G2R-1-SND-DC12(S) Omron, G2R-1-SND-DC12(S) Datasheet - Page 27

• Manufacturer Part Number: G2R-1-SND-DC12(S)
• Description: RELAY SPDT 12VDC PLUG-IN W/LED
• Manufacturer: Omron
• Series: G2RSr

Specifications of G2R-1-SND-DC12(S)

• Relay Type: General Purpose
• Contact Form: SPDT (1 Form C)
• Contact Rating (current): 10A
• Switching Voltage: 440VAC, 125VDC - Max
• Coil Type: Standard
• Coil Current: 43.2mA
• Coil Voltage: 12VDC
• Turn On Voltage (max): 8.4 VDC
• Turn Off Voltage (min): 1.8 VDC
• Mounting Type: Socket
• Termination Style: Quick Connect - .187" (4.7mm)
• Circuit: SPDT (1 Form C)
• Contact Rating @ Voltage: 10A @ 250VAC
• Control On Voltage (max): 8.4 VDC
• Control Off Voltage (min): 1.8 VDC
• Coil Voltage Vdc Nom: 12V
• Contact Current Max: 10A
• Contact Voltage Ac Nom: 250V
• Contact Voltage Dc Nom: 30V
• Coil Resistance: 278ohm
• Contact Configuration: SPDT
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: G2R-1-SNDDC12(S); G2R-1-SNDDC12(S); G2R1SNDDC12S; Z2952

If applied voltage E and the rated coil voltage of the relay are the
same, the current to the relay falls short by the quantity indicated by
the shaded portion in the following figure.
Therefore, the current must be applied to the relay as follows when
designing this driver circuit.
Time constant
When the rated voltage is applied to the relay, time A in the timing
chart below is required to turn ON the contacts. After this time has
elapsed, time B is required until the armature attraction to the mag-
net is complete.
Therefore, it is apparent that time constant T obtained as the product
of C and R must be equal to or longer than the sum of A and B. Actu-
ally, however, T should not be equal to the sum of A and B but must
be longer than that to ensure the stable operation of the circuit. Thus,
T = A + B + X
where X is the time margin.
The set time A of OMRON’s moving-loop relays (with a pickup power
of 200 mW) is rated at about 3 milliseconds. Time constant T for
them should be about three times that of A. The following graph illus-
trates this. This graph indicates that, if C is completely charged
(I
broken down into three segments. The first 1/3T equals A, the sec-
ond 1/3T, B. The remaining 1/3T is the time margin expressed as X in
the above equation. T is three times A.
Voltage drop E1 across the total resistance of the capacitance C’s
resistance and relay coil’s internal resistance is the difference
between the supply voltage E and voltage drops across two
diodes: Di1 and Di2. Hence,
E1 = E – 2VF
Assuming the supply voltage to be 5 V and VF to be 0.6 V,
E1 = 5 – 2 x 0.6 = 3.8 V
From E1 and the above graph, the required coil voltage of a relay can
be obtained. Again assuming the E, i.e., the supply voltage of a sin-
gle-winding latching relay is 5 V, the coil voltage is:
3.8 x 0.72 = 2.7 V
At this time, the capacitance of C is 246.9 μF, according to the equa-
tion shown in the above graph.
PEAK
), it takes 4.6T to discharge I to 1%. Note that time constant T is
E
Applied voltage
Contact ON
Attraction of
armature
E
Coil voltage
2VF
2VF
Pulse width and current
necessary for setting relay
A
B
Set time
Time required to set
relay completely
4.5T
Electromechanical Relays
Coil ratings and capacitance of C
In the example, the coil voltage obtained by calculation is 2.7 V,
which is 0.3 V less than the value at which the coil voltage of com-
mercially available standard latching relay is rated. The standard coil
voltages of relays at a supply voltage of 6, 9, 12, and 24 V can be
respectively calculated in the same way. Table 1 compares the
results of the calculation and the coil voltages of standard relays.
The calculated coil voltages significantly deviates from the standard
values. It is therefore necessary to determine the time constant of the
relay by adjusting the capacitance of C when the relay coil is to oper-
ate on the standard voltage.
As an example, calculate the capacitance of C and time constant T of
a relay with a rated supply voltage of 5 V. The coil voltage E
been calculated above (3.8 V). To determine how much current I
flows through the coil at 3.8 V, from Table 1, note that the coil resis-
tance is 45Ω. So,
I = 3.8/45 = 84.4 mA
Therefore, the peak current of capacitor C to be used must be
84.4 mA.
Remember, that time A of an OMRON relay is 3 ms. Capacitance C
must be a value that allows 66.6 mA to flow through 3 ms after 5 V is
applied to the relay.
Thus,
From this,
C = 280 μF
At this time, time constant T is:
280 x 10
By calculating the C of each of the relays listed in Table 1, the values
in Table 2 are obtained.
Again, these calculated capacitances deviate from the commercially
available standard capacitors. There is no problem in using standard
capacitors but, if the cost and circuit space permit, it is recommended
to use two or more capacitors so that a capacitance as close to the
calculated value as possible is obtained. At this time, pay attention to
the following points:
• Confirm that the relay operates normally even when the supply volt-
• Even if a voltage of two or three times the rated voltage is applied to
Supply voltage
Supply voltage
age is brought to 80%-120% of the rated value.
this driver circuit, the coil wire will not sever. That is why, for exam-
ple, when the driver circuit is mounted in an automobile where a
supply voltage of 12 VDC is available from the battery, it is recom-
mended to use a relay whose coil voltage is rated at 6 VDC, taking
a voltage fluctuation of 8 to 16 VDC into consideration.
12 V
24 V
12 V
24 V
5 V
6 V
9 V
5 V
6 V
9 V
–6
x 45 = 12.6 ms
Technical Information
Coil voltage
Coil voltage
(calculated)
(calculated)
16.4 V
16.4 V
2.7 V
3.5 V
5.6 V
7.8 V
2.7 V
3.5 V
5.6 V
7.8 V
Table 1
Table 2
resistance
Standard
voltage
125Ω
720Ω
405Ω
12 V
Coil
45Ω
45Ω
3 V
3 V
5 V
6 V
Capacitance
resistance
280 μF
142 μF
6.5 μF
125Ω
405Ω
720Ω
54 μF
40 μF
of C
Coil
45Ω
45Ω
27
1
has