Meanwell NES Schematic..almost complete

[cor]

I noticed on the schematic that the transformer winding feeding the 48V output does not show a center tap. I assume it is center taped else the circuit would not work as the two diodes would not pass current.

Hi Samson, Thanks!
Until yesterday the second diode did not even show on the schematic because I did not draw it for being redundant as far as I can see (I have since revised my opinion, see later in this thread), since the bottom contact of the lower secondary winding goes directly to the negative 48V output (via L100 and the shunt). But Neil questioned me where that dual diode had gone, so I suggested that he simply add the second diode in the schematic to show how it is connected on the board, even though the second diode in that package does not do any work (possibly it does in lower voltage/higher current variants that do have a center tapped secondary).

[Samson]

Hi Cor

I was just editing my post when you replied. It is possible that the circuit is as shown if the output is half wave rectified and and not full wave. D100 would simply suppress negative spikes. However it seems more likely that there is a Center tap and it is not shown IMO. If there is it probably goes to L100 and the lower tap would feed D100, I am not positive and I do not have one to check myself but it may be worth a closer look.

[NeilP]

Cor, where on the Schematic is the OVP point as printed on the board?

The short answer is - I marked it on my sketch

As Homer would say...Doh>>
added to the schematic

Cor..my theory understanding is to say the least..crap..I have been looking at the 240v / 115volt switch..and even that has got me lost..I can't see how closing the switch and feeding 115 volt AC direct into the circuit, when 240 volt setting does not need that AC input to a part of the circuit works... I can understand a basic transformer...and different primary winding count for 240 or 115..but I just can't see how this works at all

[Samson]

Cor, where on the Schematic is the OVP point as printed on the board?

The short answer is - I marked it on my sketch

As Homer would say...Doh>>
added to the schematic

Cor..my theory understanding is to say the least..crap..I have been looking at the 240v / 115volt switch..and even that has got me lost..I can't see how closing the switch and feeding 115 volt AC direct into the circuit, when 240 volt setting does not need that AC input to a part of the circuit works... I can understand a basic transformer...and different primary winding count for 240 or 115..but I just can't see how this works at all

It appears that when the 115 V switch is closed the circuit becomes a voltage doubler. It runs as a 4 diode bridge at 240 volts and at 115 volts it becomes a doubler. Cheaper than adding a transformer winding. This was common in TVs in the tube area (yes I am that old). The 115 input was doubled to get close to 300 volts. 1.414 x 115 x2 = 322 volts.

[NeilP]

Samson..just to clarify ..D100 is a single component on the board comprising of a 3 pin device PA905C4A..the two diodes are in a single package.

This is what I let the smoke out of a while back while attempting a current limiting mod. After I had converted Cor's drawing to the schematic we now have..i started looking more closely and trying to understand rather than blindly copying form drawing to software. I wondered where the D100 package that I destroyed had gone..that is when Cor pointed out where it was and I filled it in to the schematic.

On the board, there is an un populated section...D101 next to D100 that allows for a second 3 pin power rectifier with heatsink , next to D100.

Here are a couple of pics

the back of the board is reversed to allow overlay to the front image

large size front

back of board reversed

[Samson]

Samson..just to clarify ..D100 is a single component on the board comprising of a 3 pin device PA905C4A..the two diodes are in a single package.

This is what I let the smoke out of a while back while attempting a current limiting mod. After I had converted Cor's drawing to the schematic we now have..i started looking more closely and trying to understand rather than blindly copying form drawing to software. I wondered where the D100 package that I destroyed had gone..that is when Cor pointed out where it was and I filled it in to the schematic.

On the board, there is an un populated section...D101 next to D100 that allows for a second 3 pin power rectifier with heatsink , next to D100.

Here are a couple of pics

the back of the board is reversed to allow overlay to the front image

large size front

back of board reversed

I had a quick look at your pics of the board. I think the unpopulated diode would allow the circuit to form a bridge rectifier with the D100 dual diode in some power supplies (ones with different voltages) and without a center tap transformer. The single package is a dual Schottky diode with the cathode common. Again I suspect that the transformer has a Center tap. The original Meanwell supply circuit uses a CT transformer and the same dual diode. I have the same diode layout in my Meanwell clone but cannot look at it now as it is charging my pack.

Does the inductor L100 go to the D100 anode as shown?

[cor]

Does the inductor L100 go to the D100 anode as shown?

Hi Samson,
I checked that part of the circuit several times as I was surprised that they used no center tap and only one of the diodes in the dual package, but that is what it is...

[cor]

I have been looking at the 240v / 115volt switch..and even that has got me lost..I can't see how closing the switch and feeding 115 volt AC direct into the circuit, when 240 volt setting does not need that AC input to a part of the circuit works... I can understand a basic transformer...and different primary winding count for 240 or 115..but I just can't see how this works at all

Hi Neil,
The 240V setting uses the bridge rectifier as a full wave rectifier, each half of the cycle both caps get charged as a whole (two in series).
When the switch connects the AC to the center of the caps, two of the diodes in the bridge rectifier become useless (just like the second diode in the 48V output) and the two caps are alternatingly charged by half-waves, when the other AC wire is negative then the bottom cap gets charged and when the other AC wire is positive then the top diode will conduct and charge the top cap.
Each will be charged to a peak of around 160V on 110V AC.
Due to the fact that this is half-wave recification, there will be a much larger ripple on each cap because they need to sustain the current through the full cycle instead of being recharged twice each cycle. But it is a cheap and simple way to run a 240V AC design on 110V AC.

[NeilP]

Ahh that is a clever way to do it..a bit rough and ready but cheap and effective. thanks for explaining

[cor]

Wait - I have figured out that the second diode in D100 is not useless.... Because there is a current loop when the transformer is supplying 48V and the first diode is conducting, this current also runs through L100. So, as soon as the PWM pulse ends and the first diode blocks, the voltage on the +48V line stays at the same level due to the caps, but there is still a large current flowing through L100, so L100 will jump its input voltage (connected to the bottom of the secondary winding) to allow the current to flow through the second diode and also supply current into the caps. But because the reverse voltage on L100 is now very large, this state will only last a fraction of the time between the PWM pulses and the current will reduce to zero quickly after which the voltage across L100 will fall to zero again, ready for the next PWM pulse from the transformer to push the first diode open and supply current to the caps.

This also means that if you reduce the output voltage to less than half the maximum output voltage, then at maximum current output there are weird switching events that might not be appreciated by the output diodes and by the meanwell in general. Because when output DC voltage is lower than the peak PWM voltage that the transformer supplies, then the current through the L100 does not reduce to zero (discontinuous operation) but rather, the transformer starts supply current while there is still current flowing through L100 (continuous operation). This leads to interesting switching requirements, because the transformer will start to generate voltage and turn the top diode on while at the same time the bottom diode is still conducting. That leads to high current spikes because the transformer is shorted for a short period of time while the second diode switches off. This can lead to spikes and stress on the primary switching FET that was not tested or designed and the switching MOSFET can fail from this voltage.
It is even possible that other supply voltages that are derived from the transformer change in value, go very high due to high spikes and that might be a reason that components (diode, FET and even the PWM controller) have failed while modding the supply.

I was triggered to see this buck-conversion behavior of the L100/D100 combination when I reviewed the ML4800 PFC/PWM controller reference schematic. There it has the inductor in the usual spot, between diodes and + output, but it also shows a diode between the bottom of the secondary winding and the inductor and I suddenly realized that the location of the inductor in the loop does not matter.

[NeilP]

Hell, that is so logical thinking..I can see where you are coming from with the explanation if I sit there staring a the paper, drawing current lines in in pencil.

While your brain is still in high gear, something else to ponder over

One issue that has always appeared when using any of these meanwells, is the issue of parallel operation.
The more expensive ones have dedicated load balance ports to enable parallel operation.
Anyway these could be linked for equal load balancing / sharing when run in parallel?

[cor]

Hi Neil,
With the access to the voltage control on these power supplies, it is rather simple to add a circuit to allow them to share current.
The basic principle is that a supply has a current sharing input and a sharing output. If the sharing input is not connected then the supply behaves as voltage source and represents the amount of current drawn on its sharing output.
When the sharing input is connected, then the supply compares the sharing input to its own amplified current sense signal and adjusts the output voltage to match the current that is represented at its input. In other words, all connected supplies will behave as current source.
Such a board would look a lot like a mini current limiter.

[NeilP]

Maybe if there is a new run of those boards on going, then a little tweaking of the circuit before the run begins..to add that as an option, now that would be cool.

I'll..continue that line of thinking on that thread.

I have three of these boards so far, but do not use the current limiting funcition

[flanders]

hi, thanks for your work in the nes supply, I have one sitting here, has anyone actually done the current mod yet?

I think I will let some one else be the guinea pig, before I try

[NeilP]

Well Cor has the my dead one, that he is seeing if he can fix..If he gets it going maybe he will try the mods first

[cor]

Well Cor has the my dead one, that he is seeing if he can fix..If he gets it going maybe he will try the mods first

Yup - gotta fix it first, then try. THe parts that I tried until now (feedback and direct powering PWM controller) seem to work. And yes, the current mod resistor R134 is SMT (Surface Mount Technology) so you can't easily solder onto it - you either need to use thin wires or just leave the part in place (that is why I suggest to cut the trace running from this resistor) and simply solder a larger value resistor across it, from the 2.5V reference to pin 6 of U100 (or pot across the 2.5V reference and a resistor from its wiper to pin 6)
I also want to try external current limiting, so enough things to try in my scarce spare time...

[S.B.D]

Hi Cor,

What do you think about the attached current limiter circuit for the NES Power Supply?

Circuit operation:
-The current is sampled with a 1mOhm shunt resistor.
-Shunt resistor voltage is amplified by the rail2rail op amp.
-Op amp gain is set in such a way that the output reaches approximately 2.5V+0.7V when max current is reached.
-When the current reaches max value the op amp's output voltage is high enough to source current into the NPN transistor base.
-Emitter current starts flowing into the NES feedback node thus increasing the node voltage.
-The PWM controller in response to the feedback voltage rise will reduce PWM duty cycle and the output voltage will drop -> charge current will drop to max value.

The circuit's 5V supply can be taken with an LDO from the NES 12V regulator, or by using a 5V USB wall charger from Dealextreame.

NES Current Limiter.jpg

(63.64 KiB) Downloaded 3 times

S.B.D

[cor]

Hi SBD,
The 1mOhm shunt resistor is much smaller than usual, which leads to very small voltage levels.
The max NES-350-48 current of 7.3A would lead to only 7mV which means that component variation (especially DC offset in the opamp) are significant and you need at minimum a trim pot to set the input at the right level to compensate for that and choose the actual current limit, because most opamps have at least 2mV DC offset variation (may be larger over temp) so the 1mOhm shunt would make your current limit uncertain by 2A, in other words, it could limit at 5A or at 9A, which is quite a large variation. Normally shunts are 5mOhm like in Richard's design and deliver 35mV drop at 7A which is much better manageable.

Op amp gain is 600 (60k/100) so the output reaches approximately 4.2V when max current is reached (assuming 7A).

Why not drive the OVP directly from the opamp output through a diode so you can only lower the NES output voltage at current limit?
Most opamps (not all) can deliver 1 mA and the OVP point has about 1mA current flow in the NES, so you can pull hard enough with just the opamp.
In case the opamp cannot *source* current, only sink, then it only needs a resistor from the opamp output to the + supply line.

Why use 5V? The opamp can run directly from the NES 12V regulator.
With 5mOhm shunt and 7A current limit you need about 90x gain to get 3.1V from the opamp. Tie a 10k resistor from opamp output to +12V supply to make sure it can source 1mA and put a diode between opamp output and OVP point.
I did not build this circuit yet but this is how I would expect that current can be limited to 7A.
Whether it works or not, if you build it you get the guinea pig award
Note that if you look at Richard's limiter design, you will recognise a lot of this...

[S.B.D]

Hi Cor,

I got your point on increasing the shunt resistor's value. I just looked around to see what I have in my junk drawer.
This is also true for the 5V supply. I only have 5V op amps on hand...
Maybe it's time I will get some higher voltage op amps from the local store.
In the circuit the 60K resistor should have been drawn as a trimmer that sets the output current as desired.

The op amp's output would never reach 4.2V as the power supply feedback would lower the PS output voltage and decrease current.
The output should be steady at around 2.5V + 0.7V (2.5V - NES feedback voltage 0.7 - diode forward voltage drop) while in CC mode.

You are right that the op amp is able to source the feedback node without the NPN transistor. I don't like driving low impedance loads directly from the op amp's output.
I do have some concerns that the fan connected to the 12V regulator would make the 12V rail very noisy and a very high PSRR op amp will be needed. What do you think?
The op amp's voltage gain is 1+R2/R1 as it's a non inverting type. In lower gains this add 1 becomes more significant .
Do you think 1/2W of heat, if a 10mOhm shunt is used, is reasonable?

I will tweek around with the simulation later on today with your great suggestions.

Thanks.

[cor]

Hi SBD,
The opamp in your drawing is 12.6V
so if you are concerned, you can use a 100 Ohm resistor and 47uF cap to quiet the supply, but its your choice, 5V should also work.
I know that the output of the opamp should never rise above 2.5 + 0.7V, my calculation of the 4.2V output at 7Amps simply means that the current limit is seriously lower than 7A. Was that intentional?
1/2W (70mV at 7A) sounds very reasonable. The gain should only be about 45x with that setup.
Often it is best to use a series resistor with the trimmer, that allows better accuracy.
For example for 45x gain you could use a 10k trimmer and 180 Ohm to ground (between 1x and 55x gain), but it might be difficult to set the current accurately when you can vary the value so much, so it might be better to use a 240 Ohm to ground and 10k fixed with 2k pot in series. That allows you a variation from approx 40x to 50x so much easier to set the right value. You get the idea.
Success!

[NeilP]

Hi SBD

I can't comment on your schematic or the idea at all, but am wondering why you would want to do it when Cor has found a way to easily mod the NES via a much simpler approach. It just seems like re inventing the wheel with something more complicated..using extra external circuitry when a simple potentiometer will do the job..
Or is there something more to your board that will allow extra functionality. I am not knocking what you have done, just wondered what advantage doing something like that has over the R134 pot mod

Neil

[S.B.D]

Hi Neil,

I'm not trying to reinvent the wheel just make is spin a bit faster .

My motivation here is to find the easiest way get configurable current limiting with minimal tapping into the PSU internals.
If I use a 5V external regulator driven from the PSU output voltage I think I can get current limiting with only one wire going to the OVP pad (very easy task for all DIY enthusiasts like me).

I'm willing to take on the task of the PCB layout and I will be more than happy to publish the Gerber files and BOM for all to use.

I added another op amp in a comparator configuration that will show when the current falls below 3% of MAX current to indicate end of charge.
Here are the current schematics:

Current Limiter w End of Charge LED.png

(47.38 KiB) Downloaded 3 times

S.B.D

[NeilP]

Hi Neil,

I'm not trying to reinvent the wheel just make is spin a bit faster .

My motivation here is to find the easiest way get configurable current limiting with minimal tapping into the PSU internals.

OK, yes I see why you are trying to do it, and see where you are coming from with less tapping to the PSU internals, but to my mind at least, having to build and test a board, is greater complexity and less easy than just soldering two wires to the board, and maybe cutting one trace. Building up a complete board is far more work and more difficult for a complete novice. Doing one of these boards is good for someone with a bit of electronics construction experience, but I think for a complete novice, the other method is fair less likely to introduce errors ..there are two wire to go to the board, soldering to passive components, no worries about static sensitive devices, overheating IC's, transistors etc. If these boards wer to come down ready populated, and then the only work was to solder the one wire to the OVP point, then yes, even easier, but for someone new to this, having to build th board is a step to far, where jsut fitting two wires is do-able

Don't get me wrong, I use the 3 pot version of the Fetcher/Goodrum board, but for the current limiting, still mod the PSU board too, using the external board for functions that the PSU does not (yet ) have mods for. Those extra functions being, current monitoring and charge 'shutdown' once the current drops below pre set level

[cor]

Hi Neil,
I'm not trying to reinvent the wheel just make is spin a bit faster .
My motivation here is to find the easiest way get configurable current limiting with minimal tapping into the PSU internals.
If I use a 5V external regulator driven from the PSU output voltage I think I can get current limiting with only one wire going to the OVP pad (very easy task for all DIY enthusiasts like me).

Hi SBD,
But you *did* re-invent most of the 3-pot current limiter, only configured the parts a little different, but the basic operation is the same.
The only thing missing is a switch to "float" once the second opamp detects that current has fallen to somewhere in the 1A or below region, whatever set with the pot represented by R9 and R10.
One word of warning: most regulators have a 35V absolute max rating, so you can't use them on a 48V supply, for that reason Richard added a zener when you are running from a 36V or higher supply, to drop some voltage away in the zener before reaching the regulator.
BTW, there is a symbol for a potmeter that looks like a resistor with a tap from the side, that will allow you to avoid the double resistor and explanatory text around R7 and R9+R10. Note that R7 says 10 Ohm at this moment, your comment has the correct value of 10k. (or 5k will also work).

[S.B.D]

Hi,

I don't think I ever mentioned I invented anything and have never seen the circuit you are talking about. I can just be proud of the fact that my circuit is very similar as you say ( I am not sure how many ways you can go with external current limiting). My hobby is electronics and I build as much things on my own as time permits ( ebike speed controllers and rc plane electronics). I love doing things hands on. My first circuit too many years ago was an inverter driving an LED. This has probably been done before but I still was very much proud of myself and learned a lot from it.

Going back to the circuit.... I could not find a potentiometer on the LTspice sim.
What do you propose to float the battery? A MOSFET or a relay? I hate relays...

Thanks
S.B.D