Switching between two batteries.

[B0B]

Hi, I´m a student on the Electronics school, and I´m working on an assignment which I have had quite a trouble to solve, so I was hoping that somebody who knows better, could help me.=)


It is basically about:

I have to charge two batteries with one source, but while the first battery is charged the second has to be used. So the circuit has to switch charging between the battery that is full and the battery that is discharged. Also it has to have an output from the battery, that is currently full. I should say that I´m working with two Ni-mh 1,2V 300mah batteries, the charger has around 2V and the output should be around those 1,2V from the battery, but I know there will be some voltage drops so the minimal output voltage has to be around 1,05V ( the closer to 1,2V as could be).


I know it is hard to understand from the plane text so I have a prototype circuit that I´ve made, I´m not saying it would work (I was definitely hoping that it would), in the circuit opening and closing switch means full and drained battery, and multimeters”1&2” are output to the battery or better said outputs for charging the batteries, it´s all theoretical so I don´t have any specific components (tranzistors, thyristors, diodes) yet, but if it would work, I was also hoping that somebody would help me pick them out for the best efficiency. (The components which are in the circuit are just for practice and don´t have to be good at all) Thanks, for now, I´m really hoping that it could be done, and that somebody would be so good to help me with it.

 

[Harald Kapp]

Hi Bob, welcome to EP.

I dare say your circuit will not work and is imho way too complex. From your use of switches I assume (without that being explicitly stated) that you will switch between V2 and V3 being charged and being discharged alternatively. There is a simüle way using a single double pole double throw switch to achieve this without any electronics. The drawback of using a mechanical switch alone is that during switching from one source to the other the current to the load will be interrupted for a few milliseconds.

To avoid this interruption an electronic circuit can be used. It shure has to be different from your circuit.

Consider XMM1: Is it a voltmeter? If so, then no current (in practice a negligible small current only)will flow from V1 through XMM1
Is it an ammeter? If so, then Q1 will short circuit V1 when turned on - not a practical circuit.

Considering above paragraphs: how is V1 ever to charge either V2 or V3 when V1 is either short circuited by Q1 or Q2 or no current can flow through XMM1 or XMM2? Therfe is no path for current from the '+'pole of V1 to either V2 or V3 to allow a charge current to flow.

The same consideration is true for XMM2 and Q2.

What are thyristors D1 and D2 for? Their gates are tied to GND and therefore not controlled. D1 and D2 are always off and can be removed from the circuit without a change in behavior.

My recommendation:

1. draw/design a charge circuit to charge a battery V from the source V1.
2. draw/design a switch circuit that will:
   2.1 switch teh charger between two batteries V1 and V2
   2.2 switch the load inversely between V2 and V1
   The switch circuit can be an electromechanical switch or an electronic circuit using transistors.

Avoid the use of thyristors in DC circuits unless you have a definite means of interrupting current through the thyristor as you cannot turn off a thyristor by a control voltage on the gate. Use bipolar transistors or MOSFETs instead.

 

[B0B]

Harald, thanks a lot for your answer, howewer I think there is a lot of misunderstandig between us (i think it is because of my poor description of what a want or I simply don´t understand what are you saying =) But I´ll try to answer one by one to your answer.

First I cannot use mechanical switch, even when I know it would be a lot simplier.
XMM1/XMM2 are both voltmeters
Thyristors are for keeping the opposit side of the circuit closed when one batterie is not fully discharged, but the other one is charged, therefore those are not normal thyristors but thyristors GTO

I think it would be the best if I would attach two more pictures of the circuit, but with the voltmeters, so you could see, the values.

In the first picture there is switch near to XMM2 opened which should act as an drained battery, so it is charged (XMM2 shows 1,393 volts, which is voltage from the "source" or "charger"), the XMM1 shows 4,637mV which idealy should be 0 but It shows the principe. On the XMM3 there is 1,164V which is an output voltage.

In the second picture it is reversed so switch near to XMM1 is opened which simulates drained battery so it is "charged" and other switch is closed which simulates full battery so idealy it gets 0V and it is an source for the output.

As I said, maybe you know all of that and I simply misunderstood you, if so I´m sorry, but maybe I´ve helped explaining the circuit a little bit better.

In both cases, thanks for your help.

 

[(*steve*)]

You have 1.2V batteries. Do you have any idea what the forward voltage drop of the thyristors is?

 

[duke37]

I do not like relays because of reliability issues, however, in this case,a micro relay may be better than a semiconductor which needs considerable voltage across it or to turn it on (FET).

 

[Bluejets]

You need to look at how chargers work as well as ypur ideas about requirements are not correct.

 

[Harald Kapp]

There is still nopath from the '+' of V1 to the '+' of either V2 or V3 to charge it.

I repeat: thyristors, even when you use GTOs, are not suitable to simply turn a DC circuit on or off. Use nipolar transistors or MOSFETs instead.

There will be no currrent through Q1 or Q2 from collector to emitter as XMM1 and XMM2 being voltmeters have (ideally) an infinite resistance (in practice a few MΩ).

D5 and D6 have their anodes tied to GND. These diodes are always reverse biased and therefore non-conducting.

Why do you say you can't use mechanical switches when in fact you use S1 and S2?

Throw away this circuit and start a new one along the lines of my post #2.

 

[B0B]

Thanks a lot for all replies, I will consider all of advice I have got, and I´ll try to design a better circuit.