Linear Power Supply based on HP/Agilent E3610A - Page 1
Forum section: electronics / projects

author: timsu
author_rank: Contributor
timestamp: May 04, 2016, 04:02:15 pm
content: Hello,I'm trying to build a power supply based on the HP E3610A (15V/2A).You can find the Schematics as attachement to this post orhttp://sites.fas.harvard.edu/~phys191r/Bench_Notes/A1/agilent_e3610a.pdfon page 16.On the pages before there are also the component values.I think I understood most of the circuit, except Q2/CR8( (HP schematic) Q3/D7 (my schematic).The simulation seems to work fine even without it.Can somebody explain to me in what situation the transistor is passing current and when not? What is the point of it?In a service note HP is recommending to remove CR8/D7 to prevent oscillations.But herehttps://www.eevblog.com/forum/testgear/hp-agilent-e3610a/it is suspected, that this modification can destroy the pass transistors because of a too high base-emitter voltage.Also, should I include the modification, the added cap in the voltage error amplifier, from the linked thread?Did HP make more "bad" choices which I can prevent?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 04, 2016, 06:57:19 pm
content: The transistor Q3 is working as an emitter follower - so it's always in a linear range, essentially never off. The diode limits the lower output voltage to something close to -100 mV. It also provides higher current sinking capability. This should not be important with the shown BJT power stage, but is can be important with HPs MOSFET stage. The high sinking current allows to turn off the FETs fast and the 1.2 K resistor limits the speed when turning them on.The values for C1, R2 look odd - looks way to slow. I would more expect something in the 500 pF and 5 K range. It depends on the output caps and power stage too.One poor choice HP made was adjusting the voltage by changing the feedback divider. It's better to keep the divider fixed and adjust the set voltage like indicated. However with a variable ref voltage for the divider one might need an extra minimum load (e.g. constant current) to bring the output back to zero.Also the supply voltage for the OPs could have be chosen lower, not to allow the OPs to go way to high in output voltage. Limiting the voltage excursion for the regulation part that is not active can speed up the switch over from CC to CV mode and back. Especially with a BJT output stage the voltage range is well predictable, so it's easy to chose more reasonable limits. So going for a Rail-Rail OP might be one option - the LM1001 is rather slow anyway.The output caps should be a combination of a low ESR (e.g. ceramic / foil in the 100 nF to 1 µF range) and low ESR electrolytic. HP might not have modern low ESR electrolytic as an option back then. So one might get away with lower capacitance - the large caps chosen by HP might be needed to get the ESR down in the 1 Ohm or lower range, not the capacitance itself. The old style high ESR caps are usually not helping - so I would prefer lower capacitance low ESR (e.g. 0.1-0.5 Ohms range).

author: timsu
author_rank: Contributor
timestamp: May 04, 2016, 08:34:13 pm
content: So why would HP decide to remove the diode because of oscillations?That could cause harm to the transistors, right?Quote from: Kleinstein on May 04, 2016, 06:57:19 pmI thought this was also confusing, because in the high voltage model E3612 they used 22 pF and no resistor in the voltage feedback loop.Quote from: Kleinstein on May 04, 2016, 06:57:19 pmThe 330 Ohm R9 should work, right? This was also added by HP in a later revision.Quote from: Kleinstein on May 04, 2016, 06:57:19 pmI would prefer going for -5V/+5V because for other parts of the circuit (DAC/ADC) I will need 5V.Also why use a voltage divider with 1R and 500K before the current shunt?Are they compensating some error with this?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 05, 2016, 07:21:56 am
content: Removing the diode (CR8) to limit negative control voltage should not change much with oscillations - it could be an issue if the output stage (gate capacitance) is setting much of the compensation. The diode might limit the ability to go all the way to 0 V output voltage, especially if the transistors are hot. However the more logical step would be using two diodes in series instead of removing it all together. Allowing the voltage to go to negative (much below -7 V) without the diode might cause trouble for the output transistors (to negative a base-emitter voltage), so it is not a problem with only a +-5 V supply there.R9 could to some degree act as a minimum load, though it is not very efficient at low voltages and gets quite hot at high voltage. A constant current sink might be more appropriate.The divider at the shunt looks odd - though it could be just to use the same meter with different supply models.

author: timsu
author_rank: Contributor
timestamp: May 07, 2016, 06:51:49 pm
content: I think HP was limited in their ability to change something aftwerwards because they had no space on the board for two diodes, or a constant current sink, so they added just a big resistor as load.I started to draw up the schematic (attachement). The OP-Amps are just placeholders, maybe someone has a recommendation.I will have to do some more simulations to determine the right component values for the voltage error loop and the output cap.Also the digital stage is still missing, it will be 12 Bit ADC/DAC and an ATmega.I would appreciate feedback.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 07, 2016, 07:44:22 pm
content: The constant current sink could use the +5 V auxiliary supply to work all the way down to 0 V (or even a little below): could be just a PNP transistor with base to GND, collector to the neg. output and emitter through a resistor to the +5 V. The down side is that the current will flow through the shunt too, but there is base current flowing anyway. Also output voltage could go slightly below zero (a resistor/trimmer at the base could be used to adjust this to a few mV).ESR of the output capacitor might be important for stability - keep that in mind for simulation. Low ESR electrolytics have about the right ESR. The cap might be a little smaller with a low ESR version.Stability with very large caps might need a capacitor (or RC) in parallel to R29.A 10000 µF filter cap may not work very well with a relay for transformer tap switching, especially at higher voltage (e.g. 30 V). I would consider using a transistor to do the transformer tap switching. For a transformer with a split coil, like 2 x 15 V this works rather good and today a transistor might be cheaper than a relay. As a side effect no more problems with highly dynamic loads that might cause the relay so switch to often. The circuit is even simpler than relay control.For OPs something like the LF412, TL072 or TLC272 would be OK for the regulator. The readback of the current and possibly voltage might want more precision, like an OP07.The Output disable could use the second half of the LM393.

author: timsu
author_rank: Contributor
timestamp: May 07, 2016, 11:35:19 pm
content: Your solution for the constant current sink works fine (mine stopped working at around 0.4V). I can remove the error current through software.Quote from: Kleinstein on May 07, 2016, 07:44:22 pmWhat would be the advantage of this?The maximum voltage will be only at 15V. With 2 Amps this means 30W of heat in the pass transistor. If I use an large transistor (TO-247/264) maybe tap switching is not even needed.Quote from: Kleinstein on May 07, 2016, 07:44:22 pmSomething like a half bridge with two mosfets? Should I use an IC for that?Can I use the OP07 in the current feedback loop or is it too slow for that?The second half of the LM393 was planned as comparator with hysteresis for tap switching.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 08, 2016, 08:02:05 am
content: The OP07 is relatively slow. Depending on the speed planed it might work too. However current regulation is usually not needed to be as stable as the current reading.Using a second transistor for the second tap would use two filter caps get a raw voltage of something like a 0, 10 V and 20 V. Transistor stage is more like a class H audio amplifier: the 10 V connect via a diode to the collector. The 20 V via an NPN transistor with the basis connected through a resistor to the 5 V auxiliary supply. The transistor needs to be cooled as well, as it will take over much of the loss at some output voltages.For just 30 W maximum loss (even if more like 40 W since for a 15 V output the rectifier should be more like 20 V) there is no real need for tap switching, but it could permit a smaller heat sink (e.g. 60% the size) at relatively low extra costs (2nd capacitor, 1 diode, 1 power transistor - though smaller TO220 transistors could be used than, as the maximum power is about half).Having the constant current sink go down to below 0 might be an issue with the output disabled. In this case the output would go slightly negative, if the current in not turned off too. So adjusting the minimum voltage to something like -100mV instead of -500 mV might be a good idea.

author: timsu
author_rank: Contributor
timestamp: May 08, 2016, 06:15:02 pm
content: Quote from: Kleinstein on May 08, 2016, 08:02:05 amI agreee that this is not as important, but the set current limit should not be too far off from reality. With an 0.1 shunt resistor, each mA is 100uV. If the used op amp in the feedback loop has 10mV offset the current limit would be quite inaccurate.I don't know if the tap switcher is correct, I attached an simplified version. Why would the switching transistor need cooling? It is either saturated or off, the power dissipation should not be over half a watt.This solution seems nice, except it wastes quite a bit pcb space with the huge caps doubled.The negative voltage when the output is switchted off is about -200mV, limited through the output schottky diode.I could not limit this further, it would require an base resistor of several 100k , which is too large too let current flow through the transistor.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 08, 2016, 07:48:04 pm
content: The transistor to use the higher raw voltage is not switching but trying to give the "normal" transistor something like 3-4 V to work with. So at about half the voltage it will handle much of the power loss. Not as much as the single transistor, solution but still half the voltage times all the current in worst case. However its only one of the two transistors at a time that has the high power. With a low voltage a MOSFET is even better for the switchover. Attached is a crude plan to show how the second tap could be used.For the caps: you need two caps, about twice the capacitance but only half the voltage rating. So about twice the mechanical size (and price). On the other side you save on the heat sink. So tap switching might be more expensive, but could be smaller if you want it without a fan. Usually a passive heat sink is larger than the caps.Just 15 V and 2 A output could still work with a single transformer tap, even with a passive heat sink.

author: timsu
author_rank: Contributor
timestamp: May 09, 2016, 07:52:38 pm
content: Thank you for your help so far.I would prefer to have only one transistor which needs active cooling (even if the total heat stays the same), it makes mounting and cabling easier.The solution from my previous post should still work, right? I just need a comparator to do the switching.For the constant current sink I think will settle for the solution which makes the output to go to 0V if it is switched off, even if it justs works down to 0.5V.The LF412 looks good so far for the error loops. For the readout maybe OP07, does something comparable exists in a dual package?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 09, 2016, 08:10:40 pm
content: There are a few precision OPs in a dual package, e.g. LT1013, OPA2177, MCP6V12 and similar.The really high precision ones can be quite expensive though. There are many that are better than LF412 / LM358, but still relatively cheap.When doing range switching with a relay, one should not make the filter cap so large as this adds quite some load to the contacts. It might also be worth using two extra diodes and keep the lower voltage always on and only add the path for the higher voltage. With this low power there still is the option of not switching at all.

author: timsu
author_rank: Contributor
timestamp: May 09, 2016, 08:40:32 pm
content: I meant switching with one transistor as shown in post #8, not a relay.To summarize my tap switching options:no switching: (+) simple, small pcb (-) large heatsinkrelay switching: (+) filter caps only one required once (-) relay wear, dynamic loads, comparator requiredswitching as in post #8: (+) no wear, only one power transistor (-) comparator and two filter caps requiredswitching as in post #9: (+) no wear, few components (-) two filter caps and two power transistorsI still have not deciced yet, but maybe I will use two channels. In this case tap switching would be nice.Most heatsinks I looked at go down to 0,8K/W. For one channel this would not be a problem, with two it could get a bit hot.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 10, 2016, 11:16:45 am
content: For the circuit from post #8, there is still the question on how to power the comparator / driver. If powered from the auxiliary (5V) supply the drive current will flow through the shunt and an extra diode again to negative a base voltage is needed.So for switching the taps I would prefer a MOSFET over a NPN. At the rather low voltage there should be no big problem in protection against too much gate voltage. However using an N-MOSFET at the high side like drawn would need an extra supply, as the +5 V auxiliary supply for the regulator is no enough. So a slightly different circuit is needed.There is also one more difference between relay and transistor use for tap switching: the transistor solution can be fast enough to even include the ripple on the raw voltage. So the transition could be set lower and with less (no) hysteresis.For a dual channel version, the two channels will normally be fully separate, except the digital control signal going from one part to the other. It might get difficult to find a transformer with two spilit windings. So this might make tap switching more difficult.

author: timsu
author_rank: Contributor
timestamp: May 10, 2016, 08:48:11 pm
content: Is it even possible to do the switching without another floating supply circuit?It seems kind of difficult because GND has no static potential to the tap switching.I tried using a P-MOSFET with a pullup.But how to pull the gate down with a ground-referenced circuit?I know for two channels the transformer becomes a bit of a problem. I have to see what a custom made one costs.

author: StillTrying
author_rank: Super Contributor
timestamp: May 10, 2016, 10:35:41 pm
content: In case it's of any use.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 11, 2016, 04:39:46 pm
content: You could do the switching with a N_MOSFET, if you choose to switch the other half of the two voltages. It looks a little like a bridge circuit. The MOSFET goes from the most negative voltage to the negative output, with the diode from the middle voltage as an alternative. The controlling comparator is powered from the main supply and testing the voltage over the output transistor. So should be reasonably easy at low voltages (e.g. < 30 V), so the comparator can handle.Though slightly larger and less efficient, one could still use two separate power transformers for the main power and one small one for the two auxiliary supplies. For the auxiliary supply one can get a way with just a single winding, as there is no real need for a large negative supply something like a -1 V and +5 V or maybe +8 V can be enough. The -1 V could use just two diodes as shunt stabilization. The down side might be limit in the voltage between the outputs.

author: timsu
author_rank: Contributor
timestamp: May 11, 2016, 08:47:14 pm
content: I kind of got it working.At around 7V the taps should be switching, the comparator U5 does this.But I still have some trouble controlling the MOSFET. With some opamps/comparators there is some weird oscillations. The LT1077A worked, but it is not rail-to-rail, so it does not fully turn the MOSFET M1 off.

author: Kleinstein
author_rank: Super Contributor
timestamp: May 11, 2016, 09:09:15 pm
content: You don't want an rail to rail OP for the MOSFET control, as the output voltage would be to high. A comparator like LM393 with a little hysteresis should do it. The gate voltage could be from the lower 10 V supply.I would compare the collector side of the power transistor to something like 3 V higher than GND (at emitter resistor). One might need to check how the current from the hysteresis flows, but I won't expect trouble there.

author: timsu
author_rank: Contributor
timestamp: May 12, 2016, 07:41:19 pm
content: I can't quite get it to work.I used the LM393 as comparator.The inverting input is collector-3V (with a diode drop).The noninverting input the emitter voltage.Output of comparator with a pullup to the 10v low tap.Because the comparator has such a high imput impedance, a parallel resistor is required, to have a diode drop.If the comparator switches to on, the voltage goes on only 3V higher than in the off state (100R pull up), which turns the fet not complete on.Lower pullups make the circuit oscillate.Maybe it is obvious but all the different voltages which different reference to each other confuse meEdit:I tried an different circuit with an optocoupler, but even here I could not get over 3V Gate-Source Voltage.If i disconnect the gate it goes up to 10VEdit2:I got it working

author: ZeTeX
author_rank: Frequent Contributor
timestamp: May 13, 2016, 07:17:38 pm
content: You might want to have a look here:https://www.eevblog.com/forum/projects/lab-psu-design-ideaquestions/msg205205/#msg205205

author: timsu
author_rank: Contributor
timestamp: May 13, 2016, 09:37:49 pm
content: This circuit has the disadvantage of high power dissipation in the MOSFETs and an additional higher voltage to drive the gates is required.I managed to save a few components compared to my last post with inverting the comparator input.I will post an updatete schematic soon.I looked for pass transistors and found these:MJL 3281AMJH 11022GTIP142What are your opinions on these?

author: sarepairman2
author_rank: Frequent Contributor
timestamp: May 14, 2016, 03:26:02 am
content: keep in mind this was designed for mass production. you can do better O0

author: Kleinstein
author_rank: Super Contributor
timestamp: May 14, 2016, 08:24:18 am
content: The tip142 is good enough for 15 V 2 A, even without tap switching. So no need for the larger and more expensive ones.With tap switching even a smaller version might be OK, but this won't save that much and a larger transistor could alow for a higher temperature or slightly smaller heat-sink.The MJH 11022 looks a little slower - so I would not prefer that one.The MJL 3281A is faster, but you need a good layout and likely quite some real life tests to really make use of it.The FET tap switching from the old linked task is very similar to the one I suggested before, just for more taps and a regulator with emitter follower type output stage. The main disadvantage is that the FETs at some times also have quite some power - so usually the FETs and the main output transistors should be mounted on the same heat-sink. Especially the 2 tap version I showed is really simple. For the relatively low voltage (e.g. < 20 V) the MOSFET sees it is not that difficult to find MOSFETs that can work in linear mode.There is some extra effort to make the FETs switching and there can be small glitches when switching taps - so hysteresis is needed to avoid this (and possible instability) to happen just from ripple or load changes.

author: timsu
author_rank: Contributor
timestamp: May 17, 2016, 06:59:15 pm
content: Here another update:- fixed some smaller bugs (wrong values)- added tap switch- opamp model selection- beginning digital part- simpler output disableTodo:- passive and transistor selection- more protection- digital circuit- ...Edit:Should I add an RC Filter at the Output of the DAC?

author: timsu
author_rank: Contributor
timestamp: May 23, 2016, 05:42:39 pm
content: Another update, now the digital circuit is done as well, with isolated serial port.I decided to use an Arduino instead of using just an Atmega, because it does not really need that much more space.The component selection is also mostly done.Is the protection for all the op amps done right?

author: Kleinstein
author_rank: Super Contributor
timestamp: May 23, 2016, 06:10:40 pm
content: The protection at the OPs looks Ok. The current regulator might not need the diodes, as there usually will not be so much current to make the diodes work or even damage the OP.If you want really low noise, RC filtering after the DAC might be helpful, but this slows down changes in the set voltage.The TL072 can have quite some offset, this might limit the minimum current setting.

author: StillTrying
author_rank: Super Contributor
timestamp: May 23, 2016, 08:13:26 pm
content: You're asking too much of the LM393 to pull down 100R from 5V. (10mA max)I'd try to invert the sense there and make the opto part of the pull up to 5V because I hate wasted milliamps.R3 seems to have changed from 10K to 1K, needing the opto to sink ~28mA.R22 doesn't do anything.Didn't notice anything else.

author: Dave
author_rank: Super Contributor
timestamp: May 23, 2016, 09:02:50 pm
content: The serial interface isn't actually isolated.

author: timsu
author_rank: Contributor
timestamp: May 24, 2016, 08:07:51 pm
content: Quote from: Kleinstein on May 23, 2016, 06:10:40 pmMaybe the TLC 272?Quote from: StillTrying on May 23, 2016, 08:13:26 pmCan you explain this a little bit more?I thought about using something like this:http://www.mikrocontroller.net/attachment/45300/active-pullup.pngQuote from: StillTrying on May 23, 2016, 08:13:26 pmYou're right, I forgot to remove it after I changed the protection diodes.Quote from: StillTrying on May 23, 2016, 08:13:26 pmDoes 10K ensure fast enough switching of the MOSFET?LTSpice shows about 2u seconds switching time with 10k.Quote from: Dave on May 23, 2016, 09:02:50 pmTX has to use an external 5V source, right?Why did HP added resistors between GND and op amp input? (Example: R26 at the inverting input of the voltage error amp)

author: Dave
author_rank: Super Contributor
timestamp: May 24, 2016, 08:18:35 pm
content: Quote from: timsu on May 24, 2016, 08:07:51 pmYup.Quote from: timsu on May 24, 2016, 08:07:51 pmTo cancel out the offset voltage caused by the input bias currents.

author: sync
author_rank: Frequent Contributor
timestamp: May 24, 2016, 10:15:54 pm
content: Quote from: Dave on May 24, 2016, 08:18:35 pmI don't think it's about the offset voltage. It's a LF411 JFET op amp in the HP design. I think it's for the op amp compensation. Without R26 the inverting input would be grounded and R21, C22 from the op amp output to the inverting input have no effect.

author: Dave
author_rank: Super Contributor
timestamp: May 24, 2016, 11:45:36 pm
content: Quote from: sync on May 24, 2016, 10:15:54 pmYou're absolutely right. This is what happens when you just glance at schematics instead of taking the time to give it a proper look.The bias current compensation resistor like I had in mind would be R33, for example (though it should be a higher value - around 50k).

author: StillTrying
author_rank: Super Contributor
timestamp: May 25, 2016, 02:02:54 am
content: Quote from: timsu on May 24, 2016, 08:07:51 pmOn the data sheet graphs the open collector output of the LM393 it can only sink about 10ma before its saturation voltage is 1V and rising. With 100R pull up it just won't be able to pull down enough. I think you'll have to invert it and use the opto as the pullup so that all the current goes through the opto.I started to draw it and then decided to make it 2 working versions. Just to be awkward the simulations version of the LM393's output saturation seems to be about 4 times better than what's shown on the data sheets graph.QuoteI think a few uS is fine for this occasional switching of 12V at up to 2A, but there's much better expurts than me here!

author: StillTrying
author_rank: Super Contributor
timestamp: May 25, 2016, 02:17:54 am
content: Quote from: Dave on May 24, 2016, 08:18:35 pmAnd especially their drift with temperature.I'm surprised at how every everybody just sticks 10Ks, on op amp inputs, it seems like almost 50% of the time they're actually unbalancing the inputs!

author: Kleinstein
author_rank: Super Contributor
timestamp: May 25, 2016, 05:44:19 pm
content: With FET OPs like the like LF412 (HP design), TL072 or TLC272 there is no real need to balance the input impedance. This also applies to some other OPs with compensated input bias (like OP27) or those with a relatively low bias and not so high input impedance (e.g. LT1013). So there are a lot of exceptions where you don't need balanced impedance. The 10 K value is about right to prevent excessive currents that might damage some OPs and for the typical 1 MHz GBW OPs it's low enough you don't have to look at parasitic capacitance so much. Also bias currents are often not that critical at this resistance and thus no real need for balanced impedance with most OPs. So it a natural choice if not critical.The LM393 has a limited drive capacity, so one should not use the 100 Ohms to make it drive some 40 mA, which is more than it can.

author: kridri
author_rank: Contributor
timestamp: May 28, 2016, 01:09:53 pm
content: When I was looking to the schematic I don't see the point of adding resistor R15 (parallel to the Current sink). When you add a resistor parallel to it. It won't be constant anymore and at a higher voltage, the current increases. With more losses! Please can you explain me why you did it?

author: timsu
author_rank: Contributor
timestamp: May 28, 2016, 06:05:19 pm
content: To slowly discharge the output caps when the device is turned off or at very low voltage.It does not have to be exactly constant current.

author: kridri
author_rank: Contributor
timestamp: May 28, 2016, 06:41:37 pm
content: Quote from: timsu on May 28, 2016, 06:05:19 pmThanks for the reply and the explanation. If forgot that a bench supply must be turned down. Silly me!

author: Robaroni
author_rank: Frequent Contributor
timestamp: June 01, 2016, 12:46:20 pm
content: I have an E3610A that the current adjust seems to drift. That is, I can't set the current to zero. I tried to recalibrate the supply but after awhile it drifts above zero and can no longer be set to zero.I swapped out the current opamp U4 as I didn't see anything that stuck out as being bad but it didn't fix the trouble. It might be something as simple as the wire wound current pot breaking down but I was wondering if any of have seen this problem before I dig it apart.Thanks for the help,Rob

author: timsu
author_rank: Contributor
timestamp: June 01, 2016, 07:21:10 pm
content: Should the ADC/DAC should be level shifted up to get a higher accuracy at lower output voltages and allow better calibration?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 01, 2016, 07:47:01 pm
content: It depends on the quality of the ADC / DAC whether it might be worth using different ranges.As the modern sigma delta ADCs offer rather good resolution, there is usually not need for a range switching. Also many ADCs measure negative values as well, so no problem with offsets around zero. It might be a good idea to make use of the differential inputs.For the DAC it might be worth using two ranges, as really good DACs are expensive. A little shifting near zero might be needed if digital offset calibration is wanted.

author: timsu
author_rank: Contributor
timestamp: June 01, 2016, 08:14:31 pm
content: How would be the differential inputs used?What would be connected to the inverting inputs other than GND?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 02, 2016, 02:52:34 pm
content: The GND for current sensing (e.g. pin of shunt -> amplifier for current reading) can be a little different from the Ground for voltage sensing (e.g. Sensing line to output connector). For a low ohms shunt the wires/lines going to the resistor can add to the resistance. So voltage should be read close to the resistor, possibly even using a shunt with 4 wires. The difference will not be large, maybe a few mOhms times the output current, but it's possible with little effort to avoid the error.

author: mcckevin
author_rank: Contributor
timestamp: April 13, 2019, 06:34:34 pm
content: Hey Roberoni --I just stumbled across an E3610 with the same problem. I also replaced U4 (great minds ...) with little success. Did you have any luck with this? Thanks.KevinQuote from: Robaroni on June 01, 2016, 12:46:20 pm

author: Robaroni
author_rank: Frequent Contributor
timestamp: April 13, 2019, 09:44:30 pm
content: Mcckevin-I have several supplies so I just use this in situations where I don't need to control low currents. I think the trouble is the 10 turn current pot may be breaking down and not holding resistance. I haven't pulled it apart to absolutely verify it though.Rob