Lennox PRODIGY M2

15. Service Relay
Page 61
PRODIGYT M2 UNIT CONTROLLER
Example 1: Use the Service output to energize a dehumidifier at 60% RH and turn off at 55% RH. (Refer to figure 31.)
1. Determine:
ECTO 7. 22− Input Source X to option 7" (RH)
Algorithm Y to option 0" (hysteresis loop)
Inversion Z option 0" (output not inverted)
2. Apply step 1 to ECTO 7.22 formula:
ECTO 7.22 = X + (32
x
Y) + (16
x
Z)
= 7 + (32
x
0) + (16
x
0)
=7
3. Set ECTO 7.22 to 7".
4. Set ECTO 7.23 to 60% (60 counts) for 60% RH
5. Set ECTO 7.24 to 5% (5 counts) for 5% deadband.
60% RH
(ECTO 7.23)
Dehumidifier Off
Dehumidifier On
55% RH
(ECTO 7.23−7.24)
60−5=55%
Figure 31. Example 1. Service output energizes dehumidifier at 60% RH / turns off at 55% RH
Example 2: Use the service output to energize a humidifier at 30% RH and turn off at 35% RH. (Refer to figure 32.)
1. Determine:
ECTO 7. 22− Input Source X to option 7" (RH)
Algorithm Y to option 0" (hysteresis loop)
Inversion Z option 1" (output inverted)
2. Apply step 1 to ECTO 7.22 formula:
ECTO 7.22 = X + (32
x
Y) + (16
x
Z)
= 7 + (32
x
0) + (16
x
1)
= 7 + 0 + 16
=23
3. Set ECTO 7.22 to 23".
4. Set ECTO 7.23 to 35% (35 counts) for 35% RH
5. Set ECTO 7.24 to 5% (5 counts) for 5% deadband.
35% RH
(ECTO 7.23)
Humidifier Off
Humidifier On
30% RH
(ECTO 7.23−7.24)
35−5=30%
Figure 32. Example 2. Service output energizes dehumidifier at 30% RH / turns off at 35% RH
15.2. OAT Operation
Figure 33 shows the service output being energized at 80°F and off at 70°F (inverted) and the service output operation being
energized at 70°F and off at 80°F (not inverted).
IMPORTANT − This applies to input source option 9 (OAT) on ECTO 0.23, 3.21, 7.22, and 9.23.
7.23
97 COUNTS
70°F
Service Output On
Service Output Off
Service Output On
Service Output Off
7.23−7.24
81 COUNTS
80°F
7.23
97 COUNTS
70°F
Output Inverted − Input source 9 only Output NOT Inverted − Input source 9 only
7.23−7.24
81 COUNTS
80°F
Figure 33. OAT Operation − Service output energized