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Help me understand a 12V SLA charging circuit.

Buk

Jan 30, 2018
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Warning: I'm just trying to get to grips with electronics; and I'm an old dog, so its hard. :)
(Let me preempt some responses by saying that I am aware that simulators -- and especially the simple, free one I'm using -- are imperfect representations of reality.)

I've modeled the charge control circuit board from a working, 12V SLA(on its last legs) powered portable vacuum cleaner.
You can view the model here.

Here is an annotated close up of the circuit board in question.circuits.jpg
I have a bunch of what are quite possibly dumb questions about the circuit, and I hope that I can get some (appropriately simple :) ) explanations to them.

1) The "hottest" (according to the simulation) component on the board is the 22Ω R18 resistor. Why is this so low?

In the model you'll see I replaced this with a 22k value and everything still works (ie.Q4 still goes into saturation), but the resistor consumes 99% less power.

2) I'm puzzled by the use of the SCRs. I know they can be used as "latches" amongst other uses, but that does not appear to be the case here. And I think they can be used to provide low power control of high(er) power circuits, but that also does not seem to be necessary here. For example:

a) I can remove SCR T3 entirely from the simulation (with or without strapping r15/r16 to ground or c3) and the entire circuit still works. Q4 still goes into saturation and the relay still operates in concert with the on/off switch.

b) Again, I can completely remove T1 from the circuit and everything (all 4 combinations of switch position and with the battery fully charged, and depleted) continues to work. As best I can tell, exactly as it does when T1 is in place.

3) I perplexed by the need for T2 (in conjunction with Q1/Q2) to drive the green LED (actually the green half of a 3-lead bi-color LED), but no equivalent SCR for the red LED?

4) I'm confused by the purpose of C3.

C1 appears to suppress surge when the charger is plugged in (simulated with a switch).
C2 may serve a similar purpose for when the main switch is thrown.

but C3 doesn't appear to carry any current or voltage worthy of note; picoAmps and microVolts only, and no spikes at all from what I see.

Thanks for any time you can spend educating this old dog.

Buk
 

kellys_eye

Jun 25, 2010
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There is so much wrong with that schematic I don't know where to start!

Perhaps 'you' could start by posting the schematic 'as is' instead of with those distracting moving dots?
 

WHONOES

May 20, 2017
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I'm with Kellys-eye. Post a schematic without all extra ephemera as the moving dots make it very difficult to read. Also, give us some idea of what you are trying to achieve, just understand the circuit or, make a proper SLA battery charger!
 

Buk

Jan 30, 2018
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There is so much wrong with that schematic I don't know where to start!

Perhaps 'you' could start by posting the schematic 'as is' instead of with those distracting moving dots?

That isn't a schematic, but rather a full active simulation. Mouse over the components and it will tell you the voltages and currents flowing through them. Click the switches and they will operate, and the current flows will change accordingly.

But, if you press STOP on the simulator, the dots stop moving; press RESET and the residual dots disappear.

Edit: However:schematic.jpg

Edit 2: It was pointed out to me that I had Q1 (SS9012) as an NPN when it should be an PNP. The updated model is here.

I've updated the schematic above also.

I haven't yet explored if my questions above still make sense after the change, but I don't appear to be able to edit that post even if the do not?
 
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kellys_eye

Jun 25, 2010
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There's no supply for the relay. The diode across the relay is reversed. R15/16/18 are just....wrong...... and that's just for starters.
 

Buk

Jan 30, 2018
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There's no supply for the relay.

Hm. The supply for the relay is labelled Sout1. When the switch from the battery is moved to on, it energises the coil and the battery feeds directly to the motor.

Update:

You are right about the diode being reversed.

I do not understand the comment about R15/R16/R18?

Do you mean they are bad choices for the circuit (which has worked for 10+years); or I've made a mistake transcribing themto the schematic/simulation? Or?

Here are close up pictures of the circuit board front and back (though I've image flipped the back and overlayed the components.)BoardBothSides.jpg
 
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kellys_eye

Jun 25, 2010
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Hm. The supply for the relay is labelled Sout1. When the switch from the battery is moved to on, it energises the coil and the battery feeds directly to the motor.
Unclear from the schematic as drawn but if this is what you say it is.......

The position(s) of the resistors are not as per the schematic.

R18 connects from Sout2 to the cathode of D4 (to begin with) - there are other errors.
 

Buk

Jan 30, 2018
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The position(s) of the resistors are not as per the schematic.

R18 connects from Sout2 to the cathode of D4 (to begin with) - there are other errors.

Sorry for wasting your time.

My only excuse is the tool I am using is not very good! (Poor workman always blames his tools. :) ) But it was the first thing I found that was free and had a good selection of component types.

I originally drew the circuit layout exactly as on the board, but with all the crossovers it was impossible to follow, so I then tried to untangle it.

Unfortunately, the sim only allows the moving of individual components, not nodes, and I must have screwed the pooch in the process. I did check the sim still "worked" after each change and undid/ redid if it didn't, but that hasn't prevented errors creeping in.

Excuses over, to my question: Is there are better, free tool that allows the dragging of connected nodes rather than individual components?
 

kellys_eye

Jun 25, 2010
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Sorry for wasting your time.
Not a problem.

Reverse-engineering a circuit board is never straight forward however if you get it right then the operation and explanation for the circuit operation becomes relatively straight forward.

Sorry, can't help with the simulation software issue - never use them.
 

Hopup

Jul 5, 2015
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Multisim has trial version I remember, it should have that feature.
 

Buk

Jan 30, 2018
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e
Not a problem.

Reverse-engineering a circuit board is never straight forward however if you get it right then the operation and explanation for the circuit operation becomes relatively straight forward.

Okay. These are the errors I found:schematicErrors.jpg

This is the (I sorely hope) correct schematic:schematic.jpg

And this is a link to the simulation for anyone interested.

Still my original 4 questions remain:

1) Why is R18 only 22ohms, when 22k seems to work just as well (Q4 saturates), disipates less heat, and the motor runs?

2) Why does it need Q1 and T2 to drive the green half of the LED, but n active components required to run the red half?

3) What purpose does C3 serve.

4) What purpose does T1 serve?

Thanks, Buk.
 
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kellys_eye

Jun 25, 2010
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Still not right.

R18 goes in series with D4. There is no connection from D4 to the base of Q4.

The 22 ohm (R18) is the correct value and simply limits the current in the relay coil.

Although your schematic is 'close' it's not close enough - small errors make huge differences. Getting the final schematic drawn in 'conventional' form is also good practice - it reveals 'standard' structures which help in circuit understanding.
 

Buk

Jan 30, 2018
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Still not right.
R18 goes in series with D4. There is no connection from D4 to the base of Q4.

Thanks.

The 22 ohm (R18) is the correct value and simply limits the current in the relay coil.

Why? What is, what makes it "the right value"?

Although your schematic is 'close' it's not close enough - small errors make huge differences.

My errors are neither deliberate; nor would I include them, if I had seen them.

As I identified in the OP, this stuff is all new to me, and I am finding it frustratingly tough to get right.

Getting the final schematic drawn in 'conventional' form is also good practice - it reveals 'standard' structures which help in circuit understanding.

I have no idea what "conventional form" means? Nor how I would go about finding out; or converting what I have to that form.

Thank you for your time, but education by a thousand cuts may be fun for the educator, but frustrating as hell for the educatee.

I'll leave you in peace now; this place is obviously not conducive to answering the level of questions I have to ask.

Cheers, Buk.
 

kellys_eye

Jun 25, 2010
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Whilst we are fully prepared to assist with your questions you must appreciate that we can't be expected to do the work FOR you.

Yes, I could reverse-engineer the board completely and draught a schematic in conventional form but what does it do for ME? Other than take me hours of my time? Without doing this it is not possible to describe the component operation properly and we'd just be leading you astray with the description.

We have pointed out obvious errors and offered advice that allows you to progress your query but if you think this is a place where you can come to have everything done for you you're sadly wrong - and good luck finding a site that WILL.

Had you come here with a 'proper' schematic your answers would have been forthcoming without delay but you now seem 'upset' that we aren't here to do your bidding without YOU doing some work too.

Sorry you feel this way.
 

Buk

Jan 30, 2018
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Thanks to your first, borish, negative reply; there is no WE, only YOU.
Apparently simulators are beneath you; despite that the very reason for my presenting the information I have in that way was to make it trivial for responders to correct any mistakes they spotted.

Had you come here with a 'proper' schematic

If I knew how to produce a "proper schematic" in "conventional form", I probably wouldn't need to come here and ask for help.

And given that thus far you provided no help; but rather chided like an overbearing school marm; I've given up on getting any actual useful assistance from you -- and you've ensured I won't get it from anyone else here -- so I am no longer prepared to boost your over-inflated ego further by dancing the newbie jig to your command.

your answers would have been forthcoming without delay but you now seem 'upset' that we aren't here to do your bidding without YOU doing some work too.

I seriously doubt you know the answers, your comments thus far betray your lack of understanding. Having a 22 Ohm resistor burn 6.5W of power supplying current to a relay controlling a motor that dissipates 100W is asinine.

As in, stupid, foolish, ludicrous, nonsensical; but according to you "the right value".

Sorry you feel this way.

No you're not. You're only sorry that I'm cutting your newbie baiting short.

You have your head so far up your own arse -- "I don't use simulators" (like they were some kind of illegal narcotic -- that I doubt your 2456 posts have helped anyone.

As for finding fora where they actually want to help -- already found. Questions pretty much answered -- though I don't fully understand them yet --, my simulation corrected by the first responder. 5 small changes.

And an ongoing conversation about the circuit, instead of the reluctant drip-fed nudges and superior attitude that are all you've allowed to slip from your grasp.

One wonders why people like you ever bother to respond on forums; when it is apparently so taxing of your time and provides you with no reward?[/QUOTE]
 

kellys_eye

Jun 25, 2010
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Having a 22 Ohm resistor burn 6.5W of power supplying current to a relay controlling a motor that dissipates 100W is asinine.
The relay and resistor don't carry ANY of the motors current - the only current flowing is that dissipated by the relay coil (and the series resistor) when Q4 is saturated.

The circuit board you posted an image of clearly shows a 22Ω resistor and it doesn't look to be burned out - ever wondered why?

If you'd have reverse engineered the circuit CORRECTLY you would see.... perhaps.

The datasheet for the relay shows that it has a nominal coil voltage of 9V - and is being powered by 12V. Where is the other 3V going?

Through a 22Ω resistor...... which will dissipate 0.4 watts (probably less since I haven't taken Q4 Vce(sat) into account which will actually lower the voltage across R18 still further.

But since your view of the world is through your own narrow lenses you go right on believing that it's carrying 6.5W. After all - that's what the simulator is showing so it MUST be correct......????
 

Buk

Jan 30, 2018
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The relay and resistor don't carry ANY of the motors current - the only current flowing is that dissipated by the relay coil (and the series resistor) when Q4 is saturated.

Which is exactly why I wrote (perhaps a little highlighting will help you):
Having a 22 Ω resistor burn 6.5W of power supplying current to a relay controlling a motor that dissipates 100W is asinine.

So now I'm doubting your ability to read English as well as your electonic knowledge.

The circuit board you posted an image of clearly shows a 22O resistor

Firstly, if R18 is in fact a 220 Ω resistor, not a 22 Ω as I have it, then what possible motive do you have for witholding that salient piece of information? (Answer below.)

BUT..., perhaps my image was fuzzy, or your eyes are as bad as mine, so maybe this will help:

R18(400%).jpg

and perhaps you've forgotten this:https://www.xda-developers.com/wp-content/uploads/2015/09/Resistor_color_code_chart.png

and it doesn't look to be burned out - ever wondered why?
I also have the circuit, have measured the resistance directly:

P1110826.JPG

and I can see the faint scorch mark on the board under the resistor, but I can't capture it.

But you'll probably claim that my hobbyist meter isn't accurate enough, that you can't get a proper reading with the component in-situ and I should have used a 4-wire tester to be sure.

Meanwhile, I think most people would agree that R18 IS a 22 Ω resitor.

If you'd have reverse engineered the circuit CORRECTLY you would see.... perhaps.

*I'll get back to that piece of tautology!

The datasheet for the relay shows that it has a nominal coil voltage of 9V - and is being powered by 12V. Where is the other 3V going?

The datasheet shows SRD-S-109DM.jpg

That it is designed to run with coil voltages from 3V to 60V.

And when run with 12V, the coil has a resistance of 400Ω and draws 30mA.

All of which shows that the following "words of wisdom":
Through a 22O resistor...... which will dissipate 0.4 watts (probably less since I haven't taken Q4 Vce(sat) into account which will actually lower the voltage across R18 still further.

are the utter garbage I now take everything you say to be.

When Q4 is activated, the coil current flows from Sout2, through R18 (22Ω), through the coil (400Ω), through the collector-emitter junction of Q4(collector-emitter saturation voltage drop 0.16V) to ground.

At 12V nominal, that 11.84V across R18 and the relay coil.

VR18 = Vboth * 22/422 = 0.617V
Vcoil = Vboth * 400/422 = 11.22V

Ir18 = 0.617V / 22Ω = 28.06 mA.
Icoil = 11.22V / 400Ω = 28.06 mA.

Quickly and cleanly explained; and giving me a clear answer to my first question.

"Oh. Oh. Oh. But you were wrong about the current in R18." you'll say.

Yes. But as I identified in my OP:
buk" said:
possibly dumb questions
that is exactly why I came here seeking help!

Help which you steadfastly refused to give. On the basis of the little you have said, mostly since I gave up on you and this place and went elsewhere, probably because you were incapable of giving -- but still you chose to taunt-the-pre-admitted-low-knowledge--newbie with arrogant insistence that I provide "proper schematic constructed in 'conventional form'"; knowing full well that I would not be able to provide that!

And still are:
If you'd have reverse engineered the circuit CORRECTLY you would see

* Lost motorist to Kelly's Eye: "I'm trying to get to {some destination}; can you help"?
Kelly's Eye replies: "Sure, construct an accurate, 1:25000 scale map that covers here and your destination, and I'll be happy to show you".

So, thanks for nothing! Worse, for wasting 2 days of my time playing your puerile, egocentric game.

I hope you -- and the other denizens of this place -- are proud of your contribution to society.
 
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kellys_eye

Jun 25, 2010
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Firstly, if R18 is in fact a 220 Ω resistor, not a 22 Ω as I have it, then what possible motive do you have for witholding that salient piece of information?

Check your eyesight. My quote said 22Ω <----- that's the symbol for OHMS, not a zero.

That it is designed to run with coil voltages from 3V to 60V.

And when run with 12V, the coil has a resistance of 400Ω and draws 30mA.

All of which shows that the following "words of wisdom":
Wow!

The datasheet shows that the relays are AVAILABLE with coil voltages ranging from 3 to 60V and that the one on the board you have is coded as having a 9V coil.

Similarly the relay coil (for the 9V relay fitted) has a resistance of 225Ω. Along with the 22Ω resistor makes a series total of 247Ω which, if Vce(sat) was truly zero, gives a current in the resistor of around 50mA which gives a power dissipation in R18 of around 50mW - - given the resistor is rated at 250mW this is well within its operating range hence it w(sh)ouldn't be burning.

"Oh. Oh. Oh. But you were wrong about the current in R18." you'll say.
LOL
VR18 = Vboth * 22/422 = 0.617V
Vcoil = Vboth * 400/422 = 11.22V

Ir18 = 0.617V / 22Ω = 28.06 mA.
Icoil = 11.22V / 400Ω = 28.06 mA.

Quickly and cleanly explained; and giving me a clear answer to my first question.

Really?

I don't know where else you went for your advice but I suggest you try getting another opinion from elsewhere. Just to be sure...... know what I mean ;)
 
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Buk

Jan 30, 2018
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Check your eyesight. My quote said 22Ω <----- that's the symbol for OHMS, not a zero.

You may be able to modify your post, but you cannot modify the direct quote from it embedded in my post above; which clearly shows: ProofHesALiar.jpg

that you're a liar as well as a dick!

With luck this sorry story of your dickheadery will serve as a warning to other hapless souls that coming thsi way looking for help.

Buk.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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That it is designed to run with coil voltages from 3V to 60V.

No, it shows that there are 10 different coil voltages available, between 3V and 60V.
 
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