This 33 K will now become the collector resistance for the BC, so you can apply the same formula and principle for calculating the BC base resistor….. Thanks very much for your swift response and advice. Thanks for the feedback.
I am unclear how you derived the 10K ohm base resistor in the basic PNP relay driver circuit with a 12v supply and a negative trigger. Can you elaborate a little on that formula? Using the formula will identically give you the 56k value for the PNP driver as well, for a ohm relay, however, from my practical experience I have seen that a 10k usually works good without any dissipation on the BJT, along with a powerful drive for the relay contacts.
Ok, I was wanting to adapt this to drive a solenoid valve with a 24v supply and using a switched ground trigger, and the 10K in the example was confusing me… Thanks for the quick answer:. Goodmorning Sir. Hi, Kindly tell me how you calculated 68K resistance in an example in which you are using both npn and pnp transistors to operate relay. I did not calculate it, selected it randomly.
To calculate it you can use the same formula as use for the relay coil, but in this case the load are two 1k resistors at the collector of NPN. However, you can ignore the upper 1K and consider the current only through the lower 1K, in the formula. Hello Muzzam, the circuit has only a few parts, so it would be a better idea to build it practically and check how it performs. Thank you for your quick response.
I am new to electronics. I have few questions if you would answer i would be highly grateful. I have not built it practically, but it was referred from a top reputed electronic magazine, so it will definitely work. And, also can i post the picture of my simulation somewhere so that you could check where i could have made an error.
Two thoughts. First I see no throw back diode to protect the transistor when switch is opened. Second is the assumption that running a relay at lower voltage is ok. First a typical 12 volt relay will usually pull in at around volts and dropout at volts.
So using 5 volts does not allow much for variations that occur with wide temperature changes. You should also be aware that relay contacts current carrying capacity are affected by the force pressing them together which in this case is being reduced substancially. If switching currents of a milliamp or less and depending on the contact materail you can risk loss of contact over time and environmental changes. Better choice is to use a magnetic latching relay. Your approach would be fine when driving a solenoid.
Your email address will not be published. Notify me via e-mail if anyone answers my comment. You'll also like: 1. Sir, Can u suggest a simulator software which is able to do some testing, thanks.
Hi sir. Hi Fahad, which circuit are you referring to? Hi, since the two relays are in parallel will make the the result as ohms not Thanking you. A single transistor may not work efficiently unless the transistor is an Darlington type. Add a uF capacitor in parallel with the relay coil. Kind regards. Hi Swagatam, How do I place a capacitor to have a delay on? Thank you. Hi Raj, for this diagram the formula is the same as explained in the above article.
For Ub or the supply use 6V, if the input is 9V. Thank you so much. The calculated value is the accurate value but will require the input supply to be not less than 24V and relay resistance not higher than ohms A 10K will be quite fine, and will allow the relay work strongly even if the input supply is lower than 24V, or the coil resistance is higher than ohms, and also keep the transistor and R2 cool.
When auxiliary relay coil is energized by a transistor driver circuit, the path of the main relay is completed through relay 1 contacts. Thus the relay 2 coil is energized and hence it is operated to turn the heater. Similarly for turning OFF the heater relay 1 coil must be de-energized. Most of the electromechanical relays need regular checkups of their functionality for reliable performance. As the moving parts of the relay change in response to abnormal conditions there should be regular testing.
Protective relays are used in medium and high voltage power system. With a long usage, the connection of the relay gets deteriorated with carbon particles. Therefore, in order to ensure the reliable performance of the relay, it must be tested before putting into service and also after time intervals it must be checked. These types of tests include. This is performed by the manufacture at several stages during manufacture in order to check the acceptability of the unit for sale.
These tests determine the function of the relay for a particular protection scheme. These tests are carried to for checking accuracy of assembling the components in the relay, ratings, calibration and conformity with entire system. These tests are carried to identify the degradation of the service and equipment failures in the relay. These are the tests that are carried on the relays which are employed for high and medium power switching or protection system applications.
However, for low power application especially relays are that are used in electronic control systems, a multimeter is high enough to carry the relay testing. The procedure for testing the relay is as follows.
If above all conditions are met, then we can say that the relay working properly otherwise it is defective. The relay is used for the purpose of protection of the equipment connected with it. These are used to control the high voltage circuit with low voltage signal in applications audio amplifiers and some types of modems. These are used to control a high current circuit by a low current signal in the applications like starter solenoid in automobile.
These can detect and isolate the faults that occurred in power transmission and distribution system. Typical application areas of the relays include. Great Work. Is that feasible? Your email address will not be published. Electromechanical Relay. June 11, By Administrator. Your content helped me a lot to take my doubts, thank you very much. Leave a Reply Cancel reply Your email address will not be published. But microcontrollers usually can not provide enough current to drive relays.
Microcontroller pins usually provide a maximum current of mA per pin which is not enough o operate relay. The circuits which are used to derive relays are called relay driver circuits. Therefore Relay driver circuits using ULN are used to drive relays. There are many other ways to operate relays. I will also discuss other methods of relay driver circuits in the later part of this article.
While working on electronics projects which used a microcontroller, we need to use relays to control AC loads or high voltage loads. Therefore relay driver circuits IC is used to derive relays properly. There are many ways to design relay driver circuits.
But dedicated relay driver circuits integrated circuits are available which serve the purpose of relay driver IC. The relay driver circuit using an NPN transistor is given below. Transistor is used as a switch in this circuit. The microcontroller provides High or low input signals to NPN transistors. NPN transistor provides high driving current to electromechanical relay through the volt external power supply.
Following components are used in relay driver circuit using transistor:. ULN is a very famous relay driver integrated circuit. Its contains seven Darlington pair of a transistor which have high voltage and high current carrying capability. Its mean ULN can drive up to seven relays at a time. A diode is used with each pair of NPN Darlington transistor. A LED is easy enough, but large power items such as light bulbs, motors, pumps or fans required more sneaky circuitry.
You can use a 5V relay to switch the V current and use the Arduino to control the relay. A relay accomplishes this by using the 5V outputted from an Arduino pin to energize the electromagnet which in turn closes an internal, physical switch to turn on or off a higher power circuit.
The switching contacts of a relay are completely isolated from the coil, and hence from the Arduino. The only link is by the magnetic field.
This process is called "Electrical Isolation". The coil of the relay needs a large current around mA to drive the relay, which an Arduino cannot provide. Therefore we need a device to amplify the current.
The diode across the electromagnet conducts in the reverse direction when the transistor is turned off to protect against a voltage spike or the backward flow of current. A current of about. The electromagnet then pulls the switching contact and moves it to connect the COM terminal to the NO terminal.
In this example I am just turning on and off a LED. The code is very simple. Just start by defining the digital pin number 2 of the Arduino as the Relay pin. This voltage spike can damage the sensitive electronic components controlling the circuit. One when the electromagnet is on and the second one when the electromagnet is off.
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