Automatic battery disconnection or AC adaptor
The circuit is a system of automatic shutdown of the battery when fully charged, that is, it is not quite a battery charger, of course, if you complement it with a transformer and rectifier, you get a full-fledged battery charger. The original circuit was subjected to some changes the board finalized in the course of testing the final version of the board can be downloaded at the end of the article. Let us have a look at the schematic.
As you can see it is painfully simple and contains only one transistor, an electromagnetic relay and some small stuff. I also have a diode bridge on the input and primitive protection against polarity reversal (these parts are not shown in the schematic).
The circuit input is a constant voltage charger or any other power source, here it is important to note that the charging current must not exceed the allowable current through the relay contacts and the current triggered by the fuse. In my case the circuit is 8 amps.
How it works – when power is applied to the input of the circuit the battery is being charged, the circuit has a voltage divider (R2, R3, R4) which monitors the voltage directly on the battery.
As the battery charges, the voltage on the battery will rise, as soon as it becomes equal to the circuit’s trigger voltage, which can be set by turning a trim resistor, the stabilizer will trip, signaling the base of the low-power transistor and it will trip. Since the electromagnetic relay coil is connected to the collector circuit of the transistor, the latter will also be triggered and these contacts will open, and further power supply to the battery will stop. At the same time the second light-emitting diode will go off, notifying you that the charging is over.
There is another LED in the circuit, it is permanently lit, it is essentially an indicator of voltage on the board. As I said before, the divider monitors the battery voltage directly so if the battery connected to the charger is discharged to a certain value, the circuit will automatically switch off and the charging process will resume. Read also: How to connect Bluetooth instead of AUX and remove all the wires in the car
Since the divider is connected directly to the battery it will discharge it, but the discharge current is so paltry that it can be ignored. To tune the circuit, a high capacity capacitor is connected to its output, we have it as a fast charging battery. I took ionistors connected in series and connected them instead of the capacitor.
If you take a capacitor, its voltage must be 25-35 volts, first of all connect the ionic cells (in my case) or the capacitor to the output of the circuit, observing the polarity,
At the end of the charge first disconnect the charger from the network, then the battery, otherwise the relay will be false. In this case nothing bad happens, but the sound is unpleasant.
Next, we take any regulated power supply, such as a laboratory unit and set on it the voltage to which it will charge our battery and connect the unit to the input of the circuit.
Slowly turn the trimmer resistor until the red LED goes off,
until the red indicator goes off, then make one full turn of the trimmer in the opposite direction, because the circuit has some hysteresis.
And now let’s check the operation The voltage on the ionistors or the capacitor, will be shown by the multimeter when the threshold value on them is reached, the system will cut off the power.
If the voltage drops on the battery, the circuit will go off again and charge the battery to the set value again. The board can be downloaded here…
Automatic battery disconnection or AC adaptor
|Current time: Thu Sep 15, 2022 19:17:09|
Time zone: UTC + 3 hours
The attachment is a car battery charger.
|Page 1 of 2||[ Messages: 21 ]||To page 1 , 2 Next.|
The attachment is a car battery charger. Literally, it is an automatic charging switch at a certain voltage level. This attachment does not affect the charging current in any way, so it can be used with any power supply with an output voltage of 24-36V. It can be made as a separate unit and connected in the gap between the power supply and the battery. It is known that the no-load voltage is always higher than the load voltage, and in the case of a battery, it drops to the battery voltage itself, while the current in the circuit increases. It is this property that was used in the design of the circuit. How it all started. I got my hands on a Soviet battery charging block, made on the basis of a transformer. The maximum charging current was chosen by switching the primary windings of the transformer and the output voltage varied from 25V to 34V. Naturally, it was dangerous to leave the battery alone with such a charge. It was necessary to somehow disconnect the battery when it reached 14.5V, and since the charger was intended for an elderly person, protection against polarity reversal and terminal short-circuiting was needed. In addition, it had to be indestructible and have a clear indication of the operating modes. The last requirement is a small heat and size. On the basis of these considerations, a simple but sufficiently functional circuit was developed. Functions of the device: 1. indication of connection of the terminals to the battery 2. Terminal short circuit and polarity reversal protection with indication. 3. Manual start and automatic continuation of charging in case of mains voltage failure. 4. Disconnection of the battery from the charger when 14,5V is reached on the terminals, with transition to “compensation” mode of operation with a charging current of about 50mA. 5. Indication of the modes “charging” and “ready” 6. Smoothly decreasing charging current from 5A to almost zero when using a conventional transformer about 80W. How the circuit works. The transformer, 10A diode bridge and capacitor are what was inside the charging block, everything else is the actual attachment. Without the battery or in “standby” mode, the voltage of the divider R1R2R3 opens D3 and that through resistor R7 blocks the key transistor T2. Without the battery connected only LED D2 “power” is lit. The base and emitter potentials of T1 are equal, it is closed and there is no voltage at the collector, so it is not possible to turn on the set-top box. Connecting the battery in the correct polarity with voltage from 8V will light the green LED D5 “ready”, because at the drain of T2 will appear voltage of about +29V (37Vp – 8Vac.). The base potential of T1 will drop by 0.7V relative to the emitter, the transistor will open and the supply voltage appears on its collector, and the set-top box is ready to start. “Plus” of the rectified voltage goes directly to the battery, then through resistor R13 is connected to the mass, providing a charging current of about 50mA to save the battery charge in standby mode. Now, if you press button S1, the voltage from divider R8R7R6D4 (maximum 13V) will open transistor T2, LED D6 will be bypassed and extinguished, the battery will be connected to the common ground and a charging current of about 5A will flow through it. At the same time, the power supply voltage will drop from idle 36V to the battery voltage, let’s say 12V.
The voltage of divider R1R2R3 will drop below the opening threshold of D3 (2.5V) and it will stop blocking the operation of T2, so after releasing S1 the voltage on the gate will set around 10-13V, the circuit “clicks” and will wait for the battery voltage to increase to 14.5V. All this time the LED D4 “charge” will be on, indicating that the battery is charging. If the 220V mains supply fails, the device will remain on and D4 will continue to glow as the battery maintains the opening voltage on the gate via R5R7. When mains power is resumed the battery charging will continue. When 14.5V is reached, D3 will open, which shunts D4 and closes transistor T2. D4 “charge” will extinguish and D6 “ready” will light up, the so-called “standby mode” will be activated, when the charge level on the battery will be maintained by a low current. Thus, any power supply with this attachment can be left with the battery without fear. The channel resistance of T2 is fractions of ohms, and even at a charge current of 10A it generates no more than two watts of heat, so this transistor can simply be attached to the metal case of the power supply via an insulating gasket. If you use multi-turn R2 you don’t need R1 and R3, and replace R2 with 10kOhm. The capacitor C3 suppresses noise. I did not complicate the circuit automatic restart when the battery voltage drops below 12 volts because such a situation is unlikely, in addition after charging, in any case, requires human intervention and, if necessary, nothing prevents you from pressing the button “charge” again. How protection works. My power supply gave 36V, so I put the D7 at 43V, i.e. with a small margin. If you reverse polarity the battery will be connected in series with the power supply and at the drain of T2 will be about 50V, D5 will shine brighter and red D6 will be added to it, because at voltage over 43V the stabilizer D7 will open. The diode D8 protects the circuit, so T2 will be closed. When the terminals are shorted, only the green D5 will be lit, transistor T1 will also be closed. When T1 is closed it is not possible to start charging. If for any reason the circuit opens for a moment during charging, the console will go into standby mode as human intervention is required to re-establish contact. How charging works: Turn on the power supply, only the “Power” LED will light up, then connect the battery observing the polarity, additionally D6 – “Ready” will light up (If you do the opposite: connect the battery first and then plug it in, charging will start immediately). Press S1 “charge” and T2 will connect the battery to the ground bus. D6 will go out and D4 will light up, indicating that charging has started. As the battery charges, the battery voltage rises and the charging current falls. When 14.5V is reached, the console will disconnect the battery from the power supply, D4 “charge” will extinguish, and D6 “ready” will illuminate. In this state, the battery will receive a current of about 50mA through R13, compensating self-discharge, which depends on many factors, so if you fear overcharging, then eliminate R13 at all or increase it to 680 ohms (1W), just in case. I left it as it is, 470 Ohm, I got a current of about 48mA, minus self-discharge of not new battery about 20mA, the final charging current is 28mA. This current, in my opinion, will not harm the battery even in a few months. The setting is very simple.
Without connecting the mains rectifier, switch R2 to the lower position according to the diagram, connect a laboratory regulated power supply with voltage 14,5V instead of the battery, D4 will light up, now by smooth adjustment of R2 achieve switching off, i.e. when D4 goes out. Then change voltage smoothly in small range, LED should light up and go out at 14,5V. About the parts. I can not tell you the exact parameters of the transformer, in my case it was a transformer with an output of about 160W with five taps in the primary winding. One main winding is 220V, the others are 50V additive, based on the transformer ratio and the voltage increment on the secondary for one step of the switch. So the primary winding of the transformer is designed for 220V + 4x50V = 420V. The short circuit current is at least 12A. The unit is not stabilized, just a transformer and rectifier. If you are going to assemble it yourself, then put a diode bridge for 10A, better screw it to the metal case. Do not replace the red LEDs with other colors because they have a turn-on voltage of 1.8V, i.e. they are more “sensitive”. If you want up to 30V supply voltage, you can put one DB2 diode instead of D7 at 43V, since its breakdown voltage is 32-35V. All resistors except R13 are 0.125-0.25W. I took T2 from a blown converter, you can find a matching transistor on a computer motherboard, like 25N06. Our analog TL431 is 142EN19.
As you know, any battery has an internal resistance on which some of the voltage will drop. As a result, at high current the battery will be undercharged, and at low current it will be overcharged. Take your set-top box and the battery, charge it with the maximum current before you turn it off and switch to a small current and your set-top box will still be charging it for a long time. It should be set at the minimum current and charge the battery in 2 steps. First on high current until it shuts down, then charge it again on low current. I used to do something like that when I was in school. If the circuit would measure the voltage at the moment when the mains voltage goes through 0 when there is no charging current, then it would be of no value. It’s not hard to modify. And you can connect the LEDs directly to the battery just with different polarity.
_________________ If you are dissatisfied with your standard of living, the laws of our country, the level of prices, then remember all this at the next election.