Industrial Electronics is the use of electricity, electrical components and high and low voltage circuits to control behavior of electrical, mechanical and pneumatic systems.
In this example we are building a current sensing switch.
The current draw of a 120v circuit (the primary circuit) is monitored by a low voltage electrical circuit.
When the current of that primary 120v circuit reaches a predetermined level based on the low voltage logic circuit settings and parameters, a switch is closed and a second 120v circuit is activated.
The specific application we are designing for is power tool monitoring for purposes of turning on a vacuum to collect dust and debris from the power tool activities.
The power tool is the primary circuit and the vacuum is the secondary circuit.
They may be operating on the same branch circuit from the break (the same 120v source) OR they may be on two different branch circuits.
While we are designing for a power tool and vacuum, it could just as easily be a power tool and a light or a ventilation fan or anything else that may be beneficial to turn on when a power tool is turned on. OR the trigger tool could be something other than a power tool. It could be hair dryer, or a 3D printer or a phone charger.
To build this circuit we are designing for two parameters - a Trigger Current on the primary circuit that activates the secondary circuit and a post primary circuit turn off delay for the secondary circuit. (ie when the tool turns off the vacuum might run for an additional 5 seconds, for example).
Trigger Current - the amount of electrical current required to activate the switch for the secondary circuit
Post Primary Circuit Shut off Delay for Secondary Circuit - A time delay at the end of primary circuit use that extends the time the secondary circuit runs after the primary is no longer active.
NOTE: we are not currently designing for an initial time delay that would separate the Tool on time from the vacuum on time. If both the tool and vacuum (in this scenario) are on the same 120V branch circuit and not separate 120V branch circuits -- and a tool and vacuum startup demand for current trips the breaker, either two distinct branch electrical circuits will be required (two breakers controlling two different branch circuits - (one for the tool and one for the vacuum) OR an additional initial time delay component would need to be added to the low voltage logic circuit - and that is beyond the scope of these instructions at this time.
The Trigger Current (in amps) combines with a “burden resistor” and Ohms law to create a voltage that is sufficient for logic circuit control.
As trigger current goes up the burden resistor value can come down to maintain a minimum voltage required for circuit operations.
Excessive voltage is not a major issue. It can be bled off in the form of heat to optimize circuit control, but it is beneficial to change the burden resistor to match the tool for optimal performance and it may be required to match it for any performance at all for low current tools.
Time Delay in seconds is a function of a capacitor, a Bleed resistor, and a voltage equation including natural log curve and two voltages that has been empirically developed to calculate time delays.
We need to get to a consistent logic circuit voltage above 3.9 volts DC quickly for the following two reasons
To turn the vacuum on via a a MOFSET ( voltage sensitive switch) followed by a Mechanical Switch. The MOFSET closes/completes a circuit at 2.5 volts DC but it consumes 1.4 volts to operate. Thus, 3.9 volts DC is required input to result in a 2.5 v DC trigger and “coincidentally” the mechanical switch needs a 2.5v DC trigger to complete the circuit. NOTE: it’s rates as a 3 volt switch, but they will typically operate 20% below rated value.
To establish a known and consistent peak voltage sufficiently above 3.9V to prevent the switch from fluttering off and on, and with sufficient head room to allow for consistent drain time calculations and control.
3.9 volts DC can be derived from 5.46 volts AC and a rectifier. Thus our minimum goal for an AC voltage for logic circuit control starts at 5.46 volts and it actually should be a little higher for “headroom” and added assurance the switch does not flutter.
So imagine this scenario...
6.5 VAC is generated by a tool and a burden Resistor
4.6 VDC results from converting AC to DC via a rectifier ( VDC = VAC / 1.4)
4.6 VDC fills a capacitor and puts pressure of a MOFSET that opens with 2.5 VDC , and consumes 1.4 Volts
3.2 VDC makes it to the mechanical switch, which is predesigned to complete the circuit at 3VDC but will do it as low as 2.5 VDC
When the primary circuit is de-energized, the capacitor continues to provide energy to the secondary circuit. Voltage falls from 4.6 VDC to 2.5 VDC , the secondary circuit disengages and remaining energy is drained off another way.
We need 3.9 volts DC for basic MOFSET switch and mechanical switch operations. That consists of 2.5V for triggering the MOFSET and Mechanical switch, and 1.4V thats are lost using the MOSFET.
To get 3.9 volts DC we need 5.46 volts AC at a minimum to start the circuit (DC voltage x 1.4 multiplier = VAC), but that provides us with no head room for fluctuations. So lets add 20% for overhead and margin, and think about an initial goal of needing to sense or generate 6.5volts AC on this logic circuit for it to operate.
Ohm's law is V = IR
V = Voltage in volts
I = Current in amps
R = Resistance in ohm
If we think about Voltage as comparable to “inches of water” , a non intuitive way to reference pressure from water at the bottom of a bath tub, tank, or swimming pool, and Current as an amount of water that flows through a drain per second, we can think of Resistance as the size of the drain that allows a given flow for a given pressure.
6.5 Volts = Current x Resistance
and below we can see how the current and resistance can vary inversely to achieve the same outcome...
6.5 Volts = 1 amp x 6.5 ohms
6.5 volts = 6.5 amps x 1 ohm
In this case we know the desired voltage is 6.5 Volts - and we can create a range of values for tool amperage from known values for those -- and then we can solve for a compatible resistor for a given tool amperage and 6.5 volt requirement.
However, there is one final twist.
We are measuring tool current using a round magnetic sensor (an AC current transformer) with a 2000:1 down scaling ratio. If the tool was drawing 2000 amps AC we would only see 1 amp on our logic circuit. If a tool is drawing 1 amp we only see 1/2000 of an amp.
Thus our final equation for building a lookup table for a 6.5v AC initial target is:
6.5 Volts = ( Tool Current / 2000 ) x R
R = 6.5 Volts / (Tool Current / 2000 )
The following is a table that specifies a burden resistor for a given tool amperage ASSUMING a 6.5 Volt AC / 4.6 Volt DC target
Target VAC / VDC Logic Circuit Voltage for current sensing switch operation
Tool Type
Tool Amps
Secondary Amps
Minimum Burden Resistor Needed (Ohms)
(for 6.5VAC)
Recommended Standard Resistor (Ohms)
6.5VAC / 4.6VDC
Small Rotary Tool
0.5A
0.00025A
26000 Ω
33k (33000 Ω)
6.5VAC / 4.6VDC
Sander
3.0A
0.0015A
4333 Ω
4.7k (4700 Ω)
6.5VAC / 4.6VDC
Router
8.0A
0.0040A
1625 Ω
2.2k (2200 Ω)
6.5VAC / 4.6VDC
Heavy CNC / Saw
15.0A
0.0075A
866 Ω
1k (1000 Ω)
6.5 VAC is generated by a tool and a burden Resistor
4.6 VDC results from converting AC to DC via a rectifier
4.6 VDC fills a capacitor and puts pressure on a MOFSET that completes a circuit above 2.5 VDC , and consumes 1.4 Volts
3.2 VDC makes it to the mechanical switch, which is predesigned to complete the circuit at 3VDC but will do it as low as 2.5 VDC
When the primary circuit is de-energized, the capacitor continues to provide energy to the secondary circuit. Voltage falls from 4.6 VDC to 2.5 VDC , the secondary circuit disengages and remaining energy is drained off another way.