Wheatstone Bridge / Series 60 & 90 Controls

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Wheatstone Bridge / Series 60 & 90 Controls

Post by juster » Fri Jan 14, 2011 11:15 pm


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 (13.72 KiB) Viewed 712 times
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.
quarter_bridge.gif (2.88 KiB) Viewed 712 times
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 ?
* 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 (308 Bytes) Viewed 712 times
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.
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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
series901.gif (2.99 KiB) Viewed 712 times
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.
Wheatstone2.gif (3.85 KiB) Viewed 712 times
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.

Circuit #3
Wheatstone3.gif (3.86 KiB) Viewed 712 times
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.
Wheatstone3.gif (3.86 KiB) Viewed 712 times
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.
modmotor1.gif (9.19 KiB) Viewed 712 times
Todd Legere
May 6 2000

Most Valued Contributor
Posts: 278
Joined: Wed Jan 12, 2011 9:10 pm

- Wheatstone Bridge / Series 60 & 90 Controls

Post by nomadpeo » Sun Jan 30, 2011 8:57 am

been a while since i looked at these diagrams. good refresher. gives me an idea for a class i will be teaching. thanks.

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