AUTOMATIC CHARGER

.

This project is essentially to "rejuvenate" a New Electronic kits, this is the original scheme

The problem arises for two factors, the relay in the long run will ruin the integrated and become obsolete and practically impossible to find at reasonable prices.

Essential conditions of this kit were the absence of absorption from the network on standby, and full automation in the charging cycle, in practice it decides independently whether the battery needs to be recharged or not.

In case it was necessary to refill the relay is activated and feeds the transformer which supplies the constant current to the battery charging generator, once you reached the correct charge level, the circuit stops, disconnects the relay again and waits. It will repeat the cycle as soon as it detects that the battery is discharged at its terminals.

My idea

Without prejudice to the spirit of this circuit I'm going to delete the relay both for the transformer of the power control is also to manage the charge. I wanted to add a little improvement, in the kit is not present an indication of the battery status at its ends, there is only the status indication is recharged. In practice, If the battery contacts are not correctly positioned, not part of the circuit is not detecting the presence of a battery. Such a situation may give an erroneous information that the battery is sufficiently charged as not to require recharging, the lighting of the green LED READY indicates that the battery is properly connected to the circuit and does not require recharging.

I wanted to still maintain the ability to decrease the charging current simply by opening or closing a contact as in the original circuit.

Circuit description

The complete diagram is used this

Although at first glance it may seem more complex than the original circuit uses components at low cost and readily available. The mains voltage section occupies a distinct portion of printed and away from the rest of the circuit.

In it are detectable:

a comparator with hysteresis and one without hysteresis,

a delay circuit in the command,

a constant current source with selectable current

a solid state relay for the section 230V

a programmable zener to create a precise reference voltage

the indication of the state of the circuit with two LEDs

All in a circuit with few components so that fits into a base 6cm x 6cm mono-face with a single bridge.

The power to the circuit is taken from the battery terminals, also completely discharged will be able to power the circuit and to start charging, thus in the absence of battery (battery) to load the circuit will no longer be fed and will not absorb current from the network.

The battery voltage feeds the double operational amplifier U1 (LM358) and the programmable zener U2 (TL431).

U2 is used to create a stable voltage to the desired value, in this circuit the resistors R22, R9 secure the voltage value while R10 limits the current of the stabilizer, the formula to know the value of R22 as a function of this voltage is R22 = (R9 / 2,5)*(Vout-2.5V) = 880*(9-2,5) = 5720, naturally this value is not standaed and must be obtained by means of the trimmer R22. To facilitate the setting there is a test point (TP) where to connect a multimeter to calibrate the correct voltage required.

At the maximum battery charging voltage absorption will be approximately 8mA,

this voltage will also be present with completely discharged battery, if unable to reach such a voltage would have as a reference for the initial stages of charging, the battery voltage. As charging and the voltage grows, the reference value set will settle firmly on. The value is chosen to simplify their life in the comparator with hysteresis.

The resistors R14 and R15 are respectively Rx and Ry of the following formulas while the more R19 R13 series constitutes Rh

The resistors R11 and R12 // R1 form a voltage divider for detecting the voltage of the batteries, the partition ratio is 3

The maximum and minimum values ​​of the charger intervention are set at 12.5V as a minimum voltage under which to start the charger and 14.5V maximum voltage charge necessary to end the charging.

The divider will bring these tensions in

VL = 12,5V*(R12/(R 11 + R 12 // R 1)) = 4,16V equivalent to 12.5V / 3 = 4,16V

VH = 14,5V*(R12/(R 11 + R 12 // R 1)) = 4,83V equivalent to 14.5V / 3 = 4,83V

now to be considered this formula

Ry/Rx = VL/(VR-articles)

so that the two resistors are all equal the reference voltage must be

VR = VH + VL = 4,16V 4,83V = + 9V simply (12,5+14,5)/3= 9 and here found the reference voltage.

Returning to the formula

Ry/Rx = VL/(VR-articles) = 4,16V/(9V-4,83V) = 4,16V/4,16V = 1

In this way we can now use with such a reference voltage equal resistors.

Now it should be calculated Rh, knowing that

Rh / Rx = VL /(Articles-VL) = 4,16V/(4,83V-4,16V) = 4,83V/ 0,67V = 7,2

We just have to give RX values ​​to derive all other, Fixed this value to 10K and then get R14 = R15 = 10k

Rh = Rx * 7,2 = 10K * 7,2 = 72K

get this value by summing R13 and R19 calibrated to obtain the correct value.

The second half of the operational amplifier configured as a comparator inverting, It is also used to give a slight delay in switching.

We know that the U1.1 output is positioned permanently or at logic level zero ( next to 0V) or at a logical level 1 (next to the power supply) without intermediate levels.

The resistance R8 together with the capacitor C5 constituting the delay line in the commutations, C5 in stable conditions will be at the same potential output of the operational amplifier, We admit by hypothesis to 0.1V which is the typical output level of the LM358 powered single voltage. U1.2 will have its non-inverting input to this voltage while the non-inverting input to a voltage that is at least 4V coming from the divider employee input from the battery charge level, its output will be forced to zero.

At the change of state of the output of U1.1 the capacitor C5 will begin to charge through R8, after a certain time the voltage at its ends will have exceeded the voltage from the input divider and at that point the operational amplifier U1.2 output will change to logic one level.

This output has three functions, the first is to indicate the status of the battery and the second feed the mains transformer which will provide the necessary current to the charging.

Its output will be high when the battery will be charged, and under such conditions will feed the green LED READY.

When its output will be zero feed both the LED carge that VO1, this is a Phototriac which switches to zero, activating the transfer of mains voltage to the primary of the transformer which with its output will feed the charging circuit. VO1 by switching only at the zero crossing but does not introduce disturbances, for safety Having regard to the inductive load given by the transformer primary a snubber circuit has been provided formed by R5 and C1.

The last output of the operational function is the activation of the real charging circuit.

When the output is to one the mosfet Q1 P-channel is isolated through D4 and there is no transfer of energy from the transformer to the battery, the charger will not charge.

However, when the output goes to zero the mosfet transfers the voltage from the secondary of the transformer, rectified by D3 and smoothed by C2 through R4 and the mosfet to the battery.

R4 together with R6 and the transistor Q2 are used to limit the charging current. When the voltage across R4 exceeds the conduction voltage Vbe of Q2 begins to conduct this bringing the output voltage to the gate of Q1 via the diode D1.

The voltage drop in R4 will arrive at 0.7V with a current of circa1,5A, exceeded this current the circuit regulates itself becoming a constant current generator. But this threshold is too low when you need to charge a car battery, at this point shorting J3 with a normal low current switch stands the tripping current.

For closed switch R6 and R7 form a voltage divider of the voltage detected in R4, without using specific formulas now new revenue stream. With the divider of the transistor trip will occur when R7 there will be 0.7V, the current is then

I = 0,7V / R7 = 0,7/270 = 2,59mA,

this same current will pass through R6, causing a voltage drop VR6 = R6 * I = 330 * = 0.85V 2,59mA,

summing the two voltages will have the total voltage that falls in R4 + VR6 therefore VR4 = 0.7V = 0.7V + 0.85V = 1.55V,

the new intervention current time will be

I = VR4/R4 = 1,55V/0,47 = 3,3A.

For attaching the replicability of the printed images with the sizes and arrangement of the components with the dimensions of the base. Clearly printed in the lower left corner you notice the mains voltage section adequately spaced from the rest of the components.

The arrangement of components is ordered, one side dedicated to the links, on the opposite side there is the calibration trimmers and the Test-Point, side has been dedicated to the cooling fin. For fixing three screws are provided, two of which can also be used to make integral with the cooling fin with the printed.

The transistor Q1 and the bridge are the only elements that potentially warming and precisely for this reason are positioned on the edge to be fixed to a small cooling fin, the triac does not need it since the handled power is such as not to generate excessive overheating.

Amilcare Greetings