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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.

**-*modutrol.jpg*/-*

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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.

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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.

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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.

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A quarter bridge is one where only 1 resister is unknown hence 1 quarter. All other legs have fixed resistors.

**-*quarter_bridge.gif*/-*

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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 ?

* I x R

2+* 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 - (* 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 - (* 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 .

**-*ronertwo.gif*/-*

2 x 2 / 2 * 2 ohms

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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.

**-*Full_bridge.gif*/-*

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.

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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.

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**-*series901.gif*/-*

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

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\

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.

**-*Wheatstone2.gif*/-*

R1 - Stat B to R * 35 ohms

R2 - Stat R to

This Diagram is out of balance and the potentiometer on the

mod motor will move to balance out the system as fallows.

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.

*-* #*/-*

**-*Wheatstone3.gif*/-*

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

*-* Series 90 */-*

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.

**-*Wheatstone3.gif*/-*

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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.

**-*modmotor1.gif*/-*

Todd Legere

May 6 2000