uSupply - LM350T instead of LT3080 - Page 1
Forum section: electronics / beginners

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 04:04:47 pm
content: Hi, I am building a lab power supply based on the uSupply. Instead of the LT3080, I want to use LM350T, because it is cheaper and can take 3 amps. But it has 1.25V interntal reference, so I cannot bring it down to 0V using the original uSupply circuit. So my idea was to bring the adjust pin on LM350T down to -1.25V. To do that, I devised this circuit:It is basically just a summing amplifier, which will "subtract" the -1.25V. Output of the opamp should be -1.25V to 23.75V, to give 0-25V at the output of the regulator. To get the negative voltage needed for the opamp and -1.25 voltage reference, I plan to use this -Will this circuit work?Also, is the LM358A a good opamp for this application, or should I go with another one?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 19, 2016, 04:22:09 pm
content: The circuit should work as an adjustable voltage regulator. However it is not such a good idea to use the OP to implement a kind of current regulation this way.The LM358 is ok for this application, unless you need very fast changes. The offset could be an issue - so one might have to include this in the control voltage. Due to limited accuracy, there is no need to use really accurate resistors - so 1% (or even 5%) types should be good enough.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 04:45:27 pm
content: My plan is to use the same current regulation circuit as Dave used:Current will be measured by the MAX4080. Is this the correct way to do it, or not?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 19, 2016, 05:54:44 pm
content: The current regulation like Dave used in his µSupply is not that good. It is more of a bad example, or how not to build a lab supply.The voltage regulator just has very high gain (current controlled by voltage), when using it to drive current through a low impedance load. So the circuit would be either prone to oscillation or ridiculous slow.To use a regulator chip like the LM350 (or the LT3080) one would have to have a resistor behind the regulator and add your own voltage regulation loop to compensate for the drop at the resistor. So the regulator chip is more or less used as a kind of protected transistor.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 06:14:19 pm
content: Thanks for the advice!I was also thinking about making the constant current regulation software-only. Since I can measure voltage and current, I can reduce the voltage to get a set current, but I fear that the regulation might not be fast enough. I plan to use ATXMEGA32E5-AU running at 32MHz to control the whole thing. Would that be a good idea or a bad one?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 19, 2016, 06:47:58 pm
content: Software control is not a good idea - it is slow and resolution is limited. For a lab supply one would want something like a 100 kHz control loop bandwidth and maybe 16-24 Bit of amplitude resolution. SO nothing easy to do in software.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 07:14:44 pm
content: So, what would be a good solution for my power supply? Some modification of Dave's circuit, or something completely different?

author: wraper
author_rank: Unknown
timestamp: December 19, 2016, 07:28:52 pm
content: LM350T has 10 times higher dropout voltage than LT3080 has. So besides other issues, you would need to ensure that voltage on it's input is at least 2.5V higher. Then as a consequence comes huge increase of heat dissipation.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 19, 2016, 07:30:49 pm
content: I would prefer something completely different. More like the typical old HP power supplies work: Have a MOSFET or darlington transistor as a power stage, and a floating regulator with 2 OPs to control. One for the voltage and one for the current.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 07:32:06 pm
content: I know about the higher dropout and I can cope with that. My heatsink can easily take 50W and will add a fan, just to be sure.

author: David Hess
author_rank: Super Contributor
timestamp: December 19, 2016, 08:53:30 pm
content: This will work and I have done it myself but there are some issues with your implementation:1. Your LM358 is configured with positive feedback.2. The LM350 (and LM317) expects a minimum output load current whether you actively drive the adjustment pin or not. If your output is going to go to 0 volts, then this current must be drawn down to the negative bias supply. This is easy enough to do with a single resistor between the output and adjustment pin which will allow the operational amplifier to sink the constant 1.25V/R current into the negative bias supply but this would defeat your additions for current control. Fixed regulators like the 7805 do not have this problem because their quiescent current flows out their common (adjust) pin instead of their output pin (their voltage divider is internal) but I would still use the LM350.3. The reference and regulation errors of the LM350 are being added to other errors. If feedback is taken from the output of the LM350 instead of its adjustment pin, then the errors and noise from the LM350 will be removed by the external error amplifier. This adds some additional frequency compensation requirements but nothing complex.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 09:20:43 pm
content: Thanks for the feedback. Did you build your supply with current limiting? Or just the voltage regulation?1) Which one? The current limiting one is straight from uSupply schematics and I tried the other one in circuit simulator and it works as it should...2) I was going to solve the constant load the same way Dave did, with a LM334 or something similar (I know that LM350 needs about 5-10 mA, not 1mA like LT3080).3) So, more like this?

author: JoeN
author_rank: Frequent Contributor
timestamp: December 19, 2016, 09:37:38 pm
content: Quote from: IrejectYourReality on December 19, 2016, 06:14:19 pmQuote from: Kleinstein on December 19, 2016, 06:47:58 pmI don't know anything about power supplies, but if you need raw speed and the ability to read a wide (16-24 bit) ADC quickly and make adjustments, doesn't this scream out for a CPLD or something like that? Small CPLDs are very inexpensive these days and some of them have quite a few pins so you can interface a wide ADC easily and look at its output every single clock cycle.http://www.mouser.com/ProductDetail/Altera/5M160ZE64C5N

author: KerryW
author_rank: Regular Contributor
timestamp: December 19, 2016, 09:45:00 pm
content: Note that T1 can only pull the ADJ pin to 0V, so you would have 1.25V out of the supply minimum during current limit. You need it to pull down to -1.25V.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 19, 2016, 10:05:11 pm
content: What if I connect the current limiting straight to the input of the other opamp?(I have a feeling that this is not a very good idea)

author: Kleinstein
author_rank: Super Contributor
timestamp: December 19, 2016, 10:39:55 pm
content: Moving the current limit to before the other OP does not change very much. Still the same high gain, but the problem gets only worse because of extra delay of the extra OP involved.

author: David Hess
author_rank: Super Contributor
timestamp: December 20, 2016, 01:39:17 am
content: Quote from: IrejectYourReality on December 19, 2016, 09:20:43 pmI have designed them with and without adjustable current limiting. Without adjustable current limiting and maybe with it, the current limiting of the voltage regulator provides protection or I add a fixed current limit protection circuit.As you have implemented it, the added gain of the transistor will make the current control loop unstable.QuoteA 100 ohm resistor between the output and adjust pin will draw a constant 12.5 milliamps which then must be handled by the operational amplifier which will be no problem. Add a PNP emitter follower to the output of the amplifier if necessary and it will not even see the load and the constant current will provide an ideal operating point for the transistor.QuoteIC1A still has its inputs reversed.Quote from: IrejectYourReality on December 19, 2016, 10:05:11 pmThe first bench power supply I designed worked like that with the current control loop driving the voltage control loop. It worked but having two operational amplifier stages in series made frequency compensation very difficult and performance of the current control loop was poor.It is better to combine the outputs of the error amplifiers with a pair of diodes (or elegantly and cleverly two PNP emitter followers that have a high Vbe breakdown voltage) so each can pull the output down independently through the adjust pin. This works well with the single resistor current sink suggestion above.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2016, 03:01:37 pm
content: Quote from: IrejectYourReality on December 19, 2016, 04:04:47 pmThe 1.25V reference won't exactly match the LM350's reference, which can vary from 1.2V to 1.3V.http://www.ti.com/lit/ds/symlink/lm350a.pdfYou could use an op-amp to look at the LM350's voltage reference and subtract it from the output.Note, that you should use a better op-amp than the crappy old µA741 and make sure the R2 to R5 are selected from the same batch, so they're well matched.

author: David Hess
author_rank: Super Contributor
timestamp: December 20, 2016, 10:22:33 pm
content: Quote from: Hero999 on December 20, 2016, 03:01:37 pmThere is little reason to use an operational amplifier better than a 741 or 358/324. Precision will not be improved unless remote sensing is used, low frequency noise will not be improved without a low noise reference, and transient response will not be improved without careful frequency compensation. The dedicated supplies I have build using a precision OP-27 or LT1007 type of operational amplifier to control an integrated regulator had all three of these features but they are overkill for general purpose bench supply.

author: Zero999
author_rank: Super Contributor
timestamp: December 20, 2016, 11:28:09 pm
content: Quote from: David Hess on December 20, 2016, 10:22:33 pmThe main improvement will be the offset, especially in the schematic I posted that uses 470k resistors will drop a fair voltage due to the relatively high bias currents of the old 741. The values could be reduced to 10k so an old op-amp will give respectable performance.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 21, 2016, 01:32:30 pm
content: Quote from: David Hess on December 20, 2016, 01:39:17 amWhat should I change to make it stable?Quote from: David Hess on December 20, 2016, 01:39:17 amI planned to use the negative voltage inverter to supply the opamp with negative voltage. Wouldn't this mess up the negative voltage, if the opamp had to sink 12.5 mA?

author: David Hess
author_rank: Super Contributor
timestamp: December 21, 2016, 10:06:41 pm
content: Quote from: Hero999 on December 20, 2016, 11:28:09 pmThe offset current is what matters in this case because of the deliberately balanced source resistance. The errors from the 741's high bias current cancel and the offset current only contributes an error of about 5 millivolts typical into 235k which is not so bad compared to 2 millivolts of typical voltage offset error. The error from the common mode rejection is almost as great and in practice the resistor matching will create the largest error term.I agree that the 741 is the wrong operational amplifier for such high source resistances but for a different reason; Tektronix used the 741 in such high impedance circuits with balanced inputs to minimize voltage offset error from the bias currents but at very low bandwidths. The problem is that the current noise from a 741 is pretty high making it noisy with high source impedances. The typical choice in the past would be the 308 but nobody makes them anymore.I wonder what the least expensive modern alternative to the 308 would be. The OP07 is not really a low input bias current operational amplifier but it is close and inexpensive. The LT1012, LT1097, and OP97 certainly count but cost twice as much as the multiple sourced OP07. I used to use the LM11 for this sort of circuit if I needed something better than a 308 but they are also discontinued. JFET operational amplifiers lack precision.That is a clever circuit. Did you come up with that yourself or find it somewhere?

author: David Hess
author_rank: Super Contributor
timestamp: December 21, 2016, 10:37:33 pm
content: Quote from: IrejectYourReality on December 21, 2016, 01:32:30 pmThe problem with the transistor as shown is that it adds voltage gain to the error amplifier for the current control loop making it much more difficult to frequency compensate.You have two separate error amplifiers so "or" the outputs together so whichever one is active is the one pulling the LM317 adjustment pin low. Usually this is done with a pair of diodes but in the past when they were more available like the 2N404A, PNP emitter followers with high Vbe breakdown could be used. Or use diodes and PNP emitter followers. I have also seen p-channel JFETs used to do this. I have not tried it but low Vgs p-channel MOSFETs or p-channel depletion mode MOSFETs could be used also.Most of the power supply designs I would copy use diodes or diodes plus PNP transistors; the later unloads the output of the operational amplifiers for better precision but this would not matter in a general purpose bench power supply.QuoteIf you sink the current to the negative supply voltage, then it does not matter whether it goes through the operational amplifier or not except that the design is simpler if the operational amplifier does it. The problem with sinking it to ground is that your current sink has to operate with very low compliance when the output voltage is low; this might not matter except it seems like you want an output which can actually go all the way to ground. This is almost possible to do using a bipolar transistor or MOSFET as the current sink but again, it is more complicated than sinking the current to the negative supply.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 22, 2016, 03:55:30 pm
content: So, if I understand it correctly, it should look like this?I am still concerned about sinking the current to the negative supply. As I said, I want to use a PWM voltage inverter, which is a very weak supply and can generate only couple of mA. If I sink the 12.5 mA to this supply, isn't it basicaly the same process as if the supply itself sourced equivalent opposite current? I fear, that by sinking too much current into this supply, the negative voltage will drop and the supply will no longer be able to go close to 0V at the output. Is this a valid concern?Another thing, more for the sake of curiosity, about the full digital control of the constant current, as Kleinstein wrote:Quote from: Kleinstein on December 19, 2016, 06:47:58 pmI was thinking about this. I plan to use an ATXMEGA mcu running at 32 MHz. It has 12 bit DAC capable of 1 Msps and a 12 bit ADC capable of 2 Msps. I do not need more than 12 bits of precision. What if I dedicated this mcu to just running the control loop and nothing else, and handled the rest like reading rotary encoders and writing to an LCD somewhere else? I know this is probably a bad idea, but I am just curious.

author: Zero999
author_rank: Super Contributor
timestamp: December 22, 2016, 04:02:28 pm
content: Quote from: David Hess on December 21, 2016, 10:06:41 pmI designed it myself, when I noticed many people were using separate voltage references and trimmer resistors to get 0V from the LM317. I realised there's a more effective way to zero the output.Quote from: IrejectYourReality on December 22, 2016, 03:55:30 pmIt will still be unstable, I'm afraid. As long as you have another gain stage inside the op-amp's feedback loop (T1 in this case) it will be prone to oscillation.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 22, 2016, 04:05:01 pm
content: Quote from: Hero999 on December 22, 2016, 04:02:28 pmSo I need to just completely remove T1 and it will be fine?

author: StillTrying
author_rank: Super Contributor
timestamp: December 22, 2016, 04:25:43 pm
content: Quote from: IrejectYourReality on December 22, 2016, 04:05:01 pmI'd try swapping it for a PNP, and maybe taking it's collector to the -3V. You'd have to swap the two inputs to the op amp around. I don't think the large reverse b-e V would be a problem, keep the 10K.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 22, 2016, 04:34:50 pm
content: Removing T1 makes things a little better, but you still have the problem of not having a defined minium resistance at the output of the LM350. So an extra resistor at the output (before voltage feedback of cause) is more or less needed. It is possible too, to move the resistor to the GND side of the load and than use it as the shunt.Removing T1 also means the OP for current regulation should be relatively fast, as it sets the speed in preventing a big current spike. So the LM358 would be rather slow in this case.An other problem might be the shunt at the input side of the LM350. The current sensing amplifier should be relatively fast - the may408x is not that fast.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 22, 2016, 04:56:05 pm
content: What value resistor should I use to define the minimum output resistance? I would guess something around 1-10 ohm? (The maximum output of the supply should be 25V 2A)If the LM358 is too slow in this case, what opamp would be better?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 22, 2016, 05:20:32 pm
content: For a 2 A supply I would suggest something like 0.1 to 0.3 Ohms. This is not very much, but could make a big difference compared to a few mOhms parasitic resistance and the output impedance of the LM350. If you already have a negative supply, one could use something like an TL072. Many OPs to choose from in this range.The problem with the LM358 is it's very limited slew rate (e.g. 0.3V /µs) and the additional cross over distortion (add's another about 2-50 µs delay if current changes sign).

author: IrejectYourReality
author_rank: Contributor
timestamp: December 22, 2016, 06:05:42 pm
content: If I add said resistor at the output of LM350, could I use that resistor as a current shunt, or is it better to measure at the input of the regulator? Also, if I wanted to put the resistor to the lowside, what amp should I use for the measurement, would INA214 work well? (if I understand correctly the 4080 is high side only.)

author: Kleinstein
author_rank: Super Contributor
timestamp: December 22, 2016, 07:30:22 pm
content: One could use the resistor behind the regulator as a shunt, but this is not that easy, as there can be lot of common mode voltage with changing output voltage. It should be possible with a floating current regulator - though the circuit is not that simple with central control. A floating current regulator has some benefits.If the shunt on the low side is used for current regulation one could use a "simple" OP to amplify the current signal. This also needs an extra amplifier to transfer the set-point signal for the voltage regulation.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 22, 2016, 09:27:21 pm
content: So, what will be better/more accurate? Measuring the current at the input of LM350 or measuring it at lowside? I would like to get around 1mA resolution if possible. And if the max4080 is too slow for the cc to work properly, what should I use instead of it?

author: David Hess
author_rank: Super Contributor
timestamp: December 23, 2016, 01:35:00 am
content: Quote from: IrejectYourReality on December 22, 2016, 03:55:30 pmIt is a valid concern if you need the output to get all the way to 0.0 volts and the negative bias supply cannot support the sink current required. The LM350 requires a minimum load current of roughly 10 milliamps (worst case) whether that comes from the resistor between the output and adjustment pin or from a separate load on the output.If your negative bias supply is not going to support the needed current, then your only choice if you want to continue to use the LM350 will be a low compliance current sink from the output to ground and a larger resistor between the output and adjustment pin to supply the current through the diodes into the operational amplifier outputs. 1.25 volts divided by 1.2k will yield about 1 milliamp through the diode which is currently forward biased. I usually run the entire LM317/LM350 10 milliamp minimum output current through the diodes into PNP emitter followers. This is also a good place to put a pair of LEDs to show if the output is voltage or current limited.The minimum output voltage will then be limited by the load and the current sink to ground. With no load, is a minimum output voltage of about 0.3 volts enough? How close to 0.0 volts do you need?You could use a different regulator (1) which does not have such a high minimum current or replace the regulator with a discrete transistor stage. Personally I like using integrated regulators because of their built in protection circuits.While not the power supply I would emulate exactly, the well documented Tektronix PS501 and PS503 use discrete pass transistor output stages and *still* pull a current to below ground so that they can achieve a 0.0 volt output with good stability. The PS503 actually deliberately regulates to a few millivolts below ground to account for the worst case offset voltage of the error amplifier.(1) I did a quick search and did not find any direct replacements for the LM317/LM350 with a lower minimum output current.

author: StillTrying
author_rank: Super Contributor
timestamp: December 23, 2016, 04:01:02 am
content: If you're brave and rich you could try thewww.linear.com/product/LT3083Minimum load 1mA and goes down to 0V. Max 23V though.You might be able to fiddle the 1mA min load something like this, I'll try it when I get a round tuit.

author: David Hess
author_rank: Super Contributor
timestamp: December 23, 2016, 05:01:31 am
content: Quote from: Hero999 on December 22, 2016, 04:02:28 pmI do not remember seeing this idea before. It is a great solution if you do not want to add a buffered -1.25 volt reference and have a spare operational amplifier. Old application notes often use an LM329 or similar shunt regulator.

author: David Hess
author_rank: Super Contributor
timestamp: December 23, 2016, 06:30:18 am
content: The LM317/LM350 already has a series output resistance for its own current sense and emitter ballasting.An added output resistance is an ideal place for the current sense if you make provisions to handle the wide common mode voltage swing. If the regulator can be floating, then placing the current sense resistor on the ground side makes this trivial but either high or low side can work fine. The first bench supply I designed for myself did this using high side current sensing and a 4 resistor difference amplifier but the frequency compensation needed to make it stable (see below) made for poor performance.The 324/358 is slow but no slower than the 741 which is plenty fast for this application. JFET amplifiers have an advantage in slew rate which will make recovery faster. Beware of using an operational amplifier which is too fast; the extra frequency compensation needed will slow it down reducing its slew rate and and increasing its recovery time so there is little advantage. There are ways to fix this but they add complexity.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 23, 2016, 09:28:07 am
content: The LM358 and 741 have a rather low slew rate and limited bandwidth. For the shown type of regulator with a low output impedance power stage voltage regulation is easy (no problem with 358). But the cross over from CV to CC is somewhat difficult and a high slew rate and also a not so low bandwidth is a good idea. Worst case the OP for current regulation has to go all the way from around 20 V to near zero before current limiting even starts to work. At 0.3 V/µs , this is some 60 µs without active current limit and thus a larger overshoot in the current. Delays from compensation and recovery of the LM350 come on top. I know there are circuits using the LM358/324 in such a circuit - but this really compromises performance. To really take advantage of a faster OP, the compensation needs to be done in a way to keep the extra delay short. This is possible at least to a large part for the critical CV/CC transition.The easy way is having the shunt at the low side - it also gets rid of the slow max4080.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 23, 2016, 11:04:03 am
content: Quote from: David Hess on December 23, 2016, 01:35:00 amWhat if I just added a small inverting switchmode powersupply? It is cheap and I have plenty of room on the pcb.This is my first try at a lowside current sensing. Is this enough, or do I need something more complicated? I googled and used the MCP6V81UT-E/LTY beacause I do not need it to operate at high voltage and it is much more precise than TL072 (correct?). I would get the -1.25V by using a voltage divider or some sort of reference/regulator. I would like to have 1mA accuracy for current sensing. I do not really care that much for the cc accuracy, for example if I set max current at 200 mA and the supply limits it at 195 mA, it is fine.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 23, 2016, 05:06:25 pm
content: The circuit diagram still has a few points missing / wrong:There should be a compensation for the voltage drop at the shunt. This could be adding a faction of the voltage at the shunt to the + input together with the set signal.The current sensing resistors have to go to GND and the amplifier sensing from the other side.There is still the output capacitance and the compensation parts missing.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 23, 2016, 05:51:47 pm
content: Quote from: Kleinstein on December 23, 2016, 05:06:25 pmAdded it in, I supose this means that 1V from DAC no longer gets me 10V at the output, but 9.09V, correct? But this can be compensated for easily in software.Quote from: Kleinstein on December 23, 2016, 05:06:25 pmOf course, that is just o total brainfart on my side

author: Kleinstein
author_rank: Super Contributor
timestamp: December 23, 2016, 06:20:50 pm
content: There is no absolute need for R18 if you have the shunt on the low side. The shunt at the low side also works to limit current gain. The 0.1 Ohms value could be close the the lower limit though.The output capacitor should be directly across the load, not to GND. Likely a value of only 1 µF is also too small. I would more expect something like a 10-100 µF electrolytic in parallel with 100-300 nF ceramic.I am not sure about the right value for R17. My crude look at the circuit gives me more like 91 K as an appropriate value. With 100 K I would expect a little residual DC output resistance (e.g. 10 mOhm range).The part still missing is frequency compensation for the control loops. Usually this are RC series circuits from the neg input of OP1A/B to the output (or the adjust pin of the LM350). OP1B would need an extra resistor from the current signal. To improve stability with large capacitance at the output one might consider something like a 5 n in series with 10 K in parallel to R4.

author: IrejectYourReality
author_rank: Contributor
timestamp: December 23, 2016, 06:54:23 pm
content: The datasheet of LM350 suggests either 1uF tantalum or 25 uF electrolytic cap on the output.About R17, shouldnt it be 90K? Then Vout=10*(V2*1/10)+(V1*9/10), where V1 is the DAC1 IN.I tried to add the frequency compensation, but I am not 100% sure that I understood everything you said. Also, I just guessed the values for R21, C6, R20 and C5, because i have no idea what kind of values should be used for this.So, how did I do?

author: Kleinstein
author_rank: Super Contributor
timestamp: December 23, 2016, 08:14:39 pm
content: The circuit looks good now.R17=90K would be the theoretical value - to make sure the output resistance is not negative, a slightly larger value might be a good idea.The frequency compensation as shown is still rather slow, but I have not done a full simulation / calculation. I would expect C5 and C6 more in the 100pF to 1 nF range. R21 should smaller (e.g. <= 5 K) - currently the voltage loop gives a gain of about 1 at high frequencies - this is a little too much. To get a better guess for the compensation, I would use a simulation. I am not sure how different the LM350 behaves from a more normal darlington transistor / MOSFET.For a faster transition from CV to CC mode, it might be a good idea to move the right side of C5 to behind the diodes (adjust pin).

author: IrejectYourReality
author_rank: Contributor
timestamp: December 23, 2016, 09:30:14 pm
content: Thank you very much for all the advice, I will design the rest of the circuit and the pcb and will report back with results.