Non-linear CV/CC control in schematic - Page 1
Forum section: electronics / projects

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 02, 2019, 03:47:50 am
content: Hi!I modified this schematic:http://www.zen22142.zen.co.uk/Circuits/Power/1220.htmwith some info I read and found in the LM-723 datasheet and a few other LM-723 circuits I found on the web.It seems to work, but the voltage and current adjustment are not linear, the voltage jumps higher the more the potentiometer is maxed out, same with current... I was hoping for a smooth linear curve for voltage and current adjustment.In addition, I don't think the original current limiting works anymore, or at least I don't see it on the graphs in LTSpice XVII.I've attached everything in a zip file.R1 & R3 was added by me, their valves were found by experimenting with resistance vs the max & min output voltages.I re-arranged some wires from pin 2, and added in the voltage adjust potentiometer and the current adjust potentiometer, their values are just what I thought seemed to work.R14 & R15 values were found by measuring the maximum output current across the load, which I limited to ~16A. (I'd like 15A maximum with an overload trip current of 16A.)So please, pick it apart and see what I did wrong, and with some guidance hopefully I can fix it all properlyEDIT: Attached picture of the original schematic.

author: coromonadalix
author_rank: Super Contributor
timestamp: March 02, 2019, 03:59:50 am
content: would be nice to put it in a bmp format or pdfYou do know : you have logarithmic pots and linears ones

author: Kleinstein
author_rank: Super Contributor
timestamp: March 02, 2019, 08:39:13 am
content: Adjusting the voltage by using a pot as a feed back divider gives an 1/x type curve. To get it linear one could use the pot as a variable resistor only and keep the lower end constant.For the current adjustment it's unusual to include two BE junctions, just one forward biased BE junction should be enough to compensate for the drop the transistor for the current limit needs. Due to the contribution from the BE junction the curve is nonlinear. This is slightly intentional to get a larger range. a slightly better position for the pot would across the B-E junction of the Q2 (at the emitter side one could average over the 4 output stages with 4 extra resistors for a little better accuracy) - this would also eliminate the need for R14,R15. As the current limit is not very accurate, I would not worry about this so much.The horizontal connection between R4,R5..R10 is a bad idea - the separate emitter resistors are needed to balance the current. At the low cost end averaging resistors are usually easier than a separate common shunt.The BMP file formal is bloated and outdated - .PNG is the suitable choice today.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 02, 2019, 04:23:54 pm
content: Quote from: coromonadalix on March 02, 2019, 03:59:50 amSorry! Should have though of that before.... I'll add a .PNG verison to the first postYes, the potentiometers I am using should be linear if what I'm reading from potentiometers.lib is correct:The pot variables in LTSpice are: Rtot=10K wiper=.1 the other pot is Rtot=1K wiper=.1 so they should be linear right?Code:* A Collection of Potentiometers* ==============================* Helmut Sennewald, 12/23/2003 V1.1** Models:* potentiometer old style LTSPICE potentiometer* pot_lin k*x* pot_pow x^k* pot_plog exp(k*(1-x))* pot_nlog exp(kx)* pot_tab table(x)* pot_piher_plog piecewise linear, datasheet* pot_radiohm_plog piecewise linear, measured*** 1 ____ 1.0=wiper* |* | | 3* | |<---- wiper 0..1* | |* Rtap | | Tap* | |* 2 ____| 0.0=wiper*** RTOT = total resistance* WIPER = ratio of travel of the wiper* RTAP = reference resistance at wiper=Tap* It is needed only for pot_plog, pot_nlog and pot_pow.* RTAP is measured between pin-2 and wiper.* TAP = ratio of travel when Rtap is reached.func LIMIT(x,a,b) {min(max(x, a), b)}*--------------------- The Linear Potentiometer ---------------------** 1.0 <----- 0.0* |3* __V__* 1--|_____|--2** o--R1-o-R2--o*.SUBCKT potentiometer 1 2 3 Rtot=1k wiper=.5* Parameters: Rtot, wiper.param w=limit(0,wiper,1)*R1 1 3 {Rtot*(1-w)}R2 3 2 {Rtot*(w)}.ENDS*.SUBCKT pot_lin 1 2 3 Rtot=1k wiper=.5.param w=limit(0.01m,wiper,0.99999)*R1 1 3 {Rtot*(1-w)}R2 3 2 {Rtot*(w)}.ENDS*------------- The Ideal Power Function Potentiometer ---------------** It is interesting to know that the power log. curve is a* good fit to so called "log"-potentiometers, because most of* them have not true logarithm dependency in reality.** 1.0 <----- 0.0* |3* __V__* 1--|_____|--2** o--R1-o-R2--o** RTAP is resistance at travel TAP* Example: Rtot=10k, R=1k @ 0.5 (half way)* RTAP=1k, TAP=0.5* RTAP and TAP define a point of the curve resistance versus ratio.*.SUBCKT pot_pow 1 2 3* Parameters: Rtot, wiper, Rtap, Tap.param w=limit(0.01m,wiper,0.99999)*.param pwrexp=ln(RTAP/RTOT)/ln(TAP).param ratio=w**pwrexp*R1 1 3 {Rtot*(1-ratio)}R2 3 2 {Rtot*(ratio)}.ENDSQuote from: Kleinstein on March 02, 2019, 08:39:13 amOkay, I've modified the current adjust pot and hooked it up correctly from what I believe. Min current is now 3.24A with a 1 Ohm load. I am guessing adding in those 4 extra resistors you are talking about would help, but I don't quite understand where you are saying to put them? I was hoping to get currents as low as possible (250mA would be great.) I am going to modify the values of R3-R10 (Balancing resistors) to try and lower the max current as it is now upwards of 21.6A! (Changing the potentiometer value doesn't seem to affect the max current.)I also removed the horizontal connections between the load balancing resistors & relabeled all the resistors numerically.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 02, 2019, 04:57:22 pm
content: I have played a little more the circuit and changed the current limit to a way it allows adjustment to small values too.The averaging resistors are only included for 2 power stages, but it should be obvious how to include the others - it just gets messy in the diagram.

author: coromonadalix
author_rank: Super Contributor
timestamp: March 02, 2019, 05:30:12 pm
content: Adding a power diode in serial with the current shunt would help to get lower current settings, just need to adjust component valuesthis is an old edited psu schematic found on the web, a negative 1.2 volts on the voltage pot will help to start a zero volts.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 02, 2019, 05:43:44 pm
content: Okay, I added in the remaining averaging resistors and I also lower the values of the balancing resistors to increase the max current to 12.7A.That seem to work well12.5A max is fine as well, 15A was just what I was aiming for.I think the voltage adjustment is still not as smooth as I wanted, i will try to adjust the pot value for that and see if that help smooth it out.

author: duak
author_rank: Super Contributor
timestamp: March 02, 2019, 07:58:43 pm
content: If you want a linear output voltage adjustment, change it to something like the attached image where only the upper portion of the voltage divider is variable.Cheers,

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 12:54:01 am
content: Quote from: duak on March 02, 2019, 07:58:43 pmI cannot for the life of me figure this out! I've tried connecting pin 4 to the top leg of the pot & the wiper, no luck.Tried moving the pot over to pins 5&6 like I see in the LM723 datasheet, no luck - I have no control over voltage at all.I'm missing something here....I've also been reading about getting voltage control down to 0V with the LM723, it seems I need a negative voltage fed into the error amplifier of the LM723..... Not sure how that will play out.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 02:02:38 am
content: Quote from: Silver_Pharaoh on March 03, 2019, 12:54:01 amIt would be better to have a fixed divider from the output terminals to the error amp's inverting input. The Pot then is used to adjust the reference to the non-inverting input.Applying a small positive offset to the inverting input will allow the output to be adjusted to a lower voltage.

author: coromonadalix
author_rank: Super Contributor
timestamp: March 03, 2019, 04:57:33 am
content: another control variantVelleman kit k7200

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 04:55:01 pm
content: Quote from: xavier60 on March 03, 2019, 02:02:38 amthe error amp's inverting input is pin 2, which is connected to the current adjustment pot, so I'm not sure how to connect up the voltage adjust pot if the current adjust pot is connected to pin 2?

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 05:10:49 pm
content: I meant the voltage error amp's input pins, 4 and 5.

author: duak
author_rank: Super Contributor
timestamp: March 03, 2019, 05:13:48 pm
content: Silver, what is your present voltage control circuit? The linear Vout adjust I presented works very well.If you want the output voltage to go down to zero, you will have to provide a negative supply voltage to pin 7 of the 723 or figure out a way to operate the 723's core with voltages offset from the actual output voltage. The latter can be done with some reference diodes such as the LM385.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 05:27:39 pm
content: Quote from: xavier60 on March 03, 2019, 05:10:49 pmAhh, I must have misread. Thansk for clarifying, I will give it another goQuote from: duak on March 03, 2019, 05:13:48 pmCurrently, the top leg of the pot is connected directly to pin 3 which is Vout, the wiper comes from pin 4, the inverting input (which is coupled to pin 13 with a capacitor) and the other leg of the pot goes to ground where pin 7 connects.I've attached a picture of the current schematic

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 05:43:54 pm
content: I'm not certain if the idea I mentioned will get t all the way down to 0V output because of the way the voltage error amp is designed in the LM723.There needs to be enough voltage applied to pin 4 to keep Q13 turned on inside the IC but the Comp pin 13 can't pull much lower than pin 4.EDIT: I had the pin numbers all wrong from the data sheet.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 05:47:29 pm
content: Quote from: xavier60 on March 03, 2019, 05:43:54 pmDoesn't need to be perfect, I don't think I'll need anything lower than 1V. I was only asking if I need to test something low voltage like an LED or somethingI think I'm getting somewhere though, I modified the schematic again and moved the volt adjust pot over to where I think you were telling me to move it.It seems pretty linear, each 0.1 increase or decrease in the pot turn raises or lowers Vout by ~2V. That's not too bad I think.I've attached the latest revision below.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 05:54:17 pm
content: Notice that I had to correct the pin numbers in post #15. I keep forgetting that there a 2 packages with different pin outs.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 06:15:19 pm
content: It will need a resistor from Vref to the Inv pin to keep some minimum voltage on it but I'm having trouble figuring out how much it needs.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 06:30:48 pm
content: Quote from: xavier60 on March 03, 2019, 05:54:17 pmHaha I see! I better pay attention when I'm ordering the LM723.Wonder why they do that? wouldn't it make more sense to keep the pinout the same?I think the circuit is pretty much done. I'm happy with it anyway. I've attached it to the post.Can anyone see any issues with it? I added in the resistor wattages as well. They should be good I think.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 06:41:23 pm
content: My most recent experience was with the SG3525. Its error amp works with the input common mode range down to 1V.I would aim for about 1.5V minimum on the Inv pin on the LM723.I don't have the IC to experiment with.With the Ninv pin floating I would find what minimum voltage is needed on the Inv pin to keep the Comp pin pulled low.

author: duak
author_rank: Super Contributor
timestamp: March 03, 2019, 06:58:50 pm
content: To work properly the 723 requires a minimum of 2 volts between the both the inverting and non-inverting inputs and pin 7. It may work lower but performance may suffer. Your circuit violates this by allowing the non-inverting input to go to 0 volts. Try exchanging R1 and the Vadjust pot and recalculating the values. In the limit, the non-inverting input will vary between Vref (~7.15 V) and whatever your voltage divider delivers. The 723 then tries to deliver an output voltage equal to the ratio of R21/(R20+R21) times the non-inverting input voltage. With the present values or R20 and R21 it will be 11X the voltage on the non-inverting input.Does this make sense?

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 07:22:36 pm
content: I would start with a 4:1 divider, 15K/4.7K. 10K from Vref to Inv should set the 2V minimum.Use something higher like 10K for the voltage adj Pot and use it as a variable divider not as a VR

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 10:33:28 pm
content: Quote from: duak on March 03, 2019, 06:58:50 pmNot fully no...Inv & NInv pins should both have a difference of 2V between them and pin 7 correct? So those pins should be connected to Pin 7 with a voltage divider to ground? (Pin 7 is usually connected to ground no?)I tried swapping R1 and the pot leaving everything else alone and no luck. When you say recalc the values, I assume you mean R20 & R21?The last bit about Vout being X times the NInv voltage makes sense.Quote from: xavier60 on March 03, 2019, 07:22:36 pmI tried this, but I wasn't able to get anything to work....So bottom line here is Inv pin and NInv pins should be connected together? What about Pin 7 (-V?)A physical picture example would probably help me. I don't learn so well from just text unfortunately. I need to see something hands on to grasp it.

author: xavier60
author_rank: Super Contributor
timestamp: March 03, 2019, 10:58:42 pm
content: I drew this up quickly. Ill check for possible mistakes later.The values aren't supper critical.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 03, 2019, 11:49:36 pm
content: Quote from: xavier60 on March 03, 2019, 10:58:42 pmThank you.I was hooking the pot up wrong.. everything else was correct.I'm sorry you guys have to dumb things down sometimes. I appreciate it thoughCircuit has been modified and seems to function well. I adjusted the voltage divider and adjusted R1 slightly to get a bit smoother voltage control.Vdroop under a 12A load is about 0.1V. There's not a lot I can do about that eh? Something to do with VRef or current sense maybe?

author: xavier60
author_rank: Super Contributor
timestamp: March 04, 2019, 12:33:06 am
content: Good to see things work.What is the minimum regulation voltage?The lowest loaded output voltage while still being controllable with the voltage adj Pot.

author: xavier60
author_rank: Super Contributor
timestamp: March 04, 2019, 01:03:50 am
content: The cause of the voltage droop should be traceable by looking for voltage changes while loading and unloading the output.Such as, Vref, Ninv, Inv and the top of R20, all WRT pin7 -V.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 05, 2019, 01:37:35 am
content: Quote from: xavier60 on March 04, 2019, 01:03:50 amSeems Pin 2 and Pin 10 have large Voltage fluxuations:Pin 2:12A load: 5.575VLow load: 5.109VPin 10:12A load: 7.05VLow load: 5.80VI have not had nay luck much in modifying the circuit.The Vdroop gets worse after I pull 11A from the circuit. Below 11A the Vdroop is very minimal.At 11A, Vdroop is 0.02V and gets worse the more I pull after that. At 11.5A Vdrop is 0.05V...Could is be the balancing resistors?

author: xavier60
author_rank: Super Contributor
timestamp: March 05, 2019, 01:52:19 am
content: Quote from: Silver_Pharaoh on March 05, 2019, 01:37:35 amThe change at pin 10 is expected as the output transistors need more drive with increased load .The change at pin 2 is very close to causing the CL transistor to conduct especially if the LM723 is warm.I would be curious to know how much drive current pin 10 is having to source. This can be done by inserting a 10Ω series resistor and measuring its voltage drop.

author: xavier60
author_rank: Super Contributor
timestamp: March 05, 2019, 01:59:54 am
content: You would be able to tell how close operation is to the current limit threshold by moving the Pot.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 06, 2019, 12:52:03 am
content: Quote from: xavier60 on March 05, 2019, 01:59:54 amNot sure what you mean by this?Voltage drop is very low across the 10resistor.I had to change the current limit pot value back to 10K. I forget why I changed it but it didn't let me reduce the current below 10A. I must have been tinkering with it and forgot to change it back.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2019, 01:01:54 am
content: The idea is to figure out wether the CL transistor inside of the LM723 is starting to conduct or not.The simplest way is to apply a load that just starts to make the output voltage drop then make a small adjustment to the current limit Pot to see if it affects the output voltage.Measuring the voltage between pins 2 and 3 would be useful.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 06, 2019, 01:27:45 am
content: Quote from: xavier60 on March 06, 2019, 01:01:54 amAhh I see.Can't tinker much tonight, but I think you are correct in saying the CL transistor is starting to conduct.It's hard to tell because I don't have fine control with the current adjust and the voltage drops near full load, so the current adjust pot is already maxed out..I reduced the balancing resistors to 0.8each and the voltage shot back up to 5.1V like I have it set. This pulling 12A at the output.Are the balancing resistors limiting max output current too much?

author: diane
author_rank: Newbie
timestamp: March 06, 2019, 01:51:58 am
content: Wow, such a difference in if of possibility. I only did a patent on the idea, but if there is a possibility of creating without too much grevience, I will recommend that person's idea. Of course it may take another patent for idea. Read articals. About the building of assisted living quarters. 4 New ones were just finished within 5 to 10 miles of my area. They are filled and with waiting lists. Thank you for your help.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2019, 04:22:43 am
content: Quote from: Silver_Pharaoh on March 06, 2019, 01:27:45 amTo increase the max current setting, just put a resistor from the junction of the 100Ω summing resistors to the junction of the sharing resistors(power supply's output). Start with 100Ω.

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2019, 05:05:54 am
content: Quote from: diane on March 06, 2019, 01:51:58 amIs this the thread that you meant to post in?

author: xavier60
author_rank: Super Contributor
timestamp: March 06, 2019, 08:51:38 am
content: I would rather see R14 placed between B and E of Q1. Then R15 should be something lower in resistance.The low end of the current range will have a lot of thermal drift.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 07, 2019, 10:58:44 pm
content: Quote from: xavier60 on March 06, 2019, 08:51:38 amIsn't that where R14 currently is?I've added in the 100resistor. Having it around 0.5gets be a Vdrop of a mere 0.01V which is sweet! Short circuit is around 14AThat resistor has the same effect as lowing all the balancing resistors to 0.08It is better to lower this new resistor instead of the balancing resistors isn't it? I imagine finding 0.08resistors might be pricey/hard. I haven't checked yet.

author: xavier60
author_rank: Super Contributor
timestamp: March 07, 2019, 11:17:41 pm
content: That looks right except for R11 which is label as "1" but must really be 100Ω because 1Ω would result in a very high maximum setting like 100A!

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 08, 2019, 12:33:11 am
content: Quote from: xavier60 on March 07, 2019, 11:17:41 pmAt 100max current seems to be around 12.8A?Unless this is a quirk of LTSpice?

author: xavier60
author_rank: Super Contributor
timestamp: March 08, 2019, 12:43:13 am
content: Quote from: Silver_Pharaoh on March 08, 2019, 12:33:11 amThat could be right. In situations that use a transistor'a Base to Emitter threshold voltage as a reference, some try and see needs to be done.Also, the short circuited current setting will have to be a fair bit higher than the expected normal current draw to avoid voltage droop.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 08, 2019, 01:48:06 am
content: Quote from: xavier60 on March 08, 2019, 12:43:13 amMake sense, otherwise the CL transistor would start to reduce the voltage as it conducts more and more right?

author: xavier60
author_rank: Super Contributor
timestamp: March 08, 2019, 02:47:42 am
content: As the Base to Emitter voltage increases with increasing output current, the CL transistor starts turning on gradually at first. This is what would be causing the voltage droop.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 09, 2019, 05:09:31 pm
content: Well I think this is the final version!All seems to work well.I included 2 other versions with a more linear & tighter voltage adjustment.

author: Silver_Pharaoh
author_rank: Contributor
timestamp: March 09, 2019, 11:38:39 pm
content: Wait, I just thought of the inrush current.63kA? That can't be right...I know the input cap is for smoothing but 47,000uF seems like a lot. Followingthis]https://www.electro-tech-online.com/threads/how-to-calculate-the-value-of-a-smoothing-capacitor.106374/]this] formula I get about 56,000uF assuming 14A with a 2v drop. That's a lot of capacitance.Would I be able to reduce the capacitance to say 5,600uF instead?Using an NTC thermistor with a resistance of 2.5@ 25C would limit the inrush to 10A roughly which is fine, but having say a 30mOhm resistance at full load reduces the max current I can draw at 24V to 10.5A. I can get around this buy suppling the LM723 with 28V instead of 27V.Am I over thinking this too much?