content: You can't really use an off-the-shelf gate driver for the FET here, if that's what you were thinking. Those are designed for switching operation, to take the FET from fully on to fully off and back again as quickly as possible -- they are not meant for analog operation like this. An op amp will be fine, but the part choice bears some consideration depending on the FET chosen. Speaking of the FET, note that you will need to choose that carefully as well. There are a ton of parts that will appear to meet your requirements based on the first page of the datasheet, but these too are largely designed for switching operation, and may not survive for very long in a partially-on state. You will need to look at the safe operating area chart and ensure that your current/voltage operating point is in the DC region of the chart.Super helpful, thanks very much. I had in mind using three IRLz24's in parallel, to distribute heat and limit current (even though each can apparently easily handle 13A). I wasn't aware of the importance of the SOA, but the datasheet for that doesn't specify a DC region, only up to 10 ms. If I would be well-inside the SOA curve for 10 ms (with three FETs splitting the current), can I assume this would work, or should I explicitly find a FET with DC spec?Edit: It has Id = 12A continuous for Vgs = 5V. If I split up Id by paralleling three FETS, this should be fine, no?

author: ajb
author_rank: Super Contributor
timestamp: June 21, 2024, 06:21:46 pm
content: Quote from: apraxophobia on June 21, 2024, 03:40:08 pmThis is the real trick with FET specs: the numbers on the first page of the datasheet are basically useless in most applications. They assume very optimistic operating conditions, such as continuous conduction at Tj=25C, or ideal switching scenarios. As soon as you operate in between fully-on and fully-off for any significant period of time, the real numbers get a lot lower, and you have to look through the rest of the datasheet to figure out what they are. Putting three of those in parallel might be fine, but the IRLz24 is a pretty old part. The graphs scanned in from an old datasheet and the Rds(on) of 0.1R give that away immediately. You mentioned wanting to run a 6V diode from a 7V source, so you would need at least three of those in parallel just to get the Vds low enough for that to work.Spreading the thermal dissipation of the driver across three transistors isn't a bad idea, but might not be necessary with the right part, depending on the operating point. There are more modern parts that will allow you to operate with lower Vds with a single transistor, which will significantly improve the efficiency of the driver and make driving the FET(s) easier too. And a lot of those parts will have SOA charts that include DC, so you won't have to guess about how they might perform off the chart. For a single transistor operating at 13A, you probably want to start looking at parts rated for at least 20A continuous.One thing that makes this a bit trickier is that the Vf of the laser diode is not fixed, so some amount of power dissipation shifts between the driver FET, the shunt, and the diode itself depending on the VI curve (and temperature, and conversion efficiency) of the diode. Your worst case operating point for the FET might not be at full output current, but it's probably somewhat close.On the subject of driving the FET, gate charge can be an issue. The more charge has to be driven into/out of the gate to change the operating point, the longer the FET will take to respond to the driving op amp, and this directly affects the stability of the driver as well as overshoot. The driver circuit will need to be compensated to deal with this, and it's a good idea to include provisions for different compensation strategies when you build it so you can make adjustments.

author: apraxophobia
author_rank: Contributor
timestamp: June 21, 2024, 06:56:36 pm
content: This is the real trick with FET specs: the numbers on the first page of the datasheet are basically useless in most applications. They assume very optimistic operating conditions, such as continuous conduction at Tj=25C, or ideal switching scenarios. As soon as you operate in between fully-on and fully-off for any significant period of time, the real numbers get a lot lower, and you have to look through the rest of the datasheet to figure out what they are. Putting three of those in parallel might be fine, but the IRLz24 is a pretty old part. The graphs scanned in from an old datasheet and the Rds(on) of 0.1R give that away immediately. You mentioned wanting to run a 6V diode from a 7V source, so you would need at least three of those in parallel just to get the Vds low enough for that to work.Spreading the thermal dissipation of the driver across three transistors isn't a bad idea, but might not be necessary with the right part, depending on the operating point. There are more modern parts that will allow you to operate with lower Vds with a single transistor, which will significantly improve the efficiency of the driver and make driving the FET(s) easier too. And a lot of those parts will have SOA charts that include DC, so you won't have to guess about how they might perform off the chart. For a single transistor operating at 13A, you probably want to start looking at parts rated for at least 20A continuous.One thing that makes this a bit trickier is that the Vf of the laser diode is not fixed, so some amount of power dissipation shifts between the driver FET, the shunt, and the diode itself depending on the VI curve (and temperature, and conversion efficiency) of the diode. Your worst case operating point for the FET might not be at full output current, but it's probably somewhat close.On the subject of driving the FET, gate charge can be an issue. The more charge has to be driven into/out of the gate to change the operating point, the longer the FET will take to respond to the driving op amp, and this directly affects the stability of the driver as well as overshoot. The driver circuit will need to be compensated to deal with this, and it's a good idea to include provisions for different compensation strategies when you build it so you can make adjustments.Many thanks- I think the IRLZ24 is out. I have my eye on some TI models (CDS18537) that have clear DC ratings, ~20mR Rds, and low input charge at ~20 nC (although not as low as the IRLZ24).If I have three in parallel, then I need to move 60 nC around for full on/off (which is overkill). To be safe, if this happens in 1 MHz (targeting 70 khz cleanly), then I need about 60 mA gate driving current, right? The ultra-stable op amp I have in mind can't do this, so I'll need to make a gate driving circuit of some sort...