4013 On/OFF + Dual MOSFET load switch protection circuit - Page 1
Forum section: electronics / beginners

author: VEGETA
author_rank: Super Contributor
timestamp: August 12, 2025, 05:54:27 pm
content: HelloI managed to create this circuit after consulting others here, please check attached images.Basically it is a circuit which provides output protection and load switch functionality for buck regulator output. It can get remote activation signals (3.3v pulses, or just 3.3v continuously) to activate or deactivate the buck rail. Also, it must provide physical protection like totally cutting the output when off, with reverse current protection as well in case wrong voltage appeared on the output.It has 2 parts:4013B circuit:receives pulses (or continuous but mainly pulses) of 3.3v which are "ON pulse" and "OFF pulse" then outputs "Activation signal" which is 3.3v when "ON" and GND when OFF to turn the target circuit on or off respectively.Dual MOSFET circuit:takes the "activation" signal, and if it is 3.3v then the circuit is activated and vise versa. this also acts as reverse current blocking and protection if wrong voltage appeared on the output + completely physically cuts off the power when it is off.Notes:- "AUX out" is just a 3.3v regulator which acts as "auxiliary supply". it is the power source for these circuits.- Q5 is a matched PNP transistor pair (Nexperia BC856BS,115) to have the protection circuit working nicely.- Q1 and Q2 areHUASHUO HSBA90P02.- 4013 part number used isTI CD4013BPWR, which is TSSOP to save space. It is dual flip-flop, could not find a single one in package. Grounded the other unused one.- Switch (SW3) determines if the circuit is always on or depends on the 4013B circuit and its signals (from J7).I have revised it quite a lot but still wants to make sure it is correct before making the board.Kindly check and advise if it is good to go without mistakes.Thanks in advance

author: VEGETA
author_rank: Super Contributor
timestamp: August 13, 2025, 09:38:21 pm
content: another side question that i have: since i am using 2 mosfets, will they cause the same or similar effect of a linear regulator?I know they are running basically as a full on switch but still they have their resistance (so little though)... thus, I wonder if it will have even the slightest effect?? not that i would need it but still good to know

author: DavidAlfa
author_rank: Super Contributor
timestamp: August 14, 2025, 05:55:54 am
content: Q1 will always feed power through the intrinsic diode betweed D and S..Why not just use the classic back to back fet?

author: VEGETA
author_rank: Super Contributor
timestamp: August 14, 2025, 07:32:55 am
content: Quote from: DavidAlfa on August 14, 2025, 05:55:54 amWe developed this circuit here:https://www.eevblog.com/forum/beginners/reverse-current-blocking-circuit/Since both are P-Mosfets, Q1 will pass current backwards for sure, but Q2 ensures this does not happen. Q1 is used as a load switch while Q2 is for reverse current blocking and reverse protection.If Q1 is off, its own diode will not pass current as well since it is reverse biased. Q2 won't have anything as well but if a voltage appeared at Q2 output, Q2 itself will be off and its diode will not conduct backwards since it is reverse biased in that case.If Q1 is on, it will conduct and the input to Q2 will have voltage. If Q2 output is ok and does not have a voltage already, then it will conduct as usual. While if Q2 output have a voltage at its output higher than the input then it won't conduct.I have 5v regulator and 3.3v/5v regulator (depending on user choice). both regulators have this circuit... now if i shorted their outputs together, the 3.3v will be cut off immediately by the action of its Q2. Therefore, always going to be a safe operation which is my goal.Having Q1 is also essential since I need the ability to activate or deactivate the circuit via remote signals or just a continuous ON status, depending on user preference.what do you think?I am ok with this circuit as long as it works even if there is a smaller solution.I want to make sure the 4013B part works fine too.

author: DavidAlfa
author_rank: Super Contributor
timestamp: August 14, 2025, 11:26:29 am
content: Why the 4013 anyway? What's the difference between a pulse, or just setting the pin high or low?How about only using the 3.3V buck regulator, adding a resistor to the feedback loop so when pulled to ground the output increases to 5V.If you want independent 5/3.3V switching between different outputs, you can heavily simplify the circuit:

author: VEGETA
author_rank: Super Contributor
timestamp: August 14, 2025, 11:44:23 am
content: Quote from: DavidAlfa on August 14, 2025, 11:26:29 amwell, i want to integrate this with another separate system which has on pulse and off pulse separately, thus i need to make mine compatible with it.4013b was chosen instead of 555 timer because it works well with 3.3v (must be because of the separate system) more than 555 timer solution. plus it is kinda cheap and available.so my power supply can work in 2 modes: normal and remote.remote is what 4013b circuit will achieve, by taking on and off pulses from external system.while normal mode just delivers constant on signal to the mosfets circuit to keep them fully on all the time.the switching between 2 modes happens by a simple slide switch.can you re-check my circuit explained above and tell me your opinion if it will work as intended, i know it is kinda basic and should work but i love getting more than my opinion.thanks for your time and attention to my post.

author: DavidAlfa
author_rank: Super Contributor
timestamp: August 14, 2025, 11:50:22 am
content: Aha, I understand now!For 5Vmax operation, you also have the 74hc74.I guess you didn't see my edited message!Even simpler... You can control the fets straight with the 4013, no need of more parts.The fet voltage drop will be negligible if you use ones with a sensitive Vgate threshold, so they fully conduct at -3Vgs or so.

author: Zero999
author_rank: Super Contributor
timestamp: August 14, 2025, 04:26:37 pm
content: The 74HC74 is more suited to 3.3V, than the 4013.No, never connect the outputs to 0V. Always leave unused outputs open circuit.If only one flip-flop is required, then consider using a single flip-flop package, such as the 74LVC1G74, to save space.https://www.ti.com/lit/ds/symlink/sn74lvc1g74.pdf

author: VEGETA
author_rank: Super Contributor
timestamp: August 14, 2025, 05:15:54 pm
content: QuoteI already have that implemented, and a small slide switch changes it.___the target device i am powering have multiple versions but all of them have 6 pins... most important ones are 1 to 4... 1 and 2 are ground, pin 3 is either 3.3v or 5v depending on the target device, while pin 4 is always 5v.so for pin 4, I have a dedicated constant 5v regulator.and for pin 3 I have the same circuit but with changeable feedback resistors to allow the choice.now, for target devices, i have a fear that for those with pin 3 as 5v that it is internally connected to pin 4... thus i need the physical hardware protection circuit mentioned here.on the other hand, i also require ALL rails to be controlled on or off together which is why i need this circuit even for the constant 5v rail + another 9v/12v rail not mentioned here. this is so that the psu can run on or off using either remote signal or in "normal" mode via the switch mentioned.you ask if it requires a signal to turn on, then how to do it while the circuit itself does not have power? because in the PSU there is an AUX rail delivering 3.3v which is always on but this one is not delivered to the output pins but rather to the external circuit which needs to generate on and off pulses. all of this is managed ok, no need to worry about it.my circuit is already complete and layed out with BOM ready... do not want to change it unless it is wrong. price and space is kinda ok too.Quotewill this work exactly as 4013B? I mean using my circuit above.I found it here:https://www.lcsc.com/product-detail/C70285.htmlQuotewell,this commentsuggested i must hook them to 0v.so you suggest hooking the inputs to ground while leaving unused outputs floating?

author: Zero999
author_rank: Super Contributor
timestamp: August 14, 2025, 05:51:40 pm
content: Quote from: VEGETA on August 14, 2025, 05:15:54 pmNo it doesn't. Read it again.QuoteYes.

author: VEGETA
author_rank: Super Contributor
timestamp: August 14, 2025, 07:06:56 pm
content: Quote from: Zero999 on August 14, 2025, 05:51:40 pmoh yes you are correct! so the inputs are going to ground using pulldown resistors and outputs are either used or kept floating.i will consider using sn74lvc1g74 you suggested since it is very small and has only one unused output which is Q-. too lazy to do it now though.however, what do you think about the rest of the circuit?

author: VEGETA
author_rank: Super Contributor
timestamp: August 14, 2025, 08:29:13 pm
content: I saw sn74lvc1g74 and it uses active-low signals not active high as required by my circuit and 4013B uses.sn74lvc1g74 PRE and CLR are active low, thus my positive pulses won't trigger them. my default status of both inputs pulled down is considered unstable for sn74lvc1g74.is there another IC or just stick to 4013B?

author: Zero999
author_rank: Super Contributor
timestamp: August 15, 2025, 09:55:43 am
content: I suggest taking a bit more time to think about this. Rather than giving you the answers, I'll guide you, in the hope you'll learn more.Regarding outputs: why would anyone think it's a good idea to connect an output directly to either power rail? Consider what would happen if the output tried to go high, when it's connected to 0V, or low, when connected to +V?What does active low/high mean?Note preset and clear have opposite functions. And that the outputs are complementary, i.e. they are the inverse of each other.You should be able to substitute the 4013 for the 74LVC1G74, but the connections will need to be changed.I've quickly looked at the datasheet for Q5 and it doesn't appear to be a matched pair. It does say closely matched gain, but how close isn't specified, along with no mention of VBEmatching. In reality the two transistors probably will be quite well matched any they will be well thermally coupled, but there's no guarantee. You need the BCM856 which is specified to be matched within 2mV of VBEand 0.9 for hFE.https://assets.nexperia.com/documents/data-sheet/BCM856DS.pdfOther alternatives are the DMMT3906W or DMMT5401.https://docs.rs-online.com/6268/0900766b814f35bb.pdfhttps://www.mouser.co.uk/datasheet/2/115/ds30437-3214642.pdf

author: VEGETA
author_rank: Super Contributor
timestamp: August 15, 2025, 11:50:58 am
content: Quotewell, I always put pull-up and pull-down resistors so that no short circuit happens. in this case i believe 47k is the resistor value, so if it went from 0v to 3.3v, the resistor will act as a load and no short circuit will happen.but as you suggested, i will float unused outputs and connect unused inputs to ground via 47k resistors.Quotein the case of 74 single flip flop ic, it is active low. meaning it will get activated when 0v is delivered to the input... while the "regular" status will require constant 3.3v (via pull-up resistor) to always be at that pin.for my 4013b as seen, pulldown resistor is the default case making it 0v unless a HIGH pulse of signal is delivered to the pin which is exactly what I require. As I said, the external mod or system or circuit outputs active high pulses and I must deal with it.Quotewell, i thought hooking the inputs to pull-up resistors to have constant 3.3v at input pins = not active. Then get 2 small transistor with their gates/base connected to my "On" and "Off" pulses so that these positive pulses can let the transistor pull the inputs to 0v which activates them.However, this will add more parts. Yes, space can be lowered a bit but cost won't be affected really since 4013B is kinda the same price as 74.Its datasheet says this:~PRE -> Preset input - Pull low to set Q output high~CLR -> Clear input - Pull low to set Q output lowmeaning, using my suggested transistor pulldown circuit... I can hook my "On" signal to the PRE and "Off" signal to CLR (via transistor pull-down..etc) and this will achieve the same result. While everything else should be the same, like D and CLK as I layed them out in my original circuit.do you think this is correct? I've done it before in the dual mosfet circuit in the image in first post, specifically the transistor which activates the rail.QuoteNo problem, I got this one:https://www.lcsc.com/product-detail/C458150.htmlseems to also have the exact same footprint and pinout, so it is a drop-in replacement.___besides that, do you think the circuit will function as intended? i mean isolation, activation, etc..

author: VEGETA
author_rank: Super Contributor
timestamp: August 17, 2025, 08:51:52 pm
content: I tried something I never do usually which is use LTSPICE to simulate some circuits... I simulated the D-Flip-Flop circuit as seen in attachment.Looks like it works as intended. I added a pulse for ON (Green color) then waited a bit and send a pulse for OFF (Blue color). Output (Red color) follows as intended except that it is limited to just 1v regardless of using 3.3v or 5v as supply voltages. Maybe this is with the internal model itself, otherwise should work fine for real world.Tomorrow I will try the load switch + ideal diode one, unless a champ can do it and upload the file___still waiting the continuation of this discussion from my last post.

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2025, 09:21:05 pm
content: It would make it easier to follow if you labelled the nets you've plotted and avoid using dark blue, on a black background.To illustrate the point I made above: try swapping the preset and clear connections and take the output from the inverting output.

author: VEGETA
author_rank: Super Contributor
timestamp: August 17, 2025, 09:56:32 pm
content: Quote from: Zero999 on August 17, 2025, 09:21:05 pmI made the adjustments. Now On is on CLR and OFF is at PRE. output from Q- is similar to previous circuit. I guess the colors now are more clear.I think I proposed a solution to the previous post by using transistors to do the inversion but your solution seems a bit easier.You mean by using the single flip flop which has inverted inputs... I need to do the following to get the same result:- Connect ON signal pulse to CLR.- Connect OFF signal pulse to PRE.- Take the output from Q not Q-.this is from checking this table:By doing this I will get the same output result as if using CD4013B like I previously posted... meaning when ON pulse happens, Q = 3.3v. When OFF pulse happens, Q=0v... All while keeping Q- floating.is this correct?However, can this actually work like this while having both ON and OFF signals pulled down? and clock having the circuit as shown above to force off status initially?

author: VEGETA
author_rank: Super Contributor
timestamp: August 18, 2025, 06:51:46 am
content: I managed to do the circuit in LTspice, please check it.Notes:- D1 is just there to make simulation works nicely. I do not know of a better way to hook 2 sources at the same node.- E1 is to deliver 3.3v to the mosfet instead of 1v from the D-Flip-Flop model, which always outputs 1v max.- mosfet models were not specifically selected. Just a model which works. tried several values and worked.- the waveform shows the current which flows in the right side mosfet, showing that when 5v (~4.6v due to diode) exists, it turns to 0 amps.

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2025, 08:07:21 am
content: E1 isn't needed. The output voltage of the flip-flop can be set to 3.3V. Refer to A. special functions in LTSpice help.https://ltwiki.org/LTspiceHelp/LTspiceHelp/A_Special_functions_.htmIf you want a PSU which can't be back-fed, then use an ideal diode, rather than a Schottky diode.https://www.analog.com/en/resources/technical-articles/ltspice-simple-idealized-diode.html

author: VEGETA
author_rank: Super Contributor
timestamp: August 18, 2025, 08:58:02 am
content: Quote from: Zero999 on August 18, 2025, 08:07:21 amoh thanks for your awesome tips. I am really not good at ltspice, will do more learning for sure.so i think now the circuit is verified to be working correctly... do you think so? if so then i will proceed with making the prototype pcb using 4013b and the mosfets i chosen earlier + BCM857BS instead of non-M variant.I noticed a huge current spike on the mosfet m1 when it switches off, should this be worrying or is it due to simulation?anything else to worry about?

author: VEGETA
author_rank: Super Contributor
timestamp: August 19, 2025, 07:35:22 pm
content: I made a modification to the circuit in ltspice to make it 2 duplicated circuits, one for 5v and the other for 3.3v both at the same output.it worked perfectly fine as intended where 5v always gets to the output while 3.3v gets cut if both are on.i think everything is fine now, and i am ready to make the prototype!!if you guys have any last tip please share it.thanks!!P.S: I will invest more time to learn LTspice, any good resource? I found youtube videos by frez and others to be good enough for now.

author: Zero999
author_rank: Super Contributor
timestamp: August 20, 2025, 11:00:12 am
content: Reading the LTSpice documentation is an obvious start.It's also a good idea to be familiar with SPICE in general. The SPICE 3F3 manual can be helpful. Understanding the fact that the GUI is a front-end and all schematics are converted into net lists first, is very important.http://www.spiceopus.si/download/spice3_manual.pdfRegarding the 4013, note that if it's powered by 5V, it's not quite logic, level compatible with 3.3V logic. A simple diode and resistor circuit can be used to shift the signals up by 0.6V. I don't know if this is an issue in your case.

author: VEGETA
author_rank: Super Contributor
timestamp: August 20, 2025, 01:01:43 pm
content: I will invest some time in learning ltspice but generally i require only basic circuits like the one here, not very sophisticated ones with custom models and stuff.QuoteAll I need it to do is send 3.3v or so signal to BSS138 mosfet (Q6 in above picture) gate (through the SW3 slide switch) as an "Activation signal". That small mosfet pulls the gate of the load switch mosfet to ground, thus it won't require very high voltage especially that the current will be very small due to the 47k pulldown resistor.BSS138 shows a Vgs threshold of about 1.5v max, and in the graphs show that for something like Vgs=3v it will be about 1.5-2 ohms resistance at low currents. No need for any logic stuff here, just a simple n-mosfet gate driving voltage.it will have 3.3v and will be ok in my opinion, do you think so too?

author: Zero999
author_rank: Super Contributor
timestamp: August 20, 2025, 04:54:53 pm
content: All I know is you posted two circuits in the .asc file, which appear to be for 3.3V and 5V supplies. Presumably these will both be on the same board, each using one half of the 4013. I noticed you set the voltage for both 4013s to 3.3V, yet one has 5V on it's inputs; is this an error? Applying 5V directly do the input of a CMOS IC powered from 3.3V is a bad idea. They can both be powered from 5V, but if you want one to have 3.3V logic levels connected to its inputs, it won't reliably work.SPICE models are often rough approximations. Learning how to use it, involves being aware of the model limitations.

author: VEGETA
author_rank: Super Contributor
timestamp: August 20, 2025, 05:39:21 pm
content: Quote from: Zero999 on August 20, 2025, 04:54:53 pmwell, then to clarify this, the circuit uses only one part of 4013B which is 3.3v and won't take any 5v inputs, it will deliver only one output to all circuits for all rails. Each circuit is like the above one (load switch mosfet + ideal diode mosfet)... the total is 3, so that they all get activated and/or deactivated together.I have modified the circuit and attached the file, this should be accurate. I also added a 7v supply + idea diode to the load to test the functionality of protection as we did previously, but of course the final circuit won't have that.the psu will have 2 modes: normal or remote, from switch SW3. Normal mode delivers constant 3.3v to N-MOSFETs gates of all rails to keep them all always on at all time, hence the name "normal" which is how any PSU works. it will of course have a physical AC power on button at the very input, and with "normal" mode, once the user presses the physical power switch, the psu will power the device as any regular psu."remote" mode, is when the circuit developed here will deliver the 3.3v activation signal to the mosfets gates. so that the regulators will be powered on but the dual mosfet circuit will completely cut off the power from the output target device unless 4013b sends the signal to activate it.the 3.3v for 4013b and even the constant one is derived from a dedicated 3.3v small rail (AUX) which gets activated once the physical switch is pressed and does not have controls to deactivate it (but has protection).Now, the user must install an external mod board to the target device which gets powered by the 3.3v AUX rail and deliver the 2 signals "ON" and "OFF" to 4013B. this external mod is beyond the scope of this thread and we only want to take the signals from it. this mod sends pulses rather than continuous pulses, which is why we opted for the latch circuit. this cannot be changed.i think everything is clear now, right?what do you think now?

author: VEGETA
author_rank: Super Contributor
timestamp: August 20, 2025, 09:12:15 pm
content: i noticed something, that the gate of the ideal diode mosfet (with matched pnps) does not go to 0v but rather more voltage. is this ok? i mean i do not want the mosfet to dissipate any heat more than the Rdson... meaning to be fully on.i changed the mosfets and gave slightly different results but overall not 0v.

author: VEGETA
author_rank: Super Contributor
timestamp: August 20, 2025, 09:49:56 pm
content: this is the mosfet I plan to use:https://www.lcsc.com/product-detail/C22359240.htmlit has Ron of 3.5 mOhm at VGs of -2.5v, I found in LTspice this one:https://www.vishay.com/docs/69063/si7137dp.pdfwhich looks familiar specs of about 3.9 ohms at Vgs = -2.5v. The resulted drop from source to output across 2 mosfets is about < 21 mv at 2.73 amps, at 2 amps (which is the expected max) it is about 18 mv.Thus, I can make the rails 5.07v and 3.35v for example instead of 5.00v and 3.30v to compensate. especially that 15 cm wiring harness (AWG16) is coming next which i tested previously and did not decrease the voltage.my concern is heat also... but assume 21 mv absolute max drop at 2.7 amps it is going to be 3.4 degrees over ambient, which should be totally fine. at 2 amps -> 2.16 degrees without accounting for pcb copper pour (not so big).But all of this assuming the ideal-diode mosfet gets enough gate voltage as i asked in last post. maybe a pnp pair which is more current capable can help?___Another totally separated question: since these mosfets are in series and have resistance... will they inflect a similar noise reduction effect as a linear regulator? I know they won't work in that region but still even the slightest??

author: Zero999
author_rank: Super Contributor
timestamp: August 21, 2025, 07:07:23 pm
content: How much current? Those MOSFETs are pretty big.I don't think it's an issue. Note the MOSFETs don't switch sharply, but slowly, as the voltage across the emitters of the transistors increases.You could try adding a PNP and an NPN to make the switching more sharp, but I think it might be unstable and oscillate; especially on the cusp of switching.

author: VEGETA
author_rank: Super Contributor
timestamp: August 21, 2025, 08:19:36 pm
content: Quote from: Zero999 on August 21, 2025, 07:07:23 pmMy chosen mosfets are good as you mentioned, maximum expected current is 2 amps so it won't be an issue at all. I can compensate for the drop too as mentioned.However, the question was about the gate not becoming 0v of the 2nd mosfet, not about how fast since it will be turned once and that is it.your suggested circuit seems to have a chance of being unstable, so i guess i will stick with original one.I tried it in my simulation file and worked though. I used an NMOS BSS145 since I have BSS138 on board already.if using Si7137DP and 1.5R load resistor, previous circuit shows 11mv drop across the mosfet with gate voltage of 2.9v and Vgs= -2v while new circuit (with Nmos) shows around 6-7mv with gate voltage of 0v and Vgs=-5.My chosen mosfet datasheet says (VGS=-2.5V , ID=-20 --> Ron = 2.7 typical and 3.5 max) which means maybe at -2v it will be 4-5m maximum? that is if we trust simulation to be fully correct though. I never saw anyone online measuring these stuff for this circuit.I tried adding the 47k resistors before the PNPs and putting lower resistors in their place, worked but showed huge back current fed through the mosfets (like -50 amps).Also, what do you think about my question of having a similar effect as linear reg?___Perhaps gate charge has an effect?______Am I over-thinking this? should I just shut up and do the prototype board already and stop thinking all together???

author: Zero999
author_rank: Super Contributor
timestamp: August 21, 2025, 08:47:52 pm
content: Those MOSFETs are overkill.The ideal diode circuit switches the MOSFET in a linear manner. The greater the voltage drop across the MOSFET, the harder it turns it on. It isn't an on/off circuit.

author: VEGETA
author_rank: Super Contributor
timestamp: August 21, 2025, 10:03:28 pm
content: aha so it is not full on like the load switch mosfet before it? this means it will also do an effect like a linear regulator in reducing noise (low freq mainly)?so that using my mosfets (already chosen and priced nicely), how can i expect the maximum drop voltage in order to compensate for it? I was assuming it will be like 4 mOhms similar to datasheet at approx. 2v Vgsmeasuring across the 2 p-mosfets:when using Si7137DP: 6.6mv at 500 mA and 17.87mv at 2.4 amps.when using FDR840P: 17mv at 500mA and 59.4mv at 2.4 ampswhen using AON7423: 7mv at 500mA and 26.7mv at 2.4 amps.the dropout varies a lot. the 3.3v rail is expected to have a max of around 600mA but it is also adjustable to 5v if the user needs it... in 5v mode, it will draw more current than this.is there a way to make this as neat as possible? i mean to never drop below 3.3v and 5v in all situations.my current solution: making 5v rail about 5.07v and 3.3v about 3.36v or so. now for maximum of say 30 mv it will be fine and for a minimum of 7mv it will also be ok since 3.35 becomes like 3.343v or so

author: Zero999
author_rank: Super Contributor
timestamp: August 22, 2025, 07:21:00 am
content: Why not use an ideal diode controller IC?https://www.ti.com/lit/ds/symlink/lm74700d-q1.pdfThen you can use N-channel MOSFETs which can be smaller and achieve the same or lower voltage drop.

author: VEGETA
author_rank: Super Contributor
timestamp: August 22, 2025, 09:42:56 am
content: Quote from: Zero999 on August 22, 2025, 07:21:00 amI thought of this for sure, it is roughly 0.5$ from LCSC which is not too bad.I already have a load switch P-MOSFET and cost-wise it makes sense to use the same one for ideal diode. this device also says forward drop of 20mv to 50mv which is not better than what we achieved here as it seems.I think it is a good choice to stick to our circuit here and it is worth a try. i mean there is nothing wrong with it except for the drop voltage uncertainty which i will compensate for and see how it goes.

author: Zero999
author_rank: Super Contributor
timestamp: August 22, 2025, 09:52:29 am
content: Quote from: VEGETA on August 22, 2025, 09:42:56 amWhy is the requirement for volt drop so stringent?Have you taken into account the PCB traces, cables and voltage regulator, in the volt drop calculations?

author: VEGETA
author_rank: Super Contributor
timestamp: August 22, 2025, 10:24:53 am
content: Quote from: Zero999 on August 22, 2025, 09:52:29 amwell, i do not want excess heat but rather cold operation as my other similar products.the previous prototype took all what you said in consideration and worked fine. the extra compensated millivolts took care of the cable drop and so on.

author: VEGETA
author_rank: Super Contributor
timestamp: August 27, 2025, 06:46:12 am
content: I have sent the pcb gerbers to getting manufactured using the circuit of 4013b and the dual mosfets.i made 5v rail to be 5.08v and 3.3v to be 3.345v to account for drop at any point of load, and also used 0.1% feedback resistors to keep it as accurate as possible.will provide feedback + readings once I get it for sure.