LCD High/low side switching positive/negative voltage with logic levels - Page 1
Forum section: electronics / beginners

author: aeberbach
author_rank: Regular Contributor
timestamp: July 02, 2021, 04:09:27 am
content: I want to drive a LCD screen (https://www.panelook.com/HJ080IA-01E_Innolux_8.0_LCM_overview_17000.html) from a FPGA. The data sheet shows that it needs a few different supply voltages like 18.9 and 3.3V and 10V and -7.8V... I can make those with various controllers. And now appreciate why most people just use the supplied converter board. What I need to do is control the sequence of applying power at startup which needs to be in a specific order, same for shutdown. No problem to control the timing with the FPGA with a clock and a state machine.The LCD has common ground suggesting high side switching with P channel MOSFETs. For example a FPGA pin controls one P-channel MOSFET Q1 (Source = Vcc) that drives a second MOSFET Q2 (Source = VGH 18.9V) with Q2 Drain connecting to LCD panel VGH pin. FPGA says turn on, power is applied. And so on at specified time delays according to the timing diagram.What I am unsure about is the negative voltage. If the supply voltage is -7.8V does that mean two N-type MOSFETS should be used in low-side switching?This ends up using about 12 individual MOSFETs but they are pretty cheap, space is not an issue and it means not being restricted to power controllers with SHDN/EN control. Also, Vcc powers both the FPGA and the LCD - the FPGA can't shut down its own supply so at least one switch will be required.Finally to trigger shutdown I'm thinking the supply voltage, 12V, can be monitored and the FPGA trigger shutdown when that drops below a threshold. Should be enough time to jump into shutdown sequence and protect the LCD. I'm not sure how robust it is if the timing is not followed but don't want to find out.

author: aeberbach
author_rank: Regular Contributor
timestamp: July 05, 2021, 09:46:52 am
content: I came up with some solutions. I hope someone can tell me if this looks OK?Case 1 - switching 3.3V where the load is the same voltage as the FPGA.Load is powered when FPGA pin (gate) is low. Sorry for no labels here I snipped too close - gate connected to FPGA IO, source connected to VCC, load connected to drain.Case 2 - Need to switch a positive voltage higher than Vcc.NPN transistor connected to FPGA pin AVDD_EN has pulldown, stays off until FPGA drives it high, then N-channel mosfet gate is low and switches on 10V AVDD to load.Case 3 - Need to switch a negative voltage.NPN transistor is connected to FPGA pin VGL_EN, Base is pulled up so it stays on, keeping P-channel mosfet gate turned off. When FPGA drives base low, mosfet switches on, switching on -7.8V VGL to load. Actually I am pretty sure I have this wrong since the base of the transistor is being pulled down to -7.8V. Perhaps this needs to be a PNP. I am getting confused at this point.

author: rs20
author_rank: Super Contributor
timestamp: July 05, 2021, 10:07:32 am
content: You've pretty routinely got your P-FETs upside down (i.e., drain swapped with source). Keep in mind that the little body diode shown in the FET symbol is a real thing; if you look at your Cases 1 & 2, you'll see that those body diodes are oriented in such a way that the voltage will flow through the MOSFET (through the body diode) pretty much unimpeded (aside from the forward voltage drop of the diode) regardless of how the gate is driven.For case 2, you've described it as an "N-channel mosfet" in your text but drawn a P-FET in the diagram. The diagram looks correct though, aside from the P-FET being upside down.For case 3, you've done a truly confusing thing by drawing the negative supply as being above everything else on the page. This is bad practice, and I had to redraw the diagram on paper to see what you had actually drawn here. Positive voltage = top of page, (relatively) negative voltage = bottom of page. Anyway, you're going to want an N-channel FET to switch a negative supply, since it's now a low-side switch (at least, when drawn with the negative at the bottom.) It's also highly doubtful the BJT will do anything sensible in that configuration.

author: rs20
author_rank: Super Contributor
timestamp: July 05, 2021, 10:10:16 am
content: Figure 12-2 ofhttps://www.latticesemi.com/-/media/LatticeSemi/Documents/ApplicationNotes/MO/MonitoringandControllingNegativePowerSupplieswithPowerManagerDevices.ashx?document_id=6413has a reasonable circuit for your case 3.

author: jpanhalt
author_rank: Super Contributor
timestamp: July 05, 2021, 10:19:54 am
content: I went through that exercise of sequencing voltages to an LCD fro Newhaven Display years ago. Then, I came to a part in the description that effectively said the important thing was not to have negative applied before positive, As I generated my negative with the MCU and a charge pump, I haven't worried about it since.Are you sure your display requires such careful sequencing?Can you provide a link to the datasheet that doesn't require sign in?

author: aeberbach
author_rank: Regular Contributor
timestamp: July 05, 2021, 11:04:45 am
content: @rs20 thanks for that sheet - looks like exactly what I want to do. I will redraw and base on that. Also the tips on how to fix the other cases.@jpanhalt - here's a no-login link.https://datasheetspdf.com/pdf-file/1111065/Innolux/HJ080IA-01E/1There may be better LCDs around but this one shows plenty of stock and is fairly cheap at under AUD$40. Others I might have tried are over $200. I am unsure about how critical the timing might be but the datasheet is very specific about required max and min times in both startup and shutdown sequences.

author: viperidae
author_rank: Frequent Contributor
timestamp: July 05, 2021, 10:53:11 pm
content: Do your converter chips that generate these voltages have enable pins?

author: aeberbach
author_rank: Regular Contributor
timestamp: July 06, 2021, 05:11:01 am
content: Some parts I have picked have a control pin. Vcc has to be switched of course; it has to be on all the time for the FPGA while being switched on in sequence for the LCD. I was leaning toward ignoring built-in control pins. It adds roughly $1 per circuit but it means a lot more flexibility in deciding what to use for those circuits. Seems like half the popular parts are out of stock and aren't expected in a long time. Maybe I should lay out MOSFET control options for everything (reserving space for it) as well as using SHDN pins where available and make the final decision with 0R resistors on the board. There are enough FPGA pins to double up.

author: aeberbach
author_rank: Regular Contributor
timestamp: July 06, 2021, 09:24:34 am
content: Take 2. Thanks everyone for the comments so far.Case 1: Switching VccCase 2: Switch a voltage higher than VccCase 3: Switch a negative voltageAnd the LCD panel datasheet is attached.

author: rs20
author_rank: Super Contributor
timestamp: July 06, 2021, 11:24:20 am
content: Much more pedantic stuff now. Overall, it basically looks good to me.For Case 1, you're going to have a Vgs of -18.7V on the BSH202, which has a absolute maximum of ±20V. This is technically OK I suppose, but it's generally a bit scary to be so close to a limit like that. I think you should be OK. The other cases' FETs have a good margin on Vgs.For Case 3, the 10kiloohm resistor you have on the base is concerning. Did you mean to copy 10ohmsfrom the document I linked? These gate resistors are kinda sorta maybe optional (my feeling is you probably don't need them for an LCD supply); but you should probably at least be consistent in having them for all or none of your cases.Similarly R23, R20 and R16 are all doing broadly the same job, so they should probably be the same value (10k would be my random guess).

author: aeberbach
author_rank: Regular Contributor
timestamp: July 07, 2021, 01:50:07 am
content: Thank you!MOSFET DMP610DL-7 might be better than BSH202. Everything looks similar but -30V Vgs rating. Vgs th is 2V rather than 1V but that should be OK - and it's cheaper. I think I overlooked it by selecting SOT-23 and missing SOT-23-3.In the case of the gate resistors, I understand the MOSFET has no static gate current when fully on or off. A small gate resistor would allow a high current in that transition period to make the switch fast?

author: viperidae
author_rank: Frequent Contributor
timestamp: July 07, 2021, 05:46:34 am
content: This isn't battery powered is it? A 1k pull-up on an 18v power line is going to burn 18mA when ever the MOSFET is switched on.Also going to burn around 1/3 of a watt.

author: rs20
author_rank: Super Contributor
timestamp: July 07, 2021, 06:19:48 am
content: Quote from: aeberbach on July 07, 2021, 01:50:07 amYou are correct that there is (practically) no static gate current to a MOSFET unless switching. And yes, a small gate resistor does allow a high current in that transition period to make the switch fast. But the very smallest gate resistor is (approx) 0 ohms, i.e., a direct connection rather than via a resistor -- so that gives the fastest switching.

author: aeberbach
author_rank: Regular Contributor
timestamp: July 07, 2021, 06:25:42 am
content: No batteries.

author: aeberbach
author_rank: Regular Contributor
timestamp: July 07, 2021, 09:46:01 am
content: There was a question about enable pins earlier. I don't plan on using them because they don't really mean "off" as in 0V. I don't know how critical switching voltages completely off is going to be and finding out could cost one LCD panel, or around $29.I have a circuit based on LM27313 (from TI Webench). I have a PCB laid out according to the guidelines in the datasheet. With SHDN at Vin it works pretty much how I would expect. The meter says 20.10 when I wanted 18.9 but I figure that's within the bounds of tolerance and I can adjust the resistors.My circuit. I'm testing it with Vin = 12.3v.When SHDN is grounded it "stops" in that I no longer see a frequency at L1 with the scope but I still see a voltage of 11.8v at VGH. Is it right that the inductor is basically a wire at DC and I'm seeing one diode junction worth of voltage drop?I think this means that for each one of my power circuits I need to check whether off means off, and use the MOSFET switch when it doesn't.