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Before you can design an adjustable voltage regulator into your circuit, or do a redesign, you need to calculate the values for two resistors. This is not difficult in itself, but actually finding the right resistors may pose problems. Fortunately a trick is available to make it all much easier. With most adjustable voltage regulators like the LM317 and LM337, the input voltage has to be 1.2 to 1.25 volts above the desired output voltage. This is because the voltage at the ADJ (adjust) input is internally compared to a reference voltage with that value. The reference voltage always exists across R1.
Together with preset R2 it determines the current flowing through the ADJ pin, as follows: Vout = VREF [1+(R2/R1)]+I ADJ R2 If for the sake of convenience we ignore I ADJ, enter the reference voltage (1.2 V) and for R1 select a value of one thousand times that voltage (i.e., 1.2 k?) then the equation is simplified to: R2 = 1000 (Vout – 1.2) In practice, simply determine the voltage drop across R2 (output voltage minus reference voltage) and you get your resistance value directly in kilo-ohms. For example, for 5 V R2 becomes 5–1.2 = 3.8 k? which is easiest made by connecting 3.3k and 470R resistors in series. In the case of relatively low voltages, smaller resistor values are recommended. This is because sufficient current needs to flow to enable the voltage regulator to do its job. A simple solution is to choose, say, 120 ? for R1. R2 then becomes: R2 = 100 (Vout – 1.2)

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Fuse Box BMW 318i 1991 Diagram

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With just a low cost DC adapter and the circuit described here it is possible to build a low cost stabilized, uninterruptable 9V supply. On the grounds of safety and economy, a simple unstabilized 12V D.C. adapter is used as the power source, a universal adapter with its output set to 12 V will do equally well. The output voltage of an adapter under low load conditions (up to approximately 1/3 of the rated output current) is over 15 V, even at the rated output current, there will be sufficient voltage to supply a 9 V voltage regulator. The rating of the DC adapter should be chosen according to the output current required at 9V. Common values are 300mA, 500mA and 1A.

The 9V voltage regulator used in this circuit has a built in thermal shutdown mechanism so that if too much current is drawn from the device, it simply turns off as it overheats and will not supply any current until the case temperature returns to normal. If the unit is intended to supply more than say 150-200mA then to prevent thermal shutdown it will be necessary to fit a heatsink to the voltage regulator. The rule of thumb used to calculate the size of heatsink is that you should be able to touch it during operation at maximum load, without burning you finger. When choosing the DC adapter, it is always better to select one with a higher current rating than is needed this will ensure that its output voltage is high enough to be able to also charge the 12V cells.

As long as mains voltage is on the DC adapter, the voltage across C1 will be higher than the voltage of the cells. Charging current will flow through R1 and D1 to the cells. Current also flows to the voltage regulator and out to the load connected at the output. Diode D2 in this situation will not conduct because the voltage at its cathode is greater than that at its anode When the mains voltage fails or is turned off, diode D2 conducts and current will now flow from the Nickel Cadmium cells to the voltage regulator, thereby automatically keeping the output voltage at 9V. The value of resistor R1 is chosen so that a charging current to the cells is not greater than 1/10th of the cells capacity (if the cells are rated at 1100mAh, the charging current must not exceed 110mA).

From the point of view of cell longevity it is better to reduce this charging current even further (1/20 or 1/50 C). When calculating this resistor, the value of the no-load voltage should be used. This will give the highest charging current. To calculate the charging current using R1 with a value of 180 Ω. The cells measure 13.8 V when fully charged and the no-load output voltage of the DC adapter is 17V. Charging current is given by the formula: (17V – 13.8V – 0.7V) / 180 = 13.9mA. Substituting the actual measured values in this formula will enable you to calculate the value of R1 to give the correct charging current for the cells.

This circuit uses a thin piezoelectric sensor to sense the vibrations generated by knocking on a surface; eg, a door or table. Basically, it amplifies and processes the signal from the sensor and sounds an alarm for a preset period. In operation, the piezoelectric sensor converts mechanical vibration into an electrical signal. This sensor can be attached to a door, a cash box, cupboard, etc using adhesive. A 1-1.5m long shielded cable can then be connected between the sensor plate and the input of the circuit. The signal generated by the sensor is amplified by transistors Q1-Q3 which are wired as common-emitter amplifiers.


The signal is then rectified by diode D1 and amplified by transistors Q4-Q6. As shown, the output from Q6s collector is fed to pin 4 (reset) of 555 timer IC1. This is wired as an astable multivibrator. Each time Q6 turns on, its collector goes high and IC1 activates and produces an alarm tone in the speaker. The alarm automatically turns off 10s after knocking ceases - ie, the time taken for the 22µF capacitor on Q4s emitter to discharge. Finally, note that it may be necessary to adjust the 470O resistor in Q6s collector circuit to ensure that IC1 remains off in the absence of any perceptible knock. A value somewhere between 220O and 680O should be suitable.