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Hi Guys need help with a charger circuit please

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Dr Bob

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I hope this isn't too big an ask, The charger isn't actually faulty yet but I am trying to get an understanding of it for when it does. I have attached the circuit diagram and I need a clever electronics expert to give me an idea of what the different parts of the circuit do? I have a few ideas but some of it doesn't make sense, like if you look at the part bottom right consisting of IC2 C and D, because the power fet is an "N" fet you would expect pin 14 to go high when it is charging, but measuring it shows it goes down to about a volt. and up to 8.5 volts when it's not charging.

I thought maybe the two op amps make an ocillator to drive the fet, but the scope shows no waveform at pin 14 when it's charging? I am well confused

IC 2 A and B seem to be mainly concerned with controlling the dual colored led to indicate charging or not

Control of to charge or not seems to be done with IC1 A, B and C making a comparator and comparing the voltage on pin 2 of the power fet with the precision reg which measures 3 volts on pin 6.
Measurement shows pin 8 is 8.5 volts when charging, less than a volt when not.

I think IC 1D is mainly for thermal overload protection?

What do you guys think please?
 

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Welcome to the forum.

That made my head hurt at this time on a sunday morning.

I dont profess to be an expert at anything.
What makes you think the charger will go down?, if anything the fet or the bridge will be the only likely things to fail, and even thats very unlikely.
 
Hi mate, probably won't but I have been staring at this circuit on and off for some months and while I have figured out parts of it, other parts have me a bit confused and I have to know. It's like trying to remember that actors name or similar
 
if you look at the part bottom right consisting of IC2 C and D, because the power fet is an "N" fet you would expect pin 14 to go high when it is charging
I agree. Providing the charge current doesn't exceed the limit set by S1/IC2d. Are you sure you were measuring IC2d and not IC1d?
I thought maybe the two op amps make an ocillator
No. They just monitor FET current via the source resistor and pin 14 should pull the gate voltage down if the limit is exceeded.
IC 2 A and B seem to be mainly concerned with controlling the dual colored led
Agreed.
Control of to charge or not seems to be done with IC1 A, B and C
Agreed.
pin 8 is 8.5 volts when charging, less than a volt when not.
Looks right.
I think IC 1D is mainly for thermal overload protection?
Agreed.
 
Hi Alec, thanks for your help mate. Excellent, you have shifted the mental block and I can see it now. First of all, I re checked the vlotages at pin 14 of IC2D and it does appear to go lower when charging, but after what you said it's clear that it would.

When switched on the charge current tries to go through the roof with max volts on pin 14 as current flows through the r0015 resistor a voltage is measured across it and fed to pin 10 of IC2 C this op amp has gain and the output is applied to the negative input of IC2 D which then reduces the output voltage and the charging current, Thus the voltage will appear to fall when monitored and as the battery I have connected while doing these tests is practically full I guess it reduces a lot.

Well done mate, that cleared that up. Now then the circuit with IC2A and B. I have been studying that and note it also has a diode D16 for removing the gate voltage of the fet and stopping the charge, just like the comparator of IC1 A,B abd C diode 7

So I thought there must be more to it than that, then I remembered that this charger also does 6 volt batteries, note the switch top left and over on the transformer.

The voltage on pin7 of IC2 B stays around 8.5v when charging or not charging in 12 volt mode so do you think this circuit could be a comparator for the six volt mode?
Bob
 
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Oh! if anyone decides they want to build this from the circuit note that C 7 and 8 up by the bridge reg. are not fitted in the charger I have, they just use the battery as a giant capacitor, they appear to be shown connected the wrong way round anyway
 
Now then the circuit with IC2A and B
That seems to be monitoring battery voltage by comparing the voltage on the FET drain to a reference from a voltage divider and which is changed depending on the battery being a 6V or 12V type. When a 12V battery reaches ~13.8V pin 7 should go low and pull the 'charge' LED voltage down, at the same time allowing the transistor to turn off and hence light the 'full' LED.
 
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Yes but the top comparator also does that through D8 and it seems pointless having two circuits that do the same thing, looking at the switch at the top in the 12volt position it changes the bias resistors of the transistor in the top circuit, but in the 6 volt position it changes the divider resistors in the bottom circuit, which makes me think the bottom circuit is just for the 6 volt charging.
Also in 12 volts mode we have the top circuit monitoring the drain and shutting down the charge via D7 and switching the LED's via pin 8 to r25 and d8 so why the second circuit?
Plus as I said above when charging a 12 volt batter, which is all I have by the way, measuring pin 7 of IC2B shows little change from 8.5 volts between charge and no charge. I thought it seemed logical that when all the circuit voltages drop due to the transformer being switched to 6 volt, the resistor values and the precision reg setting in the top circuit would be wrong for monitoring a six volt battery, so they made another for the six volts mode?
Both circuits are controlled by the sense line from the drain of the fet. Unfortunately without a six volt battery I can't charge one and see if the top circuits output stays high when charging or not and the bottom circuit comes into play
 
it seems pointless having two circuits that do the same thing
Agreed. The circuit looks over-complicated for what it apparently achieves, if I've understood it correctly.
but in the 6 volt position it changes the divider resistors in the bottom circuit, which makes me think the bottom circuit is just for the 6 volt charging.
If it was intended just for the 6V charge I don't see why it would need to be changed; a fixed divider could have been used.:confused:
 
I'm guessing that's something to do with preventing it from working when the unit is in the 12 volt mode and why pin 7 IC2 B stays at 8.5 volts allowing the top circuit to take control and then when the switch is moved to the 6volt mode the base bios of vs1 is changed so the top circuit is inactive in the six volt mode and the voltage reference on pin 3 IC2 A is changed to enable that one, what do you think Alec?
 
Possibly. I'd have to download the cct again to check. No time this evening.
 
Ok Alec, thanks for your time I don't normally have anyone to chat with when I am doing a bit of electronics, haven't done it full time for years.
 
I got it wrong above. I've now run a sim of the circuit. With the component values as shown, IC2a,b never change state, regardless of battery/supply voltage, so do nothing useful !!
However, if R20 is changed from 5k1 to something like 200k it all begins to make sense. On the 12V setting, IC2a,b outputs then stay low if the battery voltage is below ~13.2V, i.e. the charging will only turn 'on' when the battery voltage is >13.2. Charging will continue until the battery voltage reaches ~14.4V, at which point IC1a,b,c trip and turn the charging 'off'. These figures are consistent with the ON/OFF figures shown just below the title of the schematic (a bit of a clue there :)). Similarly consistent (more or less) figures are obtained on the 6V setting.
So, in summary, it looks like IC2 does the 'ON' voltage check and IC1 does the 'OFF'. It also seems the circuit will refuse to charge a very flat battery.
If you've got LTspice, here's the .asc file to play with. If you haven't, then I recommend you get it (it's a free download from Linear Technology and /or the Yahoo LTspice User Group).
 

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Excellent, Thanks so much for your help Alec I was getting so I couldn't see the wood for the trees I wondered what those figures meant, I thought it was referring to switch settings as it's next to it :rolleyes: That's it then, sorted. Well done indeed.
 
Just had a bit of a play based on what you said, Because you said it made no sense until you changed a resistor value, well it was obviously designed to work as it is so I tried a few tests.
I find that the charger will charge a capacitor I had a 2200Uf to hand and tried that and it charged ok and auto switched off when it reaches 14.4 ish but the point is it charged a capacitor which had no volts on it, so I am wondering if this circuit is to prevent charging mode when the leads are not connected to a battery?
So I took the voltages with the leads disconnected and I find that if I switch it on not connected to anything pin 7 is low and it cant go into charge mode, which I guess means the output circuitry is protected, even if I connect a 10uf it charges it :)
 
Because you said it made no sense until you changed a resistor value, well it was obviously designed to work as it is so I tried a few tests.
Interesting. I also changed R6 for the sim. Can you confirm your actual charger circuit resistors, particularly R6,R12,R18,R20, are exactly the same as in the schematic? You've already pointed out that the two caps on the bridge have incorrect polarity, and it's not unknown for published schematics to have deliberate errors to prevent/discourage copying.

Edit: The 2 bridge caps are also shown as being in series (?) and are across the wrong bridge diagonal.
 
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Hi Alec R6 is hiding under the black wire and is 1k6 R18 = 20k R20 = 5k1 R12= 10k had to remove some hot glue to find it you won't see it in the pictures, so it seems only r6 is different surely that would only affect the top circuit?

Yes I noticed that about the caps, wouldn't smooth much without one end on the deck would they :)

DSC03869.JPG
 

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R6 is hiding under the black wire and is 1k6
Ah, my suspicions were well-founded. That would give a total collector load (with the pot at max) of 2.10k. In my sim I needed a total of 2.12k. A pretty good match!! I agree it should only affect the top circuit.
But if R20 is only 5k1 then (according to a revised sim) the bottom circuit (IC2a,b) turns charging on via the FET if the battery voltage is > 2.6V. But even with the FET off there is a current path via R18,R20 to ground. That would account for how you were able to charge a cap from 0V.
About the caps, we've both been assuming they're the main smoothing ones. Possibly the circuit is intended to charge the battery with unsmoothed 100Hz pulses? In which case the caps as shown back-to-back across the AC input to the bridge may be there for spike suppression? Seems an odd arrangement though, and the ripple current through them would be peaking at ~600mA for 220uF caps. Can you trace whether the caps are indeed across the AC side of the bridge?
 
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Good morning Alec, Nice one mate :) Yes if you look at the photo of the full board you can see on the right the caps don't exist, what isn't obvious in the same picture is top right where there is a cable tie around the black red and white cables, just to the right of the cable tie, that is the bridge rec, circuit board mounted and bolted to the heatsink. They are nowhere else and I think they just use the battery as a huge capacitor and indeed if you measure the probes you get about 11 volts and then if you put a cap across them it goes up to about 17

Edit: Few more pictures, I moved some hot glue and wires, R12 can be seen now and R6 can be seen next to the green cap top right of the chip.

DSC03881.JPGDSC03884.JPGDSC03885.JPGDSC03887.JPGDSC03888.JPG

Strange? I deleted two that were out of focus and the forum shows them as attached thumbnails?
 

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So, there are a few confirmed discrepancies between your actual circuit and its schematic. Doesn't make the circuit analysis easy :D.
 
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