Dual-rail lab PSU project - Page 1
Forum section: electronics / projects

author: francesco_97
author_rank: Contributor
timestamp: December 26, 2021, 03:03:40 pm
content: Hi, I'm trying to design a PSU to use in my tiny home setup. I have a commercial switching supply but it's full of noise and it disturbs my opamps so I decided to try to make one myself.My idea is to obtain something similar to the bench PSU showed by Dave in the lab video (min. 5:00) sotwo dual-rail independent outputs with fixed values to avoid the analog meter or the display and to increase the precision of the voltage, a very very VERY low ripple, 1 Amp max and 18 V max stabilized also with low ohm loads.After some calculations using the books of my prof. I decided to use two sets of LM317 and LM337 and two rotary switches wired to a set of calibrated resistors to obtain the various voltages. In the scheme attached below is reported only the 1,6KOhm resistor that I used to test the min. load of 14 Ohm with 1,25 Amp at 18 V. To maintain a stabilized voltage with so much low load and to compensate the anti ripple filter with the cap multiplier (THANKS DAVE) I used a 220V to 30-0-30 V transformer with two full-bridge rectifiers in combo with two 2.2mF caps for each circuit. I'll add also two 1.25 Amp fuses to each rail to the outputs to avoid flames.What do you think about it? Do you see any serious error in the project or could it be right? Feel free to give me any tip or to move the thread to another section.ThanksFrancescoEDITI added also the PNG schemeEDIT 2Ok right, let's go deep into it.First of all the transformer and the full bridge rectifier. I chose a 220V prim. 30-0-30 V sec. and two bridges for two reasons:1) the double secondary in combo with two full-bridge rectifiers lets us to use the PSU also as a single rail with the stabilization, otherwise a single load could unbalance the voltage2) the 30 V is needed to supply the LM317 at 18 V. When the load is high it isn't a problem but for very low loads (min 14 Ohm otherwise out of 1,25 Amp max) the next filter drops the voltage to 22 V. From another forum it seems that the IC needs some space up the voltage to stabilize it. I can't find this in the LM317 (Texas Instruments reference model) but it seems logical and it's confirmed by the SPICE simulation. Subtracting the 3 V max drop out of the LM317 we have a 1 V of extra space for the IC to stabilize the 18 V. Obviously when we go up the 50 Ohm load this effect is hidden but we want the right voltage also on the border.After two big 2200 uF smoothing capacitors with an anti-ripple low pass filter.Here I followed the tip of Dave putting the Cap multiplier between the filter and the stabilization chip. Low pass filter with a 100 Ohm resistor and a 100 uF cap so the cut frequency is 15 Hz. This lowers the 50 Hz over 3 dB. The transistor is a 2n2222 or equivalent (use similar but PNP for the negative rail) with ~200 hfe for a (β + 1) ~ 200 multiplier for ripple suppression. Here we enter in the world of doubt. In the class I talked about this transistor and his current limit with the prof. The datasheet says that this transistor can hold MAX between collector and emitter 800mA but the prof. says that can hold ~1.5 A. THIS IS THE REASON OF THIS POST. Obviously anything can be created so if I get 1 A at the output it MUST flow into this transistor and this value is nearly between 0.8 and 1.5 so: can it hold this amount of current? What transistor should I use for this project? Unfortunately the 2n2222 is the only physical model used in the course so I used it only for his β. Feel free to suggest a better transistor for this use (maybe a "power" transistor?).Finally the stabilization module.Here the LM317 datasheet is strict. All caps values are needed for more ripple suppression and for bypass and they are all the same as the datasheet. Two diodes here to protect the IC by inverted output. The only group calculated here is the two resistors divider. 120 Ohm the first (LM317 needs min 10 mA to work) and a set of resistors the second to obtain all the output values. As I said in the scheme there is only the max resistor of 1.6 kOhm to get the 18 V but as Strawberry said it must be calibrated to supply the 250 mV drop due to the fuses.At the end one more diode for additional protection against inverted voltage.Datasheet of main components attached belowEDIT 3It's passed a bit of time, I hope that this thread is not dead. After some research I discovered that stripboards can hold like 2.5 amp/ 3 amp. I don't believe it but I think that surely they can hold 1 amp so I decided to use them. Below is attached the (maybe) final scheme and the first stripboard layout. I added four relays to passively get the feedback from the output and to switch between the two transformers to optimize the dissipated power. There shouldn't be strange oscillations by the feedback because I'll get some 12V SPDT relays with a minimum voltage of 11.5V so the 12V position of the knobs should be over that. The stripboard layout is right stretched because I don't know the exact size of the relays of my local dealer so I left a bit more space to fit them later. Imagine the four to220 sockets bent in opposite directions above the holes and under them placed the heatsinks. Here also I don't know the effective size of the heatsinks but I assume that the dealer will sell me the tiny single to220 iron piece with one screw hole in the middle. I also tried to keep the first big capacitor away from the heatsinks of the regulators placing it at the extreme left. Attached below the single double rail module layout (v7). Let me know what do you think about it and if I can modify something or if I should redesign the entire board. Cheers.

author: Terry Bites
author_rank: Super Contributor
timestamp: December 26, 2021, 03:21:31 pm
content: ASC text File?

author: francesco_97
author_rank: Contributor
timestamp: December 26, 2021, 03:35:23 pm
content: It was generated by ltspice. What's wrong?

author: strawberry
author_rank: Super Contributor
timestamp: December 26, 2021, 08:50:08 pm
content: \two 1.25 Amp fuses to each rail to the outputs to avoid flames. \this will add to output voltage drop ~250mV@1A per fuse

author: francesco_97
author_rank: Contributor
timestamp: December 26, 2021, 09:21:40 pm
content: Ok, so I'll compensate the drop rising up the output voltage of 250 mV. Anything else?

author: Damianos
author_rank: Frequent Contributor
timestamp: December 27, 2021, 02:55:26 am
content: Did you studied anything while making this schematic diagram?For example:- Have you calculated what means a 30V AC input for the outputs needed?- Have you studied the data sheets of the components to be used?- Have you studied any rectifier circuit and the various configurations?- Have you studied what is a capacitance multiplier and how it works?- ...My opinion for this thread is to be moved to the "beginners" section, to have advice for step by step studying and implementing each part.BTW Welcome to the forum! There are many people here that will help you in learning electronics and many other things.

author: ledtester
author_rank: Super Contributor
timestamp: December 27, 2021, 03:52:54 am
content: There was a somewhat recent discussion on the use of a capacitor multiplier in a linear regulator design:https://www.eevblog.com/forum/beginners/capacitance-multiplier-video-follow-up-questions-for-a-bi-polar-power-supply/

author: francesco_97
author_rank: Contributor
timestamp: December 27, 2021, 01:44:10 pm
content: Ok, I added some explanation about the project in EDIT 2. Let me know if it is well balanced, especially if the transistor is right or if I should change it.

author: Damianos
author_rank: Frequent Contributor
timestamp: December 27, 2021, 09:19:33 pm
content: Quote from: francesco_97 on December 27, 2021, 01:44:10 pmSome notes on your "deeper" analysis:- The explanation of the use of two bridge rectifiers doesn't make any sense. Study the operation of a full wave rectifier with two diodes and a centre tapped transformer. Then add two more diodes in reverse direction and study the entire circuit. In the schematic there are eight diodes connected randomly by wires that in a real circuit will give "not good results".- The 30V AC input has a peak value of about 42V. Where is lost half of it? Do you feel that something is wrong? For 1A output the circuit consumes about 40W; for 18V output it gives 18W and spends 24W in heat. For a 10V output ... calculate it.- For the capacitance multiplier, the base of the transistor must be biased in a voltage lower than the lowest input voltage (due to ripple) to operate linearly. As it is, the 100μF capacitor will be charged in a high value and, when the input drops, it transfers this to the output... In other words it will act as a diode with an RC delay, that transfers the ripple to the output. If the target is the 3db ripple rejection, that you refer it, the regulator itself has more than 50db!- About the transistor you have misunderstood the specifications. The hFEis far from the 200 you mentioned, on the conditions that you indented to use it. Check the data sheet for the conditions that the absolute current limit is valid. It also has other limits (such as power, temperature...). Do you plan to use it over the limits?- For the regulator I'm sure that you have not taken a look even at the first paragraph of the first page of the data sheet! Placing fuses at the outputs destroys the regulation and protects nothing! The output capacitor helps the load, not the regulator and its value is "huge".- What are these "inverted voltages and outputs" and what you protect with small signal diodes? Out of curiosity: what is the role of the D1, D6, D15 and D19?- ...My advice is:- if you are in a hurry, copy a ready made and proven designelse- if you want to learn, be patient and methodic. Divide the target in sections and start studying the components one by one, analysing it's role and operation. Calculate values and note the expected results. Assemble the first part and test it. Keep notes and compare the measurements with the expected values. When any problems are solved and it operates satisfactorily proceed to the next...

author: Ian.M
author_rank: Super Contributor
timestamp: December 28, 2021, 11:30:54 am
content: Quote from: Damianos on December 27, 2021, 02:55:26 amDamianos has elaborated further in reply #8, but I also wished to note that your so-called "two full-bridge rectifiers in combo" concept is fundamentally incorrect. Here's your LTspice sim pared down to just the bridge rectifiers and reservoir capacitors with two 30V RMS AC voltage sources added to simulate the transformer secondary. As no load is connected, the secondary currents should drop to near zero as soon as the capacitors have charged. It does not, because some of your diodes are effectively shorting the transformer secondaries. Plot the current through them to see which are problematic, note *ALL* of them then remove them and see what's left!In general, its a bad idea to link either AC terminal of a bridge rectifier to either DC terminal. Also you can't 'stack' bridge rectifiers fed by the two halves of a center-tapped secondary, as they would need to be fed from separate floating secondaries for that to be permissible.

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 01:21:46 pm
content: This could be the right answer and not only "go to study boy". Unfortunately in the electronic class of Sapienza the dual rail isn't treated, so i must guess what to search for. After some quick conversation with some friends, i have discovered a better way to do this. Could a dual diodes setup work? Later i will upload a new modified scheme with i think some improvements. Let me know what do you think about.

author: Ian.M
author_rank: Super Contributor
timestamp: December 28, 2021, 01:46:43 pm
content: See Hammond Mfg. Transformer division'sDesign Guide For Rectifier Usefor circuits, rule-of-thumb design equations etc.A bridge rectifier fed by a center-tapped secondary is a special case. Consider it as two separate two diode full wave rectifiers, providing independent (but balanced) positive and negative rails, sharing a single secondary. This only works for center tapped windings, as any winding asymmetry will result in increased line frequency ripple, and as the asymmetry increases, if there are reservoir capacitors it degenerates to half-wave rectification.

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 02:24:37 pm
content: Ok this design is similar to the scheme that i've found but instead of 2 diodes it uses 4 diodes. I'll try both and after i'll post the results.

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 04:20:56 pm
content: Ok so: both rectifiers lead to 10 A spikes flowing into the diodes. What could be wrong here?

author: strawberry
author_rank: Super Contributor
timestamp: December 28, 2021, 04:27:26 pm
content: 2diode centertaped (diode drop=0.7V)(transformer power used ~60%) vs bridge (diode drop=1.4V)(transformer power used 100%)2diode rectifiers were popular back in day when transformer cost less than diode rectifier~1/5 of half waveform is conducting due to capacitor stored voltage(diode is not conducting reverse bias), load current * ~5 = diode peak currentlarger capacitor less ripple = more current through diode and poorer PF

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 04:38:49 pm
content: Ok so definitely the bridge option. In the real world which will be the most likely value?

author: Ian.M
author_rank: Super Contributor
timestamp: December 28, 2021, 04:53:07 pm
content: We don't have your symbols for U5 and U6:AutoGenerated\LM317_TRANS.asyAutoGenerated\LM337_N_TRANS.asynor do we have the associated models so we cant run your sims.A high peak current in the diodes is to be expected as the average DC current (which appears to be approx 2A total as the feedback divider resistor values look like you intend +/-18V from both sets of regulators) is concentrated at the peaks of the AC waveform as the diodes don't conduct till the rising waveform reaches the reservoir capacitor voltage, then cease conduction fractionally after the voltage peak. 10A does seem a bit excessive, but you have a total of 8800uF reservoir capacitance per rail, and the voltage sources modelling the transformer secondary do not account for the secondary resistance or the effective primary impedance referred to the secondary circuit.

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 05:14:25 pm
content: Right. I used the standard Texas Instruments unencrypted transient models.

author: mariush
author_rank: Super Contributor
timestamp: December 28, 2021, 05:58:14 pm
content: You may want to see what happens when you change the selection while something is powered from your power supply. At the moment you switch, your resistor is disconnected and a few ms later the new resistor value is connected ... what happens when there's no resistor connector, that small period between, does the voltage jump to 40v or whatever DC voltage you have?You should also look at the thermals and how much power you dissipate... you're saying 30v AC transformer ... that rectified is around 40v ... if you output 5v at 1A through your LM317, you're looking at 35w of dissipated heat... good luck cooling that.You should consider two transformers, each with two 12v/15v/18v AC windings, or a toroidal transformer with 4 x 12v/15v/18v AC secondary windings, and use a couple relays to switch between 12v AC / 24v AC depending on desired output voltage. Could have a rotating selector like on multimeters, that sets both the resistor and separately connects relay to power as needed .. or you could have some voltage comparator or something turn on the relay when you select a high enough voltage (for example anything higher than 10v DC, enable the relay to get 24v AC )

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 07:19:07 pm
content: Wait. I understand the knob problem but how is possible that the rectified voltage is more thant the input voltage? We should have something less or al least equal in an ideal world with infinite caps. Could you explain this to me?

author: Ian.M
author_rank: Super Contributor
timestamp: December 28, 2021, 07:33:44 pm
content: AC voltages are customarily specified as RMS voltages, but the reservoir capacitor will charge up to the Peak AC voltage (less diode Vf drops) of approx 1.4 times the RMS voltage. If unloaded it remains there between peaks, leading to an average DC voltage 40% higher than the measured AC voltage feeding the rectifier.See:https://testguy.net/content/%20270-Peak-vs-Average-vs-RMS-Voltage

author: mariush
author_rank: Super Contributor
timestamp: December 28, 2021, 07:41:12 pm
content: Assuming you use a bridge rectifier :Your 30v AC from the transformer is rectified using a bridge rectifier to Vdc peak = sqrt(2) x Vac - 2 x voltage drop on diodes.So for example : 30v AC => 1.4142 x 30 - 2 x 0.7v = 41vYour DC current is around Idc = 0.62 x IacThen your capacitor must be sized according to the minimum voltage you want ... approximate with formula C = Current / [ 2 x Freq AC x (Vdc peak - Vdc min desired) ]So for example for 1.5A of current and 50 Hz frequency (as you say 230v 50Hz transformer) and let's say you want at least 32v, then C = 1.5 / 2 x 50 x (41-32) = 1.5 / 900 = 0.001666 = 1666 uFOf course, you would not use 41v, but rather you would estimate the output of the transformer based on the lowest AC voltage you'll have on input, kind of a worst case scenario.For example if the transformer is 230v : 30v that's a 7.66 : 1 ratio , so let's say the power goes down to 210v AC then your transformer may only output 210/7.66 = ~ 27.4v AC which then results in a peak DC voltage of only around 37v DC.If you use center tap and two diodes, it's a slightly different story. Still ... you should aim for maximum 10-15w dissipation on a to-220 regulator, and anything higher than 10w would be a bit difficult to cool passively, without a fan...Also keep in mind the AC could be as high as 250-255v AC at some points ... which would make the rectified DC voltage get close to 50v, so you'd want to use 63v rated capacitors.Last but not least, small transformers usually output 10-15% (or even higher) voltage under no load, so your AC voltage on the secondary could be even higher when your circuit only consumes a few mA.

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 08:14:43 pm
content: Awesome. I known the teory and the proof behind the rms in this application but I wrongly thought that the voltage shown on the transformer was the peak and not the rms. This is better because i can choose a much lower value like a 30/1.42 = 20-0-20 transformer. About the heat at maximum in the worst case it should be like 40 - 1,25x4= 35 W to dissipate. Maybe is this the time to put a fan?

author: Damianos
author_rank: Frequent Contributor
timestamp: December 28, 2021, 09:33:40 pm
content: Quote from: mariush on December 28, 2021, 05:58:14 pmFor this short period, the capacitor of the adjust pin keeps the previous setting. If needed, an increase of its value will help more.The big problem is when the switch remains open...

author: francesco_97
author_rank: Contributor
timestamp: December 28, 2021, 09:44:35 pm
content: This may be easy. There won't be an off position on the rotary switch. There will be 1,25 , 3,3 , 5 , 9 , 10 , 12 , 15 and 18 V position. A general on and off switch and two separated switches to connect or to disconnect the load. The only way to be in a not defined state is to block the knob in a middle position.

author: strawberry
author_rank: Super Contributor
timestamp: December 28, 2021, 10:50:51 pm
content: open when switch change state

author: Kleinstein
author_rank: Super Contributor
timestamp: December 29, 2021, 09:43:23 am
content: Quote from: francesco_97 on December 28, 2021, 09:44:35 pmFor the switches there can still be a short time, when the knob is in between stable position that the switch is open. It depends on the switch details.Most switches are break before make, and thus for a short time there will be an open. With rotary switches there are also make before break versions, though not as common. There the center pin is temporarily connected to 2 of the outer pins.A capacitor to ground could avoid possible upsets from a temporary open situation. Even a make before breake switch may show short glitches from dirt or bouncing contacts, like a scratchy pot.

author: francesco_97
author_rank: Contributor
timestamp: December 29, 2021, 07:37:26 pm
content: Project update. This evening I've called my electronic dealer to ask about the transformer and he said that he has a 21-0-21 V + 9 V 60VA transformer for 30 euros. The first question is: do you think that it is a good price? This extra 9 volt is good news because I could use it with a separate rail in combo with another stabilizer like the 7809 to power a fan and some other stuff. With like 35 watts do you think I should add the fan or a passive cooler? Instead of using a dual 8 position knob could I use the extra 9 V to power a comparator? This could ground the ADJ pin so between the switch there could be 1,25V. What do you think about it?

author: strawberry
author_rank: Super Contributor
timestamp: December 29, 2021, 08:03:29 pm
content: https://www.tme.eu/lv/details/tmb100_400_2x24v/transformatori-ar-stiprinasanu/indel/tmb-100-010m-1/30EUR 100VA there is 3 output versions as wellfor 100VA could get about 50W

author: mariush
author_rank: Super Contributor
timestamp: December 29, 2021, 08:15:54 pm
content: It's not a particularly good price, and the voltage is not that great.So that's a 42v AC 60VA transformer ... that means it's a 60VA / 42 = 1.42 A ... when you rectify that to DC you get around 58v peak DC voltage (or 2 x 29v) and around 0.9A peak DC current.Look at sites like TME.eu, they have a lot of transformers and they're not expensive.For example: - links are in Romanian , but you can change to Italian from the top right part of the pages -toroidal transformers, 50VA or higher start from 14 euro:very long linkHere's a 230v - > 2 x 15v AC 50VA transformer for 16 euro :https://www.tme.eu/ro/details/tst50w_2x15v/transformatoare-toroidale/indel/tst-50-029/Here's a 230v - > 2 x 12v AC 60VA transformer for 17.5 euro:https://www.tme.eu/ro/details/tst60w_2x12v/transformatoare-toroidale/indel/tst-60-002/The 80VA version is 21 euro :https://www.tme.eu/ro/details/tst80w_2x12v/transformatoare-toroidale/indel/tst-80-002/100VA 2 x 24v is 22 euro :https://www.tme.eu/ro/details/tst100w_2x24v/transformatoare-toroidale/indel/tst-100-014/- If you decide to stick to just one transformer and use one winding for positive and one for negative, this would work well .. 24v AC is high enough you'd get 18v DC output without any problems.I'd get two of these toroidals with 2 secondary windings (one toroidal for positive voltages, one for negative voltages), and switch between using just one winding, or both, or maybe optionally paralleling the two windings to get double the current, but only half the voltage.Also note that with toroidal transformers, you could just loop a bunch of wire over the surface of the toroid and you get yourself a third winding, if you need a low voltage / low current third winding.Classic transformers are also reasonably priced :long tme linkex70VA 230v -> 2 x 12v AC for 23.5 euro :https://www.tme.eu/ro/details/ts70_002/transformatoare-cu-prindere/indel/ts-70-002/70VA 230v -> 2 x 16.6v for 24.5 euro :https://www.tme.eu/ro/details/ts70_16/transformatoare-cu-prindere/indel/ts-70-16/They also have PCB mounted transformers ... but these only go up to 60VA, here's link :long link to selectionSome example:55 VA 230v 1 x 12v for 10 euro :https://www.tme.eu/ro/details/tszz45_12v/transformatoare-pcb/indel/tszz55-004m/45 VA 230v 2 x 12v for 12 euro :https://www.tme.eu/ro/details/tszz35_2x12v/transformatoare-pcb/indel/tszz45-006m-5/Yeah, it will cost you a few euro in shipping... it's around 1 kg for each transformer. But basically your "dealer" isn't do you any favors, it's no great deal what he's giving you.

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2021, 08:18:14 pm
content: Quote from: francesco_97 on December 29, 2021, 07:37:26 pmHow many amps is each secondary rated for? If it only had one secondary (even centertapped) you could calculate from the VA rating, but with two, you need the specs.I'd probably add 3.3V and 5V regulated outputs powered by the 9V winding as so much stuff nowadays is a mix of analog and logic chips. Don't link their 0V with the other supplies - leave it floating so you've got the option to reference logic circuits to either analog supply rail as well as analog ground. However if the 9V winding is particularly wimpy, it may only be suitable for internal PSU use for control circuits and metering.

author: francesco_97
author_rank: Contributor
timestamp: January 03, 2022, 10:54:16 pm
content: More about thermal design. I read in the lm317 datasheet that thermal resistance junction to top is like 16 C°/W for to-220 package. This means that in the worst case of 1.25 V at 1 A the chip is dissipating like 20 Watts so if i use only a passive heatsink the temperature will rise to minimum 320 C° above ambient. This means that i MUST use a fan.It seems a bit too much but following the Dave thermal video the calculation seems this. Do you agree with this conclusion?

author: strawberry
author_rank: Super Contributor
timestamp: January 04, 2022, 12:05:06 am
content: for TI regular TO-220 LM317 Rjc=0.1K/Wfor TI cheap thin tab TO-220 LM317 Rjc-4.2K/WTransformer winding switching to reduce power loss

author: francesco_97
author_rank: Contributor
timestamp: January 04, 2022, 09:15:51 am
content: Uhmmmmm, i read in this datasheet in 7.4 section a value of 15.9. Could be my datasheet wrong?

author: Kleinstein
author_rank: Super Contributor
timestamp: January 04, 2022, 09:33:47 am
content: With a heatsink the relevant thermal resistance is from the chip to the bottom side (0.1 K/W), not the on to the plastic top, unless you mount the heat sink on the plastic side.To toal thermal impedance is than from the LM317 + the one of a insulator (mica or pad) + the heat sink (depends and can go down to about 2 K/W). So the total heat resistance may be something like 0.1 K/W + 1 K/W + 5 K/W = 6.1 K/W.In this case 20 W is no real problem. The FAN is mainly allowing a smaller heat sink - if the case thermal resistance would be the limit a FAN would not help. A FAN + relatively small heat sink can relapace a large heat sink. It is not improving on the case of the chip.With a 21 V transformer the raw voltage may be more like 25 V and thus maybe up to 25 W.

author: francesco_97
author_rank: Contributor
timestamp: January 13, 2022, 08:06:26 pm
content: Update. Linked below is the new scheme. I added the second transformer so there will always be a maximum voltage jump of like 10 V so the temperature will be 10V x 1A x 6 C/W = 60 degrees above ambient up the chips. In a critical summer scenario of 40C, there will be 100C and it should be ok (max recommended temp on lm317 is 125C) so THEORETICALLY it should work without a fan only with heatsinks. I added also new transistors looking for 1A support. BD135 and BD136 should do the job. Respect the switches I chose to use 4 double poles 8 positions make before break switches allowing the user to have also asymmetrical voltages and solving the "open knob" scenario. Now I should choose the diodes. For discharge the 1n4148 should work but to have a good rectifier and a good protection diode I was thinking to sb560: 60V and 5A (in the scheme they are marked as 1SR154-400). Finally the fuses. I read in some datasheets that the typical resistance is like 0.01 ohm so there shouldn't be too many problems with the voltage drop. What do you think about the new changes?

author: francesco_97
author_rank: Contributor
timestamp: January 29, 2022, 08:01:46 pm
content: Project update, all in the first post under EDIT 3