# Help me understand this load detecting circuit

I'm looking at this circuit:

Which has come from here

The claim is made that "it detects a voltage drop across a large bridge rectifier that acts as a high power shunt", which is referring to the transformer wired across D1.

However, that transformer appears to be wired in parallel with the rectifier, so wouldn't the coil on the transformer see a full 120V? I don't get how wiring the circuit this way causes the drop over the rectifier to become the supply for the transformer coil. What's going on?

Edit: From the description given, I am under the impression that the coil on the transformer is supplied with a very low (comparatively) voltage. I don't understand why this is the case. The transformer appears to be directly connected to the mains voltage in this circuit - so it should be supplied with 120v.

For those not familiar with woodworking, this device is intended to be installed in the power feed to a table saw, and will start a vacuum to collect sawdust when the saw is operating.

If you remove the bridge rectifier, you will see that the 6 volt winding of the transformer is connected in series with one leg of the AC.

As the bridge rectifier is wired it is effectively two pairs of series-connected diodes, with the pairs connected with opposite polarities, like so:

simulate this circuit – Schematic created using CircuitLab

When the saw is operating, the diodes will produce a voltage drop of about 2 volts which will be stepped up by the transformer to produce sufficinet voltage to operate the relay to start the vacuum.

• I still don't understand. Looking at your redrawn schematic, the coil of the transformer is still in series with the diodes. So why isn't 120V applied? Apr 14, 2015 at 5:41
• The transformator isn't applied to 120V. It measures the voltage that goes through the diodes - similar to a multimeter. complete your circuit diagramm with the power source on top (120VAC) and the motor of your saw on bottom - then you'll see that 120VAC are applied to the bridge rectifier and the motor behind it in series. so 2V Voltage fall of at the rectifier, the remaining 118V fall of at the motor Apr 14, 2015 at 6:54
• I'm sorry that just doesn't make sense to me. Are you saying if a measure over the two points circled here: imgur.com/zFSHjxs I will measure 2V? One of those points is clearly connected to the 120V supply and the other to ground - so it would measure 120V. Apr 14, 2015 at 7:10
• OK I think I understand now. Measuring between the first point and GROUND would show 120V, but we are measuring between the first point and just after the rectifier, which will of course be the voltage over the rectifier. Apr 14, 2015 at 7:36
• most of the current flows through the diodes and only a few volts is developed across the diodes with the freely conducting diodes there there cannot be a large voltage developed. Apr 14, 2015 at 10:45

It is the saw that regulates the current in this circuit, however much it wants it gets.

The diodes in the short-circuited rectifier bridge, conduct freely in both directions (one pair conducting current flowing upwards, and the other pair downwards) meaning there is only a small voltage present across the short-circuited rectifier bridge.

The transformer takes this small voltage (1.3 to 2.5V somewhere) and boosts it to get enough to drive the relay.