When does a transistor turn on

Question

I am having a very hard time understanding when exactly a transistor turns on. For example in the following circuit from Everycircuit:

Why does the transistor turn on when the push button is off, but not when it's on? The only way it seems to make sense to me is that the voltage relative to the base is "more positive" when the resistance to from the base to the positive terminal is lower than the resistance from the base to the negative terminal. For example, if I change the resistances:

From this, how would current even have anything to do with 'turning on' the transistor at all? It seems like it's only related to voltage from the below. Could someone please describe why exactly the transistor is being turned on in these conditions? Not just conceptually but with equations as well, if possible to show the current/voltage going in and such.

Answer 1

Update to DEMO Question with Analysis and Tests with variable supply.

The base R pair resistor divider ratio defines a Thevenin equivalent Voltage, Veq and Req which is not effective for limiting current when the LED is under the emitter. It is better to use the transistor as a switch and limit current with LED and Rs above the collector.

Even better is to add another transistor to act as a constant Vbe current limiting regulator called the WIDELAR Constant Current Limiter after 60's genius Bob Widlar who made Fairchild then National Semiconductor famous before the 741 and earned $1 million to retire early in Mexico, but was a "brilliant alcoholic"

This is my simulation of his brilliant design. If current is 20mA the regulator Vbe = 530mV with 100uA so the main upper transistor I= 530mV/Re= 19.6mA (Re=27V)

I applied 12 digital probes so you can see the absolute and relative voltage with current for each LED. I modeled the LEDs to match my 5mm excess stock that I have that are 20,000 mcd.

The triangle was just added to demonstrate the sensitivity to supply voltage.

Here the upper transistor is just a linear switch and the emitter R is a current sensor that when it reaches Vbe of 500mV it starts to shunt the switch off. Since the lower regulator only uses Ic=100uA, it's Vbe is not 0.7 but rather 530 mV.

This is because Transistors are Vbe exponential controlled current sinks when the emitter is grounded. So the high current gain of 2 transistors and more or less stable Vbe are used to keep the current constant. I chose it to drop rapidly when a 9V battery starts to get weak 8.5V so it lasts longer, yet it is the same current even on a car at 14.2V

Answer 2

First, let's assume the Vf for the LED is ~2V (it can vary depending upon the color of the LED, and the current, etc.). If the transistor is on, the Vbe = ~0.6V. So, if the transistor base voltage, Vb, is higher than 2.6V (2V+0.6V), the transisor and LED are on, otherwise they are off.

The two resistors bias the transistor base voltage. If Vb, without considering base current, is higher then 2.6V, the current into the base can be calculated by Ibase = (9V-2.6V)/Rupper - 2.6V/Rlower. Then Ic = Iled = B * Ibase. If Vb, without considering base current, is lower then 2.6V, the transistor and LED are off.

Answer 3

The base-emitter of the NPN requires at least 0.7V across it in the correct direction for current to flow through it. The LED is similar and in that it also requires a minimum voltage across it in the correct direction for current to flow. In this case seems to require at least 2V.

Since they are in series, that means together, they require at least 2.7V applied across them for current to flow through them. Now observe that the voltage across the 1K resistor is the same as that applied across the base emitter of the NPN and LED. If the two "diodes" (the diode-like base-emitter and the LED) are not conducting, the divider determines the voltage across them, but if they are conducting their diode drops will determine the voltage drop across them and overrides (or clamps) the voltage of the resistor divider (since the 1K resistor is in parallel with the base-emitter-LED).

Now, what is the voltage across the 1K resistor (due to the divider) when the button is pressed and when it is not pressed?

Answer 4

The emitter-base junction is a diode, with (for silicon transistors) 0.058 volts increase in voltage require to increase the emitter current by 10:1

Here is atable, of approximate currents versus Vbe.

Iemmtter Vbase_emitter

1 milliamp 0.60 volts

10 milliamp 0.658 volts

100 milliamp probably 0.8 volts or so, because of internal resistors that serve to "linearize" the transistor

1mA 0.600 volts

0.1mA 0.542 volts

0.01mA 0.484 volts

0.001mA 0.426 volts [0.600 - 3 * 0.058]