How does BJT work in this circuit?

[ToonBlue]

Homework Statement

This is not a homework question. I am trying to learn how BJT work as a switch so I come out with circuit to understand how BJT functions as a switch. I use mulitsim by the way to try out these circuit

2. Relevant diagram

The Attempt at a Solution

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I understand that if I want to operate BJT as a switch , the transistor needs to be able to fully turn OFF(cut-off region) and turn ON (saturated).

This is from BJT Note

Cut off region

• The input and Base are grounded ( 0v )

• Base-Emitter voltage VBE < 0.7v

• Base-Emitter junction is reverse biased

• Base-Collector junction is reverse biased

• Transistor is “fully-OFF” ( Cut-off region )

• No Collector current flows ( IC = 0 )

• VOUT = VCE = VCC = ”1″

• Transistor operates as an “open switch”

Saturation region

• The input and Base are connected to VCC

• Base-Emitter voltage VBE > 0.7v

• Base-Emitter junction is forward biased

• Base-Collector junction is forward biased

• Transistor is “fully-ON” ( saturation region )

• Max Collector current flows ( IC = Vcc/RL )

• VCE = 0 ( ideal saturation )

• VOUT = VCE = ”0″

• Transistor operates as a “closed switch”

In the circuit above , I am not really sure how it work? when R5 is 10k ohm , BJT is in the on state and when R1 is increased to 20k ohm , BJT is in the off state

To fully turn on the BJT , Vbe must be greater than 0.7. I apply voltage divider rule , R6/(R6+R5) * 12 v = 1.091V. which is more than 0.7 so it turns on the BJT. Am I correct?

I am still learning how BJT work so I simulate this in Multisim. I am new quite new to BJT

 

[NascentOxygen]

• The input and Base are connected to VCC

I wouldn't write it like that; it suggests a direct connection (and a direct wire is sure to destroy the transistor). I prefer something like "base needs ample current from Vcc via a resistor"

You've provided a light bulb, but show only one wire connecting to it. The key to the explanation of this BJT switching circuit hinges on that missing wire! Can you mark it in?

 

[ToonBlue]

I have insert the circuit with led in it and change the resistors location.

When R2 is 10 ohm , LED1 doesn't light up. When R2 is increased to about 40k ohm , LED1 light up.

May I know how this circuit work? Why does the increase in R2 will lead to LED1 begin light up?

 

[rude man]

View attachment 91051

I have insert the circuit with led in it and change the resistors location.

When R2 is 10 ohm , LED1 doesn't light up. When R2 is increased to about 40k ohm , LED1 light up.

May I know how this circuit work? Why does the increase in R2 will lead to LED1 begin light up?

Left circuit: current through R1 ~ 12V/220K = 0.055mA drops only 0.055 mA x 10 ohms = 0.55 mV across R2, not nearly enough to turn on the transistor which requires about 700 mV b-e volts to turn on.
Right circuit: current thru R1 ~ 11.3V/220K = 0.052 mA but the current into R2 is now only 0.7V/40K = 0.018 mA so the remaining current = 0.052 mA - 0.018 mA = 0.034 mA can flow into the transistor base.

You should study the relationship between i_c and V_be for a bjt. Essentially it says that very small i_c flows until V_be approaches at least 0.5V: ln (i_c) ~ V_be/26mV at room temperature.

 

[NascentOxygen]

I have insert the circuit with led in it and change the resistors location.

When R2 is 10 ohm , LED1 doesn't light up. When R2 is increased to about 40k ohm , LED1 light up.

May I know how this circuit work? Why does the increase in R2 will lead to LED1 begin light up?

The LED carries the collector current, the LED being in series with the collector lead. When the switch is well-designed you'll see plenty of collector current flow whenever there is plenty of base current. The base gets its current through R1. Resistor R2 can control the base current: if R2 is relatively low resistance it can divert most of R1's current to ground and leave very little to go into the base, thereby preventing collector current from flowing.