Meanwell NES Schematic..almost complete

[mxer]

This is a very interesting read, Cor and Neilp you have done a great job, But i have to confess i have no clue what's going on :? :lol: I'm ok with shaving shunt's etc but i think this goes beyond,
I'm very shortly investing in a Alloy shell 49.9 6ah £50 delivered to the UK, Shall i wait a bit to see if the NES is a cheaper solution or is it going to be beyond me :?

Great stuff guy's 8)

 

[NeilP]

Mxer..I'd be interested to see what the inside of your alloy shell charge looks like when you get it..I have just ordered an alloy shell one today, about 20 mins ago, set for 142 volt and 14 amp..the 2kW one..


if you are just going for relatively low power charger..of 49 volt and 6 amp..the Meanwell solution would be best for you, and there are plenty peple here that will walk and talk you through converting it..me included..PM me if you want and I'll send my phne number to you ..or Skype..I am in UK ( well Jersey) so it is not going to be an expensive transatlantic call

 

[mxer]

That's great, thank you neil

I like the idea of the Alloy shell and limited amp's, Look's like the NES can run away with it if not done properly ?

I only have (atm) 5ah 44.4v to charge looking to increase to 10ah soon, If the NES is not too much grief to control, i'd love to give it ago But i think 5ah is a little to small for it hence the Alloy shell option even that's over 1c ..

Might order one at the same time anyway (just to have a play with my new Turnigy watt meter) there cheap enough

Thanks again for your kind offer

 

[Samson]

Great job tracing and publishing this circuit!

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.

The center tap should go to ground to complete the circuit.

My two cents.

 

[cor]

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

The short answer is - I marked it on my sketch
Look at the voltage feedback divider with the potmeter SVR1 and then R124 connecting to R122 and to U100 pin 3. That is the OVP marked next to the LED. This has nothing to do with the Overvoltage Protection, but this is the Output Voltage divider - the output voltage will be whatever causes the OVP point to be 2.5V.
With the 1k pot at approx 430 Ohms you will get 48V at the output, because the divider is 44k2 and 2k+430. With the potmeter at zero Ohms the divider is 44k2 and 2k, so the output becomes (44.2+2)/2 times 2.5V equals 57.75V and with the pot turned to the max 1k value the divider becomes (44.2+3)/3 times 2.5V equals 39.3V.
You probably can go lower in output by increasing the value of SVR1 or feeding in an external voltage, but at some point the power supply itself is so starved for power to give to its internal circuits (which is also tapped from the same transformer) that it will simply shutdown. You may avoid the power supply shutdown by adding resistors across its output, just to bleed power away and keep the internal circuits alive. See the 5k6 resistor next to the LED at the output of the supply? That resistor has two functions - bleed the output caps down so you don't get a big spark when you connect something to the output after powering down the supply *and* to give a minimum power load to the supply to allow it to generate enough power for its internal circuits.
NOTE that this also means that the power supply should not be left connected to your batteries for months after you switched it off, because eventually that approx 10 mA bleeder current can discharge your batteries, a 5Ah pack will need approx 500 hours to be drawn down, which is around 3 weeks.

SBD..check the schematic on first post of the thread..The relay chops power to pretty much the whole board.

While it is true that the relay disconnects the AC line from the rectifier, there is still the NTC RTH1 that will still feed the power supply. The supply is not switching off due to the relay dropping out, but the relay is dropping out due to the power supply shutting down.
Note that the power supply is feeding a battery pack and the battery continues to feed the same voltage back to the power supply, whether it runs or not. That means that the voltage feedback (the OVP point that we discussed in this same post) will pretty much stay at 2.5V and the result is that the opto U2 will continue to conduct to feed back that the output voltage is regulating and around the correct voltage level.
Now look at the circuit for the COMP (pin 1) of U1. The opto pulls the COMP pin to ground and the spec of U1 says that if the COMP pin is below approx 1.2V (two diode drops to ground) then U1 will go into shutdown.
I believe that due to the 48V output voltage staying at the correct level and the output current being so low, there is no power to the internal circuits such as Q5 which regulates the power to U1 and the relay. With that power gone, the relay clicks off and U1 shuts down also (it requires a minimum Vcc voltage of around 8V, see "undervoltage lockout" of the TL3845P) Of course U1 will try to restart because it is fed through the 7k5 and 20k resistors directly off the AC line, but I believe the COMP pin will keep U1 in shutdown (will need to check once I have this NES going again) and the relay can't engage because there is a blocking diode between the Vcc pin of U1 and the output of Q5 to minimize power draw in this startup phase.

 

[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.



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!