# Battery charging circuit - can anyone explain?

• Jay_
In summary, the conversation discusses understanding how to analyze circuits and finding good sources for learning. The first circuit in the conversation involves stepping down AC mains and rectifying it to get a 15V pulsating DC. A capacitor is used to keep the voltage constant, but it is unclear how it works in this circuit. The role of the diode and transistor in the circuit is also discussed. The second circuit, a battery charger, is also analyzed and suggestions for improvement are given. The conversation also includes questions about the function of the diode and the use of two variable resistors in the voltage divider arrangement. The purpose of the relay in the circuit is explained as well.
Jay_
http://electronicdesign.com/site-files/electronicdesign.com/files/archive/electronicdesign.com/files/29/1823/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:

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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Jay_ said:
http://electronicdesign.com/site-files/electronicdesign.com/files/archive/electronicdesign.com/files/29/1823/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.

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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.

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.

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.

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?

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.

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).

Jay_ said:
Is the relay energized when the transistor is OFF or ON?

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2 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.

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?

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Jay_ said:
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

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.

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.

Jay_ said:
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.

"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?

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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.

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.

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.

Hey guys, thanks for all your patience. I am really dumb at times. I find all the posts contradictory, or I haven't understood at all.

The diagram shows the transistor in OFF (cut-off) stage, when the relay's common is closed with NC and open with NO. This is when charging is happening, right?

After a point, the voltage divider, saturates the transistor, this is when the relay common closes with NO and opens with NC. At this point the charging has stopped. And the current goes from the battery, through the resistors and all, through the base of the transistor and then through diode, to the positive plate of the capacitor, right?

What I didn't understand is from dlgoff post #9 in this thread, it seems like the current flows through the diode when the transistor is in CUT-OFF.

Shouldn't the battery be charging and no current go through the diode, when the transistor is in CUT-OFF?

If someone can explain the corresponding states of the transistor, relay position (common closed with NC or NO) and the current direction, I will probably understand. If I don't past that, forget it. I'm probably too dull to get it. Or I will give the thread another read again thoroughly.

I am probably mistaking the use of "energized" when speaking of the relay. Energized = common closed with NO and open with NC, right? And De-energized = common closed with NC and open with NO, right? So in the diagram, relay is de-energized, right? And transistor is cut-off right?

http://electronicdesign.com/site-fil.../figure_01.gif

Here is the image again

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Jay_ said:
What I didn't understand is from dlgoff post #9 in this thread, it seems like the current flows through the diode when the transistor is in CUT-OFF.
Did you read the Inductor Link? What happens is, when the transistor goes from saturation to shut-off, the stored energy in the inductor must go somewhere. So you put in that diode to shunt that energy. This diode has nothing to do with the charging of the battery.

When current through an inductor is increased or decreased, the inductor "resists" the change by producing a voltage between its leads in opposing polarity to the change.

" What happens is, when the transistor goes from saturation to shut-off, the stored energy in the inductor must go somewhere. So you put in that diode to shunt that energy. This diode has nothing to do with the charging of the battery."

The link mentions how the voltage across the inductor becomes very high by quick switching and that makes sense according to V = L(di/dt). So it makes sense that its a protective element. But what happens to it ones it flows through the diode and into the capacitor plate?

Another thing I wanted to clarify: doesn't the transistor go to saturation only when there is sufficient base-voltage due to the voltage-divider arrangement? Isn't this when the battery has reached a particular upper voltage level?

Also, isn't the transistor in cut-off because the base voltage is not sufficient while it is still charging? That was my basis to think that charging happens when the transistor is cut-off (and so current flowing through the diode as the link shows).

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When the transistor turns off, the inductor develops a very large positive voltage at the lower end where it joins the transistor's collector. This can be hundreds of volts more positive than the other end of the inductor.

The diode is forward biased by the voltage and the diode conducts a current from the bottom end of the coil into the top end of the coil. The current just goes into the inductor.

When the battery is fully charged, a current flows through the resistor chain to the base of the transistor from the battery. This keeps current flowing into the relay and stops further charging of the battery by the charging circuit.

Thanks vk6kro. Can someone explain why the direction of the current in the image dlgoff posted when the transistor is closed seems to be from negative of the battery to the positive? Doesn't current go from positive to negative (opposite direction of electrons).

Hello Jay - the arrows are just representing electrons - and IMO is misleading because from a circuit and mathematical perspective this is against convention. ( However it is technically correct, esp to a physicist, if you consider that the electrons are negatively charged (-). But confusing for this conversation nonetheless.)

Jay_ said:
Can someone explain why the direction of the current in the image dlgoff posted when the transistor is closed seems to be from negative of the battery to the positive? Doesn't current go from positive to negative (opposite direction of electrons).

Good catch. That's the convention of current direction I've always used in circuit analysis. e.i. From the highest "+" potential (voltage) to lowest "-" potential (voltage)

... it is technically correct, esp to a physicist, if you consider that the electrons are negatively charged (-).

Mathematically, those arrows in the pic you're asking about can be considered to represent "negative current". Here's an example where everything works out mathematically.

Magnetic Force on a Current

Okay. So I guess that settles it.

Thanks to all of you :) I have to read up more on inductors and their circuits.

I will probably start looking at the second link (second battery charging circuit) and post any doubts I have again.

Regards,
Jay

## 1. How does a battery charging circuit work?

A battery charging circuit is a system that regulates the flow of electricity from a power source to a battery in order to charge it. It typically consists of a power source, such as a wall outlet or solar panel, a circuit board, and various electronic components such as resistors, capacitors, and diodes. The circuit controls the voltage and current flowing into the battery, ensuring that it is charged safely and efficiently.

## 2. What is the difference between a constant current and constant voltage charging circuit?

A constant current charging circuit maintains a steady flow of electrical current into the battery, while a constant voltage charging circuit keeps the voltage level consistent. Constant current charging is typically used for fast charging, while constant voltage charging is more suitable for slower, trickle charging. Some charging circuits may use a combination of both methods.

## 3. How does a battery charging circuit prevent overcharging?

A battery charging circuit is designed with safety features to prevent overcharging, which can damage the battery or even cause it to catch fire. These features may include a voltage regulator to control the voltage level, a current limiter to prevent too much current from flowing into the battery, and a timer to shut off the charging after a certain amount of time has passed.

## 4. Can a battery charging circuit be used for different types of batteries?

Yes, depending on the design and specifications of the circuit, it can be used to charge different types of batteries. However, it is important to ensure that the circuit is compatible with the specific type and capacity of battery being charged. Using the wrong charging circuit can damage the battery or cause it to malfunction.

## 5. Is it safe to leave a battery charging circuit connected to a battery for a long period of time?

Leaving a battery charging circuit connected to a battery for an extended period of time is generally safe, as long as the circuit has proper safety features and the battery is compatible with the circuit. However, it is recommended to periodically check the battery and circuit for any signs of damage or malfunction, and to follow the manufacturer's instructions for safe and proper charging procedures.

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