Controlling A Relay With A Digital Logic Level Circuit

Description 

The schematic below illustrates 4 methods of controlling a relay with a digital logic signal. Figure (A) can probably be used in most cases where the relay coil requires 100 mA or less and the input current is 2 milliamps or more. The resistor value (R) is determined from the input voltage and the available current. For example, a 5 volt input signal supplying 2 milliamps would require (5-.7)/.002 = 2150 ohms, or a 2.2K standard value. If the transistor has a minimum current gain of 50, there will be 100 mA of current available for the relay coil. The following table shows various resistor values that can be used to obtain various relay coil currents assuming a transistor current gain of 50 such as the 2N3053. 74XX refers to standard TTL logic, 74LSXX refers to low power TTL logic, 74HC is high speed CMOS and CD40XX is the older CMOS devices. The currents given are approximate values and may not be correct for all devices but should be close. 

Input Voltage - Input Current - Relay coil current - Standard Resistor

• 4 - 74LSXX .0004 - 20 milliamps - 8.2K
• 4 - 74XX .0008 - 40 milliamps - 4.3K
• 4 - 74SXX .001 - 50 milliamps - 3.3K
• 5 - 74HCXX .004 - 200 milliamps - 1K
• 6 - 74HCXX .004 - 200 milliamps - 1.3K
• 5 - CD40XX .0003 - 15 milliamps - 13K
• 12 - CD40XX .0006 - 30 milliamps - 18K

Figure B can be used when the input voltage is the same as the relay coil voltage. The voltage on the emitter of the transistor will be about 0.7 volts less than the input, so a 12 volt relay would operate on 11.3 which should be close enough. No resistor is needed since the emitter follower configuration presents a high impedance at the input. The input current will be the relay coil current divided by the transistor gain. For example a 120 ohm relay coil will draw 100 mA at 12 volts and if the transistor gain is 50, the input current will be about 2 milliamps.

Figure C can be used to provide additional gain when the input current is very small. You can also use a Darlington transistor in place of the two transistors which is a better approach, but this idea works just as well when you dont have a Darlington transistor handy. The overall gain will be the product of the individual gains of the two transistors or about 2500 for two transistor with a gain of 50 each. This will enable supplying over 250 mA to the relay with only 100 microamps of input current. The R value will depend on the input voltage and current and gain of the first transistor. For example, using a 5 volt input and 100 microamp current and transistor gain of 50, the R value will be 5 minus two diode drops (5 - 1.4) divided by the input current times 50, or about (5 - 1.4) / (.0001 * 50) = about 750 ohms. So this setup can be used when controlling heavy duty relays with low power CMOS logic signals.

Figure D can be used to reverse the relay action so that it engages when the input is low and disengages when the input is high. The R value is determined the same as in Figure A. The R1 value should be high enough to ensure saturation of the first stage and low enough to saturate the second stage. For example, if a 12 volt relay coil requires 100 mA and the driving transistor gain is 50, then the base current will be 100/50= 2 mA and the R1 value must be less than 6000 ohms so that 2 mA does not drop more than the supply voltage of 12. If the first transistor gain is 50 and the input current is 100 microamps, the collector current will be 5 mA and the R1 value must be greater than 2400 ohms so that 5 mA drops the entire supply voltage of 12. So we need to select something between these two limits of 2.4K to 6K, something around 4.3K would be near the midrange.

Circuit Diagram

Source http://www.bowdenshobbycircuits.info/r_ctrl.htm

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Digital Radar Speedometer Schematic

This circuit is a Digital Radar Speedometer. It allows us to evaluate the speed of any object moving, especially cars and other vehicles. The speed is calculated in kilometers per hour (KPH). Its display has three digits. This radar works with the laser reflexion. It sends laser radiation to the object and this object reflects the laser radiation to the radar. To evaluate the speed of a vehicle, we must be in front of it. In other words, the vehicle must come in our direction. The front of the radar must point the front of the vehicle. The radar has the shape of a pistol. In this radar, it has a laser LED and a laser diode. Both have a lens.

Digital Radar Speedometer Circuit Diagram

The laser LED can send a spot of light to a distance of 90 m (295 ft). Its very important that the distance range of the laser LED is 90 m, if not, the speed will not be calculated properly. The laser diode, which receives the light signal by the laser LED, must be able to detect the light which is same color as that emitted by the laser LED. The laser diode and the laser LED must be placed one beside the other. They are protected by a tinted pane. They must be placed at the front of the radar and point the outside. The radar is powered by a 9V battery and it has a SPST switch to control its power state.

The display, or the speed indicator, is placed at the rear of the radar, just on the right of the overload LED indicator. All the logic components of the circuit must be of the 74AS series and TTL type. Because they have short time of response (less than 1.7 ns) and have high frequency supports (more than 200 MHz). The radar can evaluate the speed of an object moving between 0 to 999 km/h. After this speed, the overload LED indicator will turn on and the "999" will still displayed. The radar displays the speed during 3 seconds, after this time, it displays "zero" (0).

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4 Bit Analogue to Digital Converter

The operation of the converter is based on the weighted adding and transferring of the analogue input levels and the digital output levels. It consists of comparators and resistors. In theory, the number of bits is unlimited, but each bit needs a comparator and several coupling resistors. The diagram shows a 4-bit version. The value of the resistors must meet the following criteria:

• R1:R2 = 1:2;
• R3:R4:R5 = 1:2:4;
• R6:R7:R8:R9 = 1:2:4:8.

The linearity of the converter depends on the degree of precision of the value of the resistors with respect to the resolution of the converter, and on the accuracy of the threshold voltage of the comparators. This threshold level must be equal, or nearly so, to half the supply voltage. Moreover, the comparators must have as low an output resistance as possible and as high an input resistance with respect to the load resistors as feasible. Any deviation from these requirements affects the linearity of the converter adversely.

Circuit diagram:

4-Bit Analogue to Digital Converter Circuit Diagram

If the value of the resistors is not too low, the use of inverters with an FET (field-effect transistor) input leads to a near-ideal situation. In the present converter, complementary metal-oxide semiconductor (CMOS) inverters are used, which, in spite of their low gain, give a reasonably good performance. If standard comparators are used, take into account the output voltage range and make sure that the potential at their non-inverting inputs is set to half the supply voltage. If high accuracy is a must, comparators Type TLC3074 or similar should be used. This type has a totem-pole output. The non-inverting inputs should be interlinked and connected to the tap of a a divider consisting of two 10 kΩ resistors across the supply lines. It is essential that the converter is driven by a low-resistance source. If necessary, this can be arranged via a suitable op amp input buffer. The converter draws a current not exceeding 5 mA.

 

 

Source :www.ecircuitslab.com

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Digital code lock

This is a simple but effective code lock circuit that has an automatic reset facility. The circuit is made around the dual flip-flop IC CD4013.Two CD 4013 ICs are used here. Push button switches are used for entering the code number. One side of all the push button switches are connected to +12V DC. The remaining end of push buttons 2,3,6,8 is connected to clock input pins of the filp-flops.

The remaining end of other push button switches are shorted and connected to the set pin of the filp-flops.The relay coil will be activated only if the code is entered in correct sequence and if there is any variation, the lock will be resetted. Here is correct code is 2368.When you press 2 the first flip flop(IC1a) will be triggered and the value at the data in (pin9) will be transferred to the Q output (pin13).Since pin 9 is grounded the value is “0” and so the pin 13 becomes low. For the subsequent pressing of the remaining code digits in the correct sequence the “0” will reach the Q output (pin1) of the last flip flop (IC2b).

This makes the transistor ON and the relay is energised.The automatic reset facility is achieved by the resistor R11 and capacitor C2.The positive end of capacitor C2 is connected to the set pin of the filp-flops.When the transistor is switched ON, the capacitor C2 begins to charge and when the voltage across it becomes sufficient the flip-flops are resetted. This makes the lock open for a fixed amount of time and then it locks automatically. The time delay can be adjusted by varying the values of R11 and C2

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Sooper Digital Volume Control

This digital volume control has no pot to wear out and introduces almost no noise in the circuit. Instead, the volume is controlled by pressing UP and DOWN buttons. This simple circuit would be a great touch to any home audio project.

Schematic

Parts:

Part	Total Qty.	Description
C1	1	0.1uf Ceramic Disc Capacitor
U1	1	DS1669 Digital Pot IC (See Notes)
S1, S2	2	Momentary Push Button Switch
MISC	1	Board, Wire, Socket For U1

Notes:

1. U1 is available from Dallas Semiconductor.

2. S1 turns the volume up, S2 turns it down.

3. The input signal should not fall below -0.2 volts.

4. Using a dual polariity power supply (+-5V works fine) will cure most clipping problems. You will have to check the data sheet for the correct pins to connect your voltages. Link

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