Circuit Wiring Solution: charge

Showing posts with label charge. Show all posts

Friday, September 26, 2014

How to Charge a Lead Acid Battery Circuit Diagram

Batteries should not be discharged below 1.8 volts per cell, as it can cause permanent damage to the cells.
A 40 ampere-hour battery when discharged at 4—ampere rate will take 40/ 4, i.e. 10 hours to be completely discharged. Suppose the same battery is discharged at 10—amperes rate. Theoretically, it should take 40/ 10, i.e. 4 hours to get fully discharged. But in practice it 1S observed that the battery will get fully discharged within about 3% hours itself In other words, the higher the discharge rate, the lesser will become the capacity of the battery.
A fully discharged battery needs to be charged about 1% times its full ‘capacity’, to reach a fully charged, state. In other words, a 40-AH battery being charged at 4 amperes rate will take 15 hours and not 10 hours to be fully charged Charging a partially discharged battery =· It is not possible to estimate the time needed to charge a partially discharged battery.
The charging or discharging of a battery is ideal at 10 hours rate, which means that a 40 ampere hour battery is to be charged or discharged at 40/ 10, i.e. 4-ampere rate.
Lead—acid battery has a voltage of 2.1 volts per cell (on load) when fully charged, which will rise up to 2.7 volts per cell while on charge. When the voltage per cell (on load) drops to 1.8 volts, the battery is considered to be fully discharged.
However, the battery may be charged till such time as the following are observed: (a) Free gassing of the cells. (b) Battery voltage reaches its maximum value, and remains steady. (c) The specific gravity of the electrolyte (as measured by a hydrometer) reaches 1.240
That’s why is engine starting, when the battery drains 200 to 300 amperes, the battery becomes dead’ within a few seconds of use.

How to charge a lead acid battery, circuit diagram

Saturday, September 20, 2014

Circuit diagram capacitor charge control

The scheme is designed to protect against inrush current when the battery uncharged capacitor on-board network. Who has not tried to include uncharged faradnik network without limiting resistor - better not ... At a minimum, get burnt contacts.

When you turn the discharged capacity in network capacity C1 is discharged, T1 (n-MOSFET switch with low channel resistance) is closed.Capacitance C2 (the same faradnik) is charged through a low-resistance R5. T2 opens almost instantly, the shunt to ground C1 and T1 gate.When the potential negative terminal C2 falls below 1V (charge to Uakb - 1B), T2 closes smoothly C1 is charged to about 9/10 Uakb opening T1. The time constant R2C1 is large enough so that the current surge T1 (pre-charge C2 +1 V to Uakb) does not exceed the rating for T1.

In the future, the negative terminal of C2 constantly shorted to ground through T1, regardless of the direction CURRENT T1 (both literally - from drain to source, and in the opposite direction). Nothing wrong with "rollover" OPEN TIR transistor not. When choosing a good enough conductive transistor entire reverse current flow through the channel, and a built-wheeling diode will not open because the voltage drop across the channel at times less than required for the opening of 0.5-0.8 V. By the way, there is a whole class of TIR devices (eg FETKY ), designed specifically to work in the opposite direction (synchronous rectifiers), they have built a diode is shunted by an additional force Schottky diode.

Calculation: for transistor IRF1010 (Rds = 0.012 ohms) voltage drop of 0.5 ohms will only be achieved with the current channel 40A (P = 20W).For four of these transistors in parallel and the same discharge current of 40A - on each transistor will dissipate 0,012 * (40/4) ^ 2 = 1.2 W, ieradiators they are not required (the more that will dissipate 1.2W only when differential current consumption but not consistently).

Dense installation (you have plenty of space for extra radiator?) - Advisable parallels small (body TO251, DIP4) transistors, generally do not provide radiators, based on the ratio of current (power) consumption of the amplifier - Rds - limit power dissipation. Since Pds max is typically 1W (800 mW for DIP4), the number of n transistors (c Rds each) for the amplifier with an output power Pout must be at least n> 1/6 * Pout * sqrt (Rds) at 12V supply (dimension in the formula I omitted). In fact, given the short duration current pulses, n can be easily reduced by half compared with a given formula.

Resistor R5 is selected from the charge compromise heat output and charging time. When these 22 ohms - charge time of about 1 minute at power dissipation 7 watts. R5 can instead include 12V bulb, say, indicator. Resistors R1, R3 - reinsurance (discharged capacity when disconnected from the network).

Connect to indicate activation of additional inverter (reducing R2). Attention! The scheme is efficient at using npn transistors T2, T3 with h21e> 200 (KT3102). Depending on the brightness of the LED, R1, select the range of 200 ohms - 1k.

But the view of the circuit in which the key shutter control signal REMOTE (And transistor). The non-connected or off REMOTE key transistor guaranteed closed. D3-D4 LEDs indicate charging C1, D5-D6 - the open state of the key.

Accurate indication of the threshold voltage is provided easiest IP TL431 (KR142EN19) in a typical mode voltage comparator (with the corresponding subgroup in the input circuit and current-limiting circuit cathode R).

Loss schemes largely depend on the installation. Ensure that the minimum resistance (and corresponding current thickness of the wires) in the power circuit (terminal + / C2 / T1 /-terminal). In amateur practice, I think, make outgoing terminals impractical - it is better to unsolder the short wires AWG8, which binds to the terminal block diagram of the amplifier.

Original article source cxem.net