INTRODUCTION
In
the HVAC industry their are several types of control systems,
0-10 VDC and 4 to 20ma are soon becoming the most widely used. Even
with their popularity the series 60 and 90 control systems are still
the most common types. The words series 60 and 90 are actually trade
names referring to products with 6* and 9* in their model number.
Series 60
A series 60 control system is the most basic type. Series 60
represents the products with the numbers 6* in them.
The T675A thermostat and ML6161A1001 mod motor are examples of series
60 controls, these controls are also referred to as 2 position. Series 60 controls are either open or closed.
Series 90
A series 90 control system is of the type referred to as
modulating. The T991F thermostat and M9185D mod motor are examples of
series 90 controls. The series 90 control system uses 135 ohm potentiometers
connected with the whetstone bridge configuration. With this style of
control system we have one that is fully modulating instead of just
open and closed.
Whetstone Bridge
The Wheatstone bridge was invented by Samuel Hunter Christine a British
scientist. Even though he invented it, the Wheatstone
bridge gets its name from Sir Charles Wheatstone a British physicist
and inverter who first applied it for measuring the
resistance in a electric circuit.
A Wheatstone bridge is used to determine an unknown resistance.
This is accomplished by adjusting a known resistance until the measured current
is equal across the bridge. The Wheatstone bridge consists of four
resistors connected together in a diamond orientation. The resistors
are arranged so that the electric current is split into two paths,
each of these paths consist of 2 resistors. Path 1 R1
& R2 and path 2 R3 & Rx. If R2 and R3 are
equal in value.
Quarter Bridge
A quarter bridge is one where only 1 resister is unknown hence 1
quarter. All other legs have fixed resistors.
Problem #1
If the total current draw on the 12 volt power supply was 5 amps
what would the resistance of R4 be ?
R1 = 2 ohms
R2 = 2 ohms
R3 = 2 ohms
R4 = X
Applying ohms law, what would be the current flowing through
R1+R2 be ?
E= I x R
2+2=4 ohms
4/12 volts = 3 amps
3 amps from 5 amps leaves us with 2 amps, Therefore our second branch
has 2 amps of current flowing through it. If we apply Ohms law
again we can calculate Rx.
2 amps / 12 volts = 6 ohms - (R3= 2 ohms)= 4 ohms
Rx = 4 ohms.
This would be called an unbalanced bridge since one side of the
bridge has 3 amps of current and the other only has 2.
If Rx was a potentiometer (variable resistor), at what setting would
the bridge be balanced ?
If out first path has fixed resistors and the current is 3 amps we
would have to adjust the potentiometer in the second path to also equal
3 amps.
What would the resistance of the potentiometer be set at ?
3 amps / 12 volts = 4 ohms - (R3= 2ohms) = 2ohms.
Since we knew we needed a balanced bridge we could of saved some time
and just applied this formula. But then again I wanted you to work it
out :).
2 x 2 / 2 = 2 ohms
Half Bridge
Wheatstone bridges may also be formed by utilizing two legs
of the bridge with potentiometers and this is called a half bridge. In
a Half bridge 2 resistors are unknown.
As I just explained in a half bridge there are 2 known resistors and
2 unknown resistors. Are goal for the next problem will be to
balance the bridge.
Leg #1 R1 = 6 ohms
R2 = X1
Leg #2 R3 = 8 ohms R4 =
X2
Total Current = 2 amps
To balance this circuit
both path must have the same current therefore 2 / 2 = 1 amp in each
branch.
1amp / 12 volts = 12 ohms
Leg 1 R2 = 6 ohms
Leg 2 R4
= 4 ohms
If you do not feel that you understand the current resistance
relationship please review these before proceeding.
Full Bridge
A
full Bridge is one where all the resistances are unknown.
To Make understanding the series 90 controls a little easier
I will redraw the above diagram in more similar fashion to the HVAC
industry. I am also going to change the wording that I use as
well. Keep in mind, that the main propose of series 90 circuit is
to balance the current across the bridge. To accomplish this we must
adjust the resistance of resistors.
Circuit #1
With
these controls the resistors are potentiometers. These
Potentiometers have a resistance from B to W of 135 ohm's and the
resistance from either B or W to R will vary. The way this diagram
is drawn right now the Wheatstone Bridge is balanced the Resistance
between R to B and R to W is equal. The resistances are equal as well
and all 4 are 68.5 ohms.
R1
- Stat B to R = 68.5 ohms
R2 - Stat R
to W = 68.5 ohms
R3 - Mod Motor B to
R = 68.5 ohms
R4 - Mod Motor R to W =
68.5 ohms
Circuit #2
If
the temperature was to change the thermostat would also change position
and the fallowing situation would accrue. As you can see the circuits
are now out of balance, The blue circuit has far less resistance than
the red one.
R1
- Stat B to R = 35 ohms
R2 -
Stat R to W = 100 ohms
R3 - Mod Motor B to
R = 68.5 ohms
R4 - Mod Motor R to W =
68.5 ohms
Circuit
#1- R1 + R3 = 105 ohms
Circuit #2-
R2 +R4 = 170 ohms
This
Diagram is out of balance and the potentiometer on the mod motor will
move to balance out the system as fallows.
Circuit #3
Since
the 2 circuits were out of balance the logic module on the mod motor
noticed it and adjusted the feed back potentiometer on it to balance the
circuit.
R1
- Stat B to R = 35 ohms
R2 - Stat R
to W = 100 ohms
R3 - Mod Motor B to
R = 100 ohms
R4 - Mod Motor R to W =
35 ohms
Circuit
#1- R1 + R3 = 135 ohms
Circuit #2-
R2 +R4 = 135 ohms
Actual Series 90 Circuit.
In this circuit when the thermostat changes resistance due to a
change in the sensing area. The logic module in the mod motor
notices the imbalance and the motor will begin to power either the CW or CCW windings of the motor until the feedback potentiometer
that is located on the motor shaft balances out the Whetstone bridge.
Trouble
Shooting
Trouble shooting a
series 90 circuit is quite simple. We could pull out our multimeter
and take some measurements to determine the balancing point of the
motor, but that would take to long. As you just learned
any imbalance in the circuit will cause the logic module on the mod
motor to to power the motor and try to find a balance point. Use this
to your advantage. Disconnect the 3 wires and jump them out. If you
were to jump out all three the motor would travel to its mid
position, jump out R and B it will travel fully one direction
and jump out R & W it will travel fully in the other.
Todd
Legere
May 6 2000
