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Battery charging circuit - can anyone explain?

by Jay_
Tags: battery, charging, circuit
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Jay_
#1
Jan11-13, 10:03 PM
P: 135
http://electronicdesign.com/site-fil.../figure_01.gif

So let me just start by saying that, I want to understand how to analyze circuits and I would like to find a good source for that. If some one could point to online tutorials, or a website it would be great.

In the above circuit (link) for example, I get the part that we step-down the AC mains, we rectify it and get a 15 V pulsating DC.

Now, I assume capacitor C is to keep it at a constant 15V, right? Though I don't understand how a simple capacitive filter is enough for that. When does it charge, and when does it discharge in this circuit?

Next, could someone explain the role of the diode D1, and the transistor Q1. I mean like what exactly happens with them in the positive and negative half-cycles of AC mains?

The diode will be likeshort if forward biased, and like open if reverse biased. But if we have a rectified DC, so when will the diode be forward biased?

Then after I am done understanding that, we could go to more complicated circuits like this one:

http://www.circuitstoday.com/wp-cont...using-l200.JPG

If someone could elaborate/discuss these circuit (starting with the first) so that I can try and understand it, I would really appreciate. Thanks.
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dlgoff
#2
Jan12-13, 01:28 PM
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Quote Quote by Jay_ View Post
http://electronicdesign.com/site-fil.../figure_01.gif

So let me just start by saying that, I want to understand how to analyze circuits and I would like to find a good source for that. If some one could point to online tutorials, or a website it would be great.

In the above circuit (link) for example, I get the part that we step-down the AC mains, we rectify it and get a 15 V pulsating DC.

Now, I assume capacitor C is to keep it at a constant 15V, right? Though I don't understand how a simple capacitive filter is enough for that. When does it charge, and when does it discharge in this circuit?
As to the capacitor, this little applet might help.

http://www.falstad.com/circuit/e-fullrectf.html
Averagesupernova
#3
Jan12-13, 01:28 PM
P: 2,529
The transistor turns the relay coil on and off. The diode is there to protect the rest of the circuit from the voltage spike generated by the relay coil when the transistor turns off. I would guess that you didn't recognize the relay coil. That is one of the first steps in being able to understand schematics. Know what each part is.

hisham.i
#4
Jan13-13, 01:34 PM
P: 177
Battery charging circuit - can anyone explain?

Hi Jay
First the 220v ac voltage is stepped down and rectified by bridge B1, then filtered to be almost dc by using the capacitor C1.
Now, I assume capacitor C is to keep it at a constant 15V, right? Though I don't understand how a simple capacitive filter is enough for that. When does it charge, and when does it discharge in this circuit?
Of course there will be ripple in the voltage but this ripple voltage can be calculated and ignored if found very small.
The idea of this circuit is that the resistors (vr1,vr2, r1, r2 ,r3) are forming voltage divider which apply a voltage to between the base and the emitter of the transistor Q1.
When the voltage of the battery is greater than some value (assume 15) the transistor is switched on then the relay is energized so switching the relay switch from closed position to open position. Thus stop charging the battery.
You now can see that when the switch is connected to NO (normally open of the relay) the rheostat vr1 is short circuited by vbat thus making the turn on voltage of the charging state different from that of the turn off of the charging.

But i think there is a problem in the circuit which is when the relay switch is at NC then the voltage applied to the resistors is 15v which means it will switch on the relay and disconnecting the charger.
So i think it should take into consideration the charging current also, so after the voltage battery is 15v and the current is less than some value in mA then disconnect the charger.
vk6kro
#5
Jan14-13, 12:35 AM
Sci Advisor
P: 4,032
There is a clever hysteresis function in this circuit.

When the battery charges up to 12.39 volts, the transistor conducts and does not stop conducting until the battery voltage drops below 11.24 volts. This is assuming the rheostats are at their maximum values.

This stops the relay rapidly switching on and off.

However, there should be some current limiting for the battery and there should be a diode to stop the battery powering the relay if the charging voltage is removed.
Jay_
#6
Jan15-13, 11:51 AM
P: 135
Thanks guys few more questions though ,

I think I understood the capacitor's work, its got to do with it charging till the rectified sine reaches peak and then discharging (relatively slowly) when the rectified sine starts to drop. So that makes it DC almost.

Averagesupernova,

Could you explain how the diode protects the remaining circuit? Which components does it protect. I understand the point of connecting a diode is that we want it to act as a short in some instances, and an open in other.

Could you explain in which instances the diode forward biases, and in which instances it reverse biases in this particular circuit?

Hisham.i,

Can you explain why do we need two variable resistors in the voltage divider arrangement? I don't understand that. With one itself we can vary the voltage right?

Vk6kro,

So, if I understand correctly - The relay is in the NC position first, and the circuit is completed. So the battery charges, and past a point the voltage divider arrangement sends current to the base of the transistor. This npn then acts as a short and relay is connected between positive and ground, right? And so, does the relay energize and switch to NO position which breaks the circuit? Is this right?
Windadct
#7
Jan15-13, 02:41 PM
P: 564
As a basic BAT Charger - it does not need very "clean" DC to work with. The rectifier output will be DC with ripples and the CAP filters them, in the peaks of the ripple the rectifier supplys energy to the capacitor AND the charger, in the valleys, then the capacitor supplies the energy.
D1 is a small free-wheeling diode to dissipate the inductive kick from when the transistor turns off - and de-energizes the relay coil. This diode is only forward biased for an instant, each time the transistor turns off ( you can not instantaneously change the current in an inductor - the relay coil is an inductor).
The dual adjustable resistors - are a little trick to create a Hysteresis zone, when fully charged ( for a car could be 13.8V) the Transistor will turn ON (stop charging the Battery) at a higher voltage then when it turns Off ( de-energizing the relay and starting the charging again). Without this the circuit would "chatter" turning on-off-on-off too quickly as the battery comes to full charge - remember that the battery voltage is higher when charging than when not charging - even when fully charged.
Jay_
#8
Jan15-13, 08:15 PM
P: 135
Is the relay energized when the transistor is OFF or ON?

What I understood (if I did correctly!) is that the transistor turns ON after a point of voltage build-up from the voltage divider arrangement. This means charging has happened, and the circuit breaks. Is the relay energized or de-energized when the transistor is ON?

Can you explain the how/why the diode gets forward-biased when the transistor is OFF and what exactly its protecting? If we didn't have the diode what would happen (so that I can understand its purpose).
dlgoff
#9
Jan15-13, 09:33 PM
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Quote Quote by Jay_ View Post
Is the relay energized when the transistor is OFF or ON?
Transistor Operation depends on the amount of current entering it's base. With no current, it's said to be in "cut-off" (i.e. there's no collector current or relay current in your case).



With a large enough base current, it's said to be in "saturation" (i.e. there's a high collector current; causing the relay to operate or close the switch in your case).



See: Transistor Switches

Can you explain the how/why the diode gets forward-biased when the transistor is OFF and what exactly its protecting? If we didn't have the diode what would happen (so that I can understand its purpose).
Probably easiest explained this way:

Current flowing through a relay coil creates a magnetic field which collapses suddenly when the current is switched off. The sudden collapse of the magnetic field induces a brief high voltage across the relay coil which is very likely to damage transistors and ICs. The protection diode allows the induced voltage to drive a brief current through the coil (and diode) so the magnetic field dies away quickly rather than instantly. This prevents the induced voltage becoming high enough to cause damage to transistors and ICs.
From: http://www.kpsec.freeuk.com/components/relay.htm

The transistor and relay in your circuit are acting like the switch and inductor respectively in these images.

When your transistor is in saturation,



When you transistor is in cut-off,



From: INDUCTORS

Note that when the switch is opened, current is induced (hence the name inductor which is what the relay's coil is) lighting the lamp. By having a diode in the place of the lamp, this induced current is shunted. Hence protecting the rest of the circuit from damage.
Jay_
#10
Jan15-13, 10:19 PM
P: 135
Okay that was clear. Now when we say the relay is energized is it at NO or NC? (relay energized = when the transistor is in cut-off and the current going through the diode to the capacitor's positive place).

The charging is happening when the relay is connected to NC, right? So this should be when the transistor base current is low and transistor in cut-off right? So the relay is energized, and current flows from diode to the positive place of capacitor? That doesn't make sense to me.

I understood :

Situation 1: Transistor cut-off --> Relay energized --> diode carrying current
Situation 2: Transistor saturation --> relay just conducts --> diode not carrying any current.

Which of these situations is when charging, and which is when discharging?
davenn
#11
Jan16-13, 01:37 AM
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Quote Quote by Jay_ View Post
Okay that was clear. Now when we say the relay is energized is it at NO or NC?
The relay has 3 contacts ( as looking at the diagram) a NO = Normally Open on the left, a common one in the middle and a NC = Normally Closed one on the right

NO and NC always refer to the contacts in the relay's unenergised state
when the relay is energised the NO contact closes with the common and the NC contact opens from the common contact


Dave
hisham.i
#12
Jan16-13, 06:30 AM
P: 177
Can you explain why do we need two variable resistors in the voltage divider arrangement? I don't understand that. With one itself we can vary the voltage right?
The first rheostat is used to set the upper voltage to which you want to charge the battery (voltage at which the charging stops), while the other (var2) is used to set the low voltage at which charging is started.

You can see that var1 is disconnected when charging stops using the relay.
Jay_
#13
Jan19-13, 10:37 PM
P: 135
Okay, so when the relay is energized it (yes I meant the common) is at NO, right?

So again, in which of these 2 situations is charging happening and in which situation discharging?

Situation 1: Transistor cut-off --> Relay energized --> common closed with NO --> diode carrying current

Situation 2: Transistor saturation --> relay just conducts --> common closed with NC --> diode not carrying any current.
vk6kro
#14
Jan19-13, 11:00 PM
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P: 4,032
Quote Quote by Jay_ View Post
Okay, so when the relay is energized it (yes I meant the common) is at NO, right?

So again, in which of these 2 situations is charging happening and in which situation discharging?

Situation 1: Transistor cut-off --> Relay energized --> common closed with NO --> diode carrying current

Situation 2: Transistor saturation --> relay just conducts --> common closed with NC --> diode not carrying any current.
Charging happens through the NC contacts while the transistor is not conducting.

When the battery is sufficiently charged, the transistor turns on and this causes the relay to energise and the relay contacts cut off power from the rectifier to the battery.
Jay_
#15
Jan20-13, 02:19 PM
P: 135
"NC contacts while the transistor is not conducting."

I thought when the common is closed with the NC the transistor IS conducting? Are my two "Situations" correctly described?
Averagesupernova
#16
Jan20-13, 03:30 PM
P: 2,529
A correctly drawn schematic will show relay contacts in the position that they are if the relay were pulled from the circuit and laying on a desk. In other words, unenergized.
vk6kro
#17
Jan20-13, 05:39 PM
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Neither of your situations is correct.

For this circuit to work, there has to be some current limiting at the right of the relay. Otherwise, the battery could draw a very large current and destroy the diodes.

The battery voltage will rise as it is charged and the voltage at the base of the transistor will also rise.

When this voltage reaches 0.7 volts, the transistor will turn on, activating the relay, moving the centre contact from right to left.

This stops the battery charging.

A fault with the circuit is that the battery supplies current for the relay and the transistor. So, if the charging power was removed, the battery would eventually be fully discharged.
Windadct
#18
Jan21-13, 09:18 AM
P: 564
I would say the relay has 3 TERMINALS - not 3 contacts, can be confusing. The NC and NO indication refer to the relay in the de-energized state, in this case this means the the transistor not conducting or "off". The confusion here is that the transistor turns ON - to stop the charging circuit.
The Diode only conducts for an instant - as the relay is being de-energized - the coil of the relay is an inductor and you can not instantly stop (or start) the current in an inductor.


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