The relays are commonly used for switching purpose. There are many types of relays like electromagnetic relays, thermal relays, reed relay, solid state relay and hybrid relay. A relay is basically an electronically operated mechanical switch meant to be used for switching. The electromagnetic relays are also of many types like attraction type relay, induction type relay and magnetic latching relay. This tutorial is concerned about the magnetic latching relays which are to be used to switch the load between two DC power sources. The switching from one DC source to other takes some time, so it cuts the supply to the load for a short duration of time. This tutorial discusses about reducing the switching time, so that, the load does not experience any interruption during the transition phase of the relay.
Components Required –
Fig. 1: Components required for Relay Switching Time Reducer
Circuit Connections –
A latching type relay is taken. The relay is meant to switch between two power sources. Both the power sources are taken to be 12V. So, a battery of 12V rating can be taken as the power source. The power from the batteries is regulaated to 5V DC using 7805 IC. For this, the input power is applied at the pin 1 of the IC and pin 2 is grounded. The power is then drawn from pin 3 of the IC. There are two 7805 ICs used in the circuit to regulate the power supply from the two sources. One of the regulated power source is connected to the Normally Open (NO) point of the relay while the other regulated power input is connected to the Normally Closed (NC) point of the relay. A load resistance (shown as R1 in the circuit diagram) is connected at the common (C) point of the relay. For fast switching the relay response, a capacitor (shown as capacitor C5) is connected in parallel to the load R1.
While assembling the circuit following precautions must be taken care of –
1. Do not exceed the input voltage limit of the voltage regulator IC as this can damage the IC. Check out the datasheet of the IC for reference.
2. The voltage rating of the input DC supply (12V in this circuit) should be greater than the maximum required output voltage. This is due to the fact that the 7805 IC takes voltage drop of around 2V.
3. The voltage used for energizing the relay (12V in this circuit) should be equal or greater than the voltage rating of the relay. Otherwise, the relay will not be activated.
4. Always use the capacitor at the input of regulator IC (like capacitors C1 and C3 in the circuit diagram) since it can handle mains noise and reduces the unwanted ripples. Also use a ceramic capacitor (like capacitors C2 and C4 in the circuit diagram) in parallel with a high value of the polarized capacitor. This will decrease the overall ESR.
5. The capacitors used in the circuit must be of higher voltage rating than the input supply voltage. Otherwise, the capacitors can start leaking the current due to the excess voltage at their plates and can burst out.
6. The Diode D1 and D2 should be used as Protection diode so that they can prevent the external capacitor C5 from discharging through the voltage regulator ICs during an input short circuit. This will save the 7805 ICs from the back current.
7. A Flyback diode (Shown as Diode D3 in the circuit diagram) should be used across the relay for preventing the circuit from any back current.
8. Use an appropriate watt of resistor as load for preventing the resistor from any damage.
9. If a high current has to be drawn at the output then the voltage regulator ICs will get heat up. So mount a proper heatsink to the regulator ICs to dissipate the excess heat.
Fig. 2: Prototype of Fast Switching Magnetic Latching Relay
How the circuit works –
The type of relays under consideration are the magnetic latching relays. In this type of relay, the common pin (C) stays at Normally Closed (NC) pin when its input pins are not energized. When power is provided to both the energizing pin or input pin of the relay then the relay gets activated. The common pin of the relay is then switched to the Normally Open (NO) pin and it takes some time to switch from its NC to NO pin or vice-versa.
In the inverter circuits this relay is used for switching the load from one power source to other. Our power source of 12V energize the relay as well as provides input to the 7805 IC (shown along diode D1 in the circuit diagram). In this state, the load gets connected to the 7805 IC. Whenever the input 12V supply goes out then the relay is de-energized. Now the load gets connected to the other 7805 IC (shown along diode D2 in the circuit diagram) which is powered by another 12V supply but the load experiences some interruption in switching from one 7805 IC to another. This cuts the supply from the load for a short duration of time which is equal to switching time of the relay. But this switching time can be reduced by the use of capacitors.
A capacitor has mainly two basic functions. The most known function of the capacitors is for filtering the DC signal from unwanted voltage spikes and ripples. So, these type of capacitors are called filtering capacitors. Another function of the capacitor which is of great significance but least discussed is the use of capacitor as a Reservoir. The term reservoir explains all about it. As the capacitor stores some charge so this accumulated charge can provide some power. So the capacitor is also used to hold electrical charge or to act as a power source. Due to this fact, the capacitor used as a reservoir of charge is called reservoir capacitor.
For fast switching the relays, reservoir capacitor can be used which can handle fast transient response at the output. This capacitor if is connected in parallel to the load then it will store some charge. The stored charged then can be used to provide the extra power to the load during the transition period of the relay from NO to NC or vice-versa. So, a capacitor is used in parallel to the load (shown as resistance R1 in the circuit diagram). The value of the required reservoir capacitor can be calculated as follow –
Maximum current drawn by the load (I)
This can be calculated by using ohms law.
V = I*R
Where
V = output voltage
I = current drawn by the load
R = load resistance
In the circuit, a load of 50 ohms is used so maximum current drawn by the load is
I = V/R
I = 5/50
I = 100mA
Maximum ripple voltage at the output (dv)
The maximum ripple voltage is the maximum allowable voltage change at the output for which the capacitor is able to fulfill the desired voltage at the output.
In the circuit, maximum ripple voltage is assumed to be 100 mV. So,
dv = 100mV
Transient response time of the capacitor (dt)
This is the time period for which the power to the load should be supplied through the capacitor. This transient time must be equal to the switching time of the relay so that it could be compensated by the use of the capacitor.
For calculating the relay switching time, the output waveform of the load (without using capacitor) can be observed on the CRO. A waveform as shown below is observed –
Fig. 3: Graph showing Relay Switching Time in Normal Conditions
As seen from the above waveform, at one point the load voltage is zero. So this switching time must be reduced by the use of capacitor. As observed from the CRO, the switching time of the relay is around 3ms.
So by using above values of Maximum load current(I), Maximum ripple voltage (dv) and transieent response time (dt) the value of the capacitor required for reducing the switching time can be calculated as follow –
I = C*(dv/dt)
Where,
Maximum allowable ripple voltage , dv = 100mV
Transient Response Time, dt = 3ms
Desired current at the output, I = 100mA
C = Capacitance
By putting all the values,
C = (0.1*0.003)/0.1
C = 3mF
As per the avaliablity, three 1000uF capacitor connected in parallel can be used to make up 3mF in total.
Testing the Circuit –
After connecting a 3 mF capacitor in parallel with the load and observing the output waveform on the CRO, the following voltage waveform is observed –
Fig. 4: Graph showing Reduced Relay Switching Time After Connecting Reservoir Capacitor
This graph clearly shows that the switching time of the relay is now negligible that it cannot even be measured. Now the load will not experience any kind of discontinuity in terms of power supply and it always gets a constant voltage supply.
Limitation of the capacitor method –
As every circuit has some flaw along with its advantages. There are some drawbacks of this circuit as well. As seen for drawing only 100mA current, the value of the required capacitor is 3 mF which is very large. If a high current has to be drawn then the value of the required capacitor will be very large. So this makes the circuit costly. Also, the high value capacitors are not easily available.
In the next tutorial, learn about testing MOSFET.
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