Battery charging circuit - can anyone explain?

Windadct

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.

Jay_

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?

Jay_

Here is the image again

dlgoff

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.

Jay_

" 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).

vk6kro

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.

Jay_

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).

Windadct

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.)

dlgoff

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)

But like Windadct says,
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

Jay_

@dlgoff, Windadct

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