A physical explanation on how a HE trigger switch changes the output on a NPN transistor

Thread Starter

Fast Rickshaws

Joined Jun 4, 2022
Hello Everyone,

I found the forum while investigating how to understand electronic symbols in an electronic circuit that I am puzzled as to how they work in a physical action sense.

Items that may help in the explanation or understanding as to what is going on in my sensors.

I have a three wire Hall Effect sensor with a 5K pull up resistor in front of the NPN transistor that produces an output that I believe starts out as Vcc voltage and then goes to "ground" or zero volts.

The sensor tip has a Hall Effect mechanism in it that is affected by the "South" pole of a magnet. The "Digital output" switches on when the South pole of the magnet is in front of the HE sensor tip and switches off when it moves away.

The HE mechanism I believe is acting as a switch to provide a “saturation voltage low” (I sink = 20 mA) with a maximum voltage of .4 volts.

I believe, but I do not understand how or why this changes the physical path of the voltage/current in the transistor circuit which then changes the output.

The transistor I believe is a three leg NPN transistor, the Base leg I believe is connected to the HE mechanism. Or the HE mechanism is connected to the Base leg of the transistor, not sure if the wording or interaction dictates the flow of voltage or current.

I have a Vcc = either 5 volts DC or 12 volts DC (Two different applications, same sensor)

Note there is always Vcc across pins A and C

My first question, can I test the change or the signal output on pin B with a DMM or do I need an oscilloscope?

I have powered the sensor with a 24 volt capable DC power supply at 5 and 12 volts and I get either Vcc or zero volts across Pins B & C depending on where the magnet is in relationship to the sensor. I believe it is doing something. Note it does the opposite on Pins A & B.

In the sensor diagram there is a 5K resistor that is in parallel to the HE item and the top side of the resistor is connected to Vcc.

This is I believe to be the Collector side of the NPN transistor.

It also has the Pin B (Vout) after the 5K resistor and before the top of the NPN device.

Is the Pin B "output" actually the voltage that is at the Point "C" also referred to as the Collector side of the three-legged transistor?

Is there a way to trace the physical action of the HE sensor/magnet action to see how it acts (either open or closed) and then how that affects the three legs (physical action) of the NPN transistor to understand how the physical path of current and or voltage moves in the circuit? Thus changing the output.

For example does the voltage some how go directly to Pin B say Vcc and then something changes and the voltage out of Pin B goes to Zero volts, how does this physically happen and how does the voltage or current flow change in the line drawing of the circuit?

Is this case where you can label all the items in the circuit and then use electronic formulas to solve for a value at each point in the circuit based on switches opening and closing?

It also does not physically explain how the HE “switch” then affects the transistor in a physical way.

Is this the “Base” leg, does it act as a physical switch some how changing the voltage or current across the collector and emitter?

To make matters more complicated, what if wanted to install a LED which would turn OFF when the South pole of the magnet was in front of the HE tip.

Where physically would it be placed in the circuit and how does the physical path of voltage or current flow then?

Would changing to a PNP transistor make any difference? Is this maybe why one of my four sensors acts differently, three are NPN, one is PNP?

The sensor provides one digital output that is normally OFF and turns ON when a magnetic field is present. The HE switch triggers from the South pole of the magnet.

Note: The output ON means low for an NPN sensor.

Some of this makes sense theoretical sense, but physically or “action” wise does not to me.

My problem is that the sensor seems to be “ON” when I want it “OFF” and vice versa.

I thought if I could physically understand how the action of the three-legged transistor worked I could better trouble shoot how I have the system setup incorrectly.

The other question would be is the actual “output” on pin B actually a voltage signal or is it also just acting like some type of a switch that turns on and off? If it is a "switch" what then does it physically "output"? Is it considered something called a "ground trigger"?

I have no problem purchasing an oscilloscope again no idea what type or capabilities that one would need to test these items.

I have currently four different sensors, three seem to act in the same manner (NPN) and one acts the opposite way (PNP?), just trying to understand how or why this happens.

I really appreciate any help, I have tried spending hours watching videos and reading articles online, the proverbial “Light bulb” in my head will not turn on. I have a new found appreciation of electronics and electrical engineers, it is truly fascinating.

Thank you




Joined Nov 6, 2012
I'm sorry, I couldn't read the whole thing ..............
You need to study-up on how Transistors work.
Make sure that You define every single term or concept that You don't understand.

Here's an explanation of how the Output of your Hall-Sensor works ............
Hall Sensor 1 .PNG

Thread Starter

Fast Rickshaws

Joined Jun 4, 2022
Thank you,

I truly apologize for the length of my questions, sometimes written words can be too much information.

I also apologize I have spent over 80 hours researching the vocabulary and the symbols and the actual definitions of these items.

I believe I am trying to understand but a piece of aluminum that is cut with a 1" endmill with a 4" radius is as concrete as you can get, that is the base that I am operating from.

I apologize for being stupid, but you have to ask stupid questions until the actual light bulb turns on, I mean no disrespect.

I have continued to research my issue and I found this website with an example of the NPN switch use that I believe I have.

It also has symbols, labels and a lot of electronic information.

If you would scroll down to the topic "NPN Transistor as a Switch"

There is a diagram, it has what I could conceive as the HE switch on the left with Vcc voltage.

Question #1 If the HE sensor has Vcc, is the magnetic field from the South pole magnet "CLOSING" this switch?
Question #2 Given the Saturation voltage in the electrical specs of .4V is this the voltage that is being applied to the Base?
Question #3 How does the current which seems to be getting higher affect the Base?

If I may clarify what I am trying to understand in voltage terms only as a physical road map to follow:

If the South pole magnet is away from the tip of the sensor, is the switch open feeding Zero volts to the Base?

In that scenario we will for clarification have 12 Volts Vcc going thru the 5K pull up resistor where does the voltage now go?

Does the Vcc voltage go thru the 5K pull up resistor and go straight out what I believe to be the Pin B Vout and send Vcc 12 volts to what ever is connected to Pin B? Or does this get converted into a "on/off" switch output

Then what is the Voltage going from "C" collector thru "B" Base to the "E" Emitter which seems to be ground?

In the same concept what then changes when the South pole of the magnet is in front of the HE sensor tip, how does the road map or paths of voltage then change thus changing the output on Pin B?

Again I am sorry that I do not understand this and I am sorry if I am making a mountain out of a mole hill.

I am reading thru your answer and trying to trace your diagram to try to understand what is happening physically to create the two different outputs on Pin B.

Thank you


Joined Nov 6, 2012
It would appear that You have missed some basic concepts ........
The differences between Voltages and Currents is first.
Voltage can be equated to Water-Pressure in a pipe.
Current can be though of as how large the pipe is in diameter.
Current is also referred to as "Amperage", or just "Amps".
Amps and Current are the exact same thing.

So You have Pressure and Volume, or Voltage and Current, (or Voltage and Amps).

When You combine the 2 together in a mathematical equation,
you get an answer that equates to the amount of work being done,
this is called Watts.
It's a very basic Algebra Math problem.
Watts is also directly convertible into several other different "Units of Work Measurement",
such as "Horsepower" or "Joules" .
Watts or Joules are both used worldwide, and are well understood, and used in every country.
Watts, is by far, the most commonly used measurement of Power in the Electrical and Electronic fields,
although when reading Specification-Sheets for Electronic Components,
it's not unusual to see other units of measurement in use.

Electronics is around 75% Math Equations,
and around 25% inefficiencies, physical-limitations,
and other quirkiness of particular individual components.
That last 25% is why Component-Spec-Sheets are SO important in Electronics.

Here I'm adding in "Resistance-to Flow" in the above mentioned "Water-Pipe".
Resistance is measured in "Ohms".
In Ohms-Law, (for some bizarre reason), the Symbol for Voltage is "E",
the symbol for Current or Amps is "I",
the symbol for Resistance is "R", ( or the Greek "Omega" symbol Ω ).
and the symbol for Watts or Power is "W".

Ohms-Law is made easy to remember by a catchy phrase ...... "E-over-I-times-R".
If You know any 2 of the values, You can derive the 3rd missing value.

Ohms Law FLAT .png

This particular part of Ohms-Law does not include a formula for Power, or Watts,
which is simply, "Volts times Amps equals Watts".

There are other Math-Formulas that help You to quickly find other answers as well .........
Power-Watage Formulas FLAT  .png.
If You do these types of Calculations everyday, there are other convenient formulas as well ............
Ohms-Law-Pie-Chart .png
Semi-Conductors ..........
The simplest form of Semi-Conductor is the Diode.
It can be thought of as a "One-way-Valve" for Electricity,
but when a very small Voltage is applied to a Diode, nothing happens,
until that Voltage exceeds the "Forward-Voltage-Drop" rating of the Diode.
For Silicon-Based-Diodes, this is usually right around ~0.7-Volts,
this .7V will vary, and will generally slightly increase
as more and more Current is pushed through the Diode.
Some Diodes will "drop" as much as ~2-Volts under a very heavy Load, (very high Current).
This 0.7-Volts has to have a place to go, so it is converted into Heat, ( the same as a Resistor ),
so, if there are 2-Amps flowing through the Diode, and the Forward-Voltage is 0.7-Volts,
( Volts X Amps = Watts ), 0.7 X 2 = 1.4-Watts of Power will be "Dissipated" by the Diode.
Depending upon the physical construction of the Diode,
the Diode may get too Hot if it can't get rid of the Heat fast enough,
and it may fail from excessive Heat.

All Diodes, and all Resistors, have a maximum Wattage rating.

Transistors also have a "sort-of" Wattage-Rating,
but since they are only one part of a more complex Circuit,
several other factors have to be taken into account to determine what
the probable Wattage-( Heat )-Dissipation will be.
They all have a Temperature that must not be exceeded to lessen the chances of a catastrophic failure.

The Bipolar-Transistor ..............
is basically a large Diode, that can be turned on or off, by a much smaller Diode.
There are quite often shades of gray between "On" and "Off",
so the Transistor may "act-like" a Variable-Resistor.
Increasing the amount of Current going into the "Base" Pin,
proportionately increases the Current that is allowed to flow into the "Collector" Pin.
The Current allowed to flow into the Collector may be ~5 to ~40 times greater than the
Current flowing into the Base, therefore the Transistor acts as an Amplifier of Current.

This should be enough for You to chew on for a while.

Suggested YouTube viewing ...................