It works, but how?  There is something I am missing - Page 1
Forum section: electronics / beginners

author: Rick Law
author_rank: Super Contributor
timestamp: March 31, 2013, 02:10:45 am
content: I order a cheap CC/CV board to charge my phone battery. It worked for a few minutes but it blew out; but that is not the point of this post. I am trying to learn something here.I often try to learn by seeing how others do things. So, I traced the PCB to see how things are done. Attached is the schematic I managed to trace. Looking at the schematic, I cannot see how it possibly limits current.The 78L05 looks like it is responsible for the CC part. I can see that if Iadjusttrimpot 2, the virtual ground created by the OpAmp will change the feedback voltage to the LM2596 thus changing the LM2596’s output voltage. That I can understand. What I don’t see is: whencurrent flow increaseddue to load change,how would the OpAmp or the 78L05know how much current is flowing out of the LM2596 and change the virtual ground to pull the voltage down. Trimpot 1 looks like the only connection to/from the CC subsystem. The current through Trimpot 1 (V-adj) should be a pure function of the voltage out of the LM2596 regardless of the current supplied to the load.So, I am missing something. There is something there that I don’t see or don’t understand.Unless some components are stacked and hidden under another component, all the components are accounted for in my traced schematic; so I am not missing any components that I failed to trace. Just prior to the Output soldering point, the trace on the PCB for the output did a zigzag from the capacitor to the output soldering point. That makes me think that zigzag may be serving as a shunt. But I see no components there. So, I am missing how the 78L05 or the OpAmp detects when current increased beyond limit and pull the voltage down. Before the board blew, I know it worked. When I connected it to a small resistor, the Max-Current LED came on and the voltage came down. The higher the current I attempt to overdrawn, the brighter the LED. So it does work. Something feeds back to the CC subsystem to pull the voltage down. I just don’t see how.Can someone spot something I am not seeing? How would the current flow get feedback into the CC subsystem? What could I be missing or not understanding here.I am considering de-solder things to see if a component could possibly be underneath another component. But I need to return the board to get a working one. Failing in 10 minutes is as bad as DOA. Before I do that, I want to first see if someone spots something that can help me understand.Thanks for any suggestion.Rick

author: c4757p
author_rank: Super Contributor
timestamp: March 31, 2013, 03:31:21 am
content: I am 99.999% certain something is missing there. The entire circuit around that op amp/comparator is bizarre. Even ignoring the fact that there is no visible sense mechanism, the comparator is set to activate for a negative input - but there's no source of a negative voltage in that circuit.Edit: Not a comparator, it's an op amp configured as an integrator, if you drew the schematic right. What the hell it's integrating is beyond me, though. Looks like it's just integrating a constant DC, which makes absolutely no sense.

author: peter.mitchell
author_rank: Super Contributor
timestamp: March 31, 2013, 04:08:21 am
content: Thin zigzag traces are often used as fuses.

author: Rick Law
author_rank: Super Contributor
timestamp: March 31, 2013, 05:12:16 am
content: Quote from: c4757p on March 31, 2013, 03:31:21 amThanks - Good to confirm that an experienced hand like you also think there is something missing. I thought there may be something I grossly misunderstood about how the stuff works.I think I got it - perhaps:The non-inverting input connected to a trace that went under the chip. It may not be a clean pass-through to the other side of the board which is ground there. I surmised that it was ground by measuring the resistance. Given how close to the start of the ZigZag (that ends at the output solder join output and I hypothesize it being a shunt), that may be that the feedback mechanism.So,perhaps your first thought is right - it is a comparator, the non-inverting input of the Op Amp could be the voltage of the shunt to the constant voltage from the 78L05 verses the voltage across the shunt.See the modified schematics with the shunt in RED. That could have worked? Don't you think?(Perhaps only the capacitor is tied to the shunt, the diode directly grounded and the LM2569 may be direct-ground)Oh... and... there must be different configurations for this board that the Op Amp could have integrated. R1 is not installed. R1 is next to another two un-labelled open spaces with component holes but no component. R1 would have connected to the OpAmp which would have given it something to integrate.I have to decide now, should I keep this board for return, or should I totally strip it just to learn and confirm.Thanks for your input- it is educational and thought provoking!Rick

author: Rick Law
author_rank: Super Contributor
timestamp: March 31, 2013, 05:12:58 am
content: Quote from: peter.mitchell on March 31, 2013, 04:08:21 amThanks! That is something I didn't know!(Thanks for the knowledge - always glad to learn)Rick

author: c4757p
author_rank: Super Contributor
timestamp: March 31, 2013, 05:31:26 am
content: Yes, that could work, though it's a bit unusual. As long as the return current goes through that sense resistor, the op amp output will go up when the voltage drop in the resistor exceeds the set threshold. It will then push the feedback pin on the 2596 high and decrease the output voltage.I don't really like it because it is a low-side current sense - if you use this device to powerpartof a circuit rather than the whole thing, that sense resistor will be separating the grounds.Quote from: Rick Law on March 31, 2013, 05:12:16 am1) Convince them I'm an experienced EE.2) ??3)Take over the wor-er... I mean...Profit!. Yeah, "profit". Not "take over the world", why would anyone want to do that?

author: Rick Law
author_rank: Super Contributor
timestamp: March 31, 2013, 05:54:39 am
content: Yeah, the output ground is the other end of the ZigZag, and the capactor is near the start of the ZigZag. All the output current must go (return) via the zigzag.I feel good understanding it. Seeing no feedback mechanism at first really bugged me.Thanks again!Rick

author: c4757p
author_rank: Super Contributor
timestamp: March 31, 2013, 05:56:46 am
content: Quote from: Rick Law on March 31, 2013, 05:54:39 amIt's a good feeling, isn't it? Eureka!

author: c4757p
author_rank: Super Contributor
timestamp: March 31, 2013, 06:01:36 am
content: One quick, kind of off-topic thing I noticed and thought I'd mention - your capacitors are all upside down in that schematic. The curve is the negative side.

author: Rick Law
author_rank: Super Contributor
timestamp: March 31, 2013, 11:28:17 pm
content: Quote from: c4757p on March 31, 2013, 06:01:36 amAh ha, little things one forgets! Those capacitors must be getting rather dizzy by now being upside down all these time! Thanks for letting me know! I have been away from electronics for too long.At least my gut feel that the zigzag has something to do with it was right!This evening, I took an extra careful look at the return-ground soldering point: Besides being the end of the zigzag PCB trace, there is a tiny trace that is partially covered by the capacitor socket which goes under the OpAmp -and the OpAmp non-inverted input trace went under the OpAmp also. My initial assumption that OpAmp+ directly to ground (based on resistance) was wrong. A 30-40cm long by 2mm wide zigzag PCB trace must give some kind of resistance but too small for the UT-61E to tell reliably. So indeed that resistance of the zigzag (between return-ground and the board's ground) was the current sensing mechanism which feeds the OpAmp which in turn controls the reference current for the LM2596 when the Voltage Comparator triggers.Your comment points me in the right direction. Not seeing any components or ways it did feedback, I was confused and thought perhaps I misunderstood something silly like perhaps how a voltage divider works. Learning is always rewarding. Thanks for your helpful comment.Ahem, on to the next puzzle.Thanks again!Rick