Blown-Fuse Indicator

 

This is a blown-fuse indicator for a dc power supply. When the fuse is intact, the transistor is base-biased into saturation. This turns on the green LED to indicate that all is OK. The voltage between point A and ground is approximately 2 V. This voltage is not enough to turn on the red LED. The two series diodes (D1 and D2) prevent the red LED from turning on because they require a drop of 1.4 V to conduct.

When the fuse blows, the transistor goes into cutoff, turning off the green LED. Then, the voltage of point A is pulled up toward the supply voltage. Now there is enough voltage to turn on the two series diodes and the red LED to indicate a blown fuse 

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Battery Charger, Volatge and Current Control

This project is not so important for those who ride a car every day. In my case, a car is not used every day. In that case, the voltage of a battery is falling and a car sometimes cannot be put into operation.

If the battery itself is left for a long period of time, the electricity of it will be lost by natural electric discharge. Furthermore, in the case of the latest car, various kinds of electric devices are always backed up more often with a battery. Therefore, even when the car is not being used, the electricity of a battery is consumed little-by-little.
I was creating and using the battery charger which used the solar battery before. However, it is not effective if the bad day of the weather continues.
Then, I decided to create a small charger. Because I assumed always charging during parking, the charging current is controlled by the current control circuit. Moreover, in order to prevent over charge, it can be made to perform a setup of maximum voltage. General-purpose 3 terminal voltage regulator is used for control of voltage and current by each.

Voltage control circuit

This is the circuit which controls maximum charge voltage, in order to prevent over charge of a battery.
For the control circuit, 3-Terminal Adjustable Regulator (LM317) is used.
A left figure is the basic circuit of the regulator. The voltage between Vout and ADJ is fixed and is 1.25V as standard.
Control of output voltage is performed by the value of R2.
Output voltage (Vout) is calculable by the following formula.
Vout = 1.25 ( 1 + R2/R1) + I_ADJ(R2)
I_ADJ is current which flows from an Adj pin and it is several 10µA. Therefore, this can be disregarded.
In LM3xx, there is a condition to decide the resistance for voltage control. It's Load Regulation. The load current 10mA or more is required for normal operation of a device. Therefore, it is recommended to set the value of R1 to 120ohms or less.

R1 is set to 100 ohms in this circuit. R2 in the above explanation turns into VR1+R2 of a schematic.
In an actual circuit, R2 is 560 ohms and VR1 is 2k ohms.
In case VR1 is 0 ohms, the output voltage is as follows.
Vout = 1.25 ( 1 + 560/100 ) = 1.25 x 6.6 = 8.25V
In case VR1 is 2k ohms, the output voltage is as follows.
Vout = 1.25 ( 1 + 2,560/100 ) = 1.25 x 25.6 = 32V
Therefore, the output voltage of this circuit can be controlled from about 8V to 32V.
Because a current control circuit is inserted behind this circuit, the final output voltage of a charger declines by 2-3V.

Current control circuit

7805 is IC circuit for making voltage regularity. However this time, this IC is used as a circuit which makes current regularity.
The left figure is drawn in the style of voltage control in order to make an understanding easy.
Even if it changes input voltage, 7805 operates so that the voltage between an ground terminal (G) and an output terminal (O) may be set to 5V. If a resistor R3 is connected between O-G, the current which flows into R3 will be set to I = 5V/R3. Therefore, the current which flows into R3 becomes fixed.
Because the current which flows into R3 flows also into load, if the value of R3 does not change, the current which flows into load is fixed. Conversely, if R3 is changed, the current which flows into load is changeable.

This figure is the circuit used this time.
First, I decided the value of R3. In this charger, because the maximum current is set to 500mA, as R3, it is made 5V/0.5A = 10 ohms. When 500mA current flows into a 10 ohms resistor, the power consumption of a resistor is I²xR = 0.5²A x 10ohm =2.5W. I am using the cement resistor of 5W in consideration of safety.
Next, I calculated the value of VR2. I assumed controlling in about at least 80mA. Therefore, R3+VR2 is 5V/0.08A = 62.5 ohms. R3 was 10 ohms, so the value of VR2 was set to 50 ohms. When 80mA current flows into 50 ohms, the power consumption of a resistor is 0.08² x 50 = 0.32W. I am using the variable resister of 2W in consideration of safety.
It is also possible to use LM317 for a current control circuit. However, there is a fault. In LM317, the voltage between O-G is 1.25V. In this case, the resistance for setting a current value to 500mA is 1.25V/0.5A = 2.5 ohms. It is 15.6 ohms for making it 80mA. Compared with 7805, it is a small value. Current control will become difficult if the error of resistance is taken into consideration.
Moreover, if a regulator with high output voltage is used, the power consumption of the resistor for control will increase more. For example, when the regulator for 12V is used, the resistance for making it 500mA current is 12V/0.5A = 24 ohms. And the electric power consumed by the resistor is 6W. For the above reason, I am using 7805 for current control.

R4 and C3 may not have necessity. In this circuit, the diode for preventing the reverse current from a battery is used. As for diode, ON state (state where current flows), and the OFF state (state where current does not flow) have clarified. If the voltage of a battery rises by charge and becomes higher than the voltage of a charger, current will not flow from a charger. Then the voltage of a battery falls by that cause and current begins to flow from a charger again. It will be oscillated if such a thing occurs for a short time. So, in order to suppress a rapid voltage change of a charger, I put C3. R4 is put in order to make C3 discharge. However, it seems that the voltage of a battery does not change so quickly in fact. Therefore, I think that it is satisfactory even if C3 and R4 do not use.

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INFRA RED TEST RECEIVER by: Dmohankumar

This circuit can operate a load such as LED, Buzzer or Relay for a period of  3 minutes through a Remote hand set. It uses the IR sensor module TSOP 1738 which operates in 38 KHz Infrared pulses.

The circuit is a Short duration Monostable  using 555 timer IC. Its trigger pin 2 is connected to the output of the IR sensor through R2 and LED. Normally the output of  IR sensor is low. But the timer IC will not be triggered because its trigger pin remains high through R3. When the remote hand set is focused on to the IR sensor and any one of the buttons is pressed, out put of IR sensor goes low and triggers the 555 IC. With the given values of the timing elements R4 and C2, output of the timer remains high for 3 minutes. By changing the value of R4 or C2 this time can be changed.

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