Imagine this circuit and ignore the resistor values, they were just for testing purposes. V2 is a bipolar source. I have two questions.

I get how this works for the cut-off region, I mean, when V2 is 0 VBE is 0 < 0.7, therefore the transistor is off, so the collector is an open circuit, therefore 10V. So, later we have 10V at V2, all good, it activates the switch, however I don't get how it activates saturation mode, since there's 10V in both base and collector, being VBC = 0, when it should be 0.7. What am I not getting here?

The other one is, putting a motor in the collector, wouldn't it affect the 10V, or imagine, a resistor, wouldn't it act as a voltage divider? Wouldn't this require a buffer?

Thanks in advance.

 

So, later we have 10V at V2, all good, it activates the switch, however I don't get how it activates saturation mode, since there's 10V in both base and collector, being VBC = 0, when it should be 0.7. What am I not getting here?

The base isn't at 10V. It's around 0.7V which forces 9.3mA into the base. Assuming a beta of 10 for saturation mode, that will cause Ic to want to be 93mA, but with a 1k load, it can only be a tenth of that.

In saturation, the CB junction is forward biased.

 

In saturation the collector-emitter voltage can less than the base-emitter voltage.
So, in saturation, Vbe will be about 0.7V and Vce can be 100mV or less.

The other one is, putting a motor in the collector, wouldn't it affect the 10V, or imagine, a resistor, wouldn't it act as a voltage divider? Wouldn't this require a buffer?

Don't understand your question.
A motor load would be in place of the collector resistor, not in addition to it, so when the BJT saturates, nearly 10V will appear across the motor, turning it on (assuming the supply can provide 10V with the motor drawing current).

 

Imagine this circuit and ignore the resistor values, they were just for testing purposes. V2 is a bipolar source. I have two questions.

I get how this works for the cut-off region, I mean, when V2 is 0 VBE is 0 < 0.7, therefore the transistor is off, so the collector is an open circuit, therefore 10V. So, later we have 10V at V2, all good, it activates the switch, however I don't get how it activates saturation mode, since there's 10V in both base and collector, being VBC = 0, when it should be 0.7. What am I not getting here?

When the transistor is in either the active mode or in saturation, the Vbe will be around 0.7 V, not 10 V. You will drop about 9.3 V across the base resistor, which is what will set the base current.

In normal (i.e., active) operation, the BC junction is reverse-biased. As this junction begins to become forward biased, the transistor begins moving into saturation. As the transistor becomes more and more saturated, this junction becomes more forward biased, but it doesn't get so forward biased for current to flow from the base to the collector. In deep saturation the Vce will be in the 100 mV to 300 mV range, meaning that the base-collector junction is forward biased by that much LESS than the base-emitter junction is.

The other one is, putting a motor in the collector, wouldn't it affect the 10V, or imagine, a resistor, wouldn't it act as a voltage divider? Wouldn't this require a buffer?

The transistor, in the active region, will moderate the collector-emitter voltage in order to maintain a collector current that is roughly beta times the base current. When it can't lower the collector-emitter voltage any further the transistor will move into saturation.

Yes, you can think of the motor (or resistor) in the collector circuit and the collector-emitter junction as forming a voltage divider, but when intentionally using it as a switch, you are going from a very high resistance (when off) to very low resistance (when on).

 

Some ref material for your library -

https://www.google.com/url?sa=t&rct..._Edition.pdf&usg=AOvVaw2f7XXcjqmJ9jcprgAQGE7P

https://www.electronics-tutorials.ws/transistor/tran_4.html


Regards, Dana.

 

In saturation the collector-emitter voltage can less than the base-emitter voltage.
So, in saturation, Vbe will be about 0.7V and Vce can be 100mV or less.
Don't understand your question.
A motor load would be in place of the collector resistor, not in addition to it, so when the BJT saturates, nearly 10V will appear across the motor, turning it on (assuming the supply can provide 10V with the motor drawing current).

Why is that though? Why is it 100mV or less, since it works as a diode? I thought saturation mode meant base had more 0.7V than collector.

About the motor, I meant, 10 V would connect to a resistor which connects to a motor, which goes to ground. In this case the motor is an inductor, but imagine a resistor, it would divide the voltage. I don't know much about inductors, don't know what it would do in this scenario.

Thank you all for the answers!

 

Why is that though? Why is it 100mV or less, since it works as a diode?

When the transistor is in saturation mode, the CB junction is forward biased.

In your circuit, the emitter is grounded; so the emitter voltage is 0V. The CE saturation voltage depends on how much the CB junction is forward biased. If Vce is 100mV, that means the CB junction is forward biased by Vbe-100mV.

I thought saturation mode meant base had more 0.7V than collector.

It only requires that the CB junction be forward biased.

You should breadboard your test circuit and gradually increase Ib while monitoring the CB voltage so you can better understand what's going on.

 

Why is that though? Why is it 100mV or less, since it works as a diode? I thought saturation mode meant base had more 0.7V than collector.

About the motor, I meant, 10 V would connect to a resistor which connects to a motor, which goes to ground. In this case the motor is an inductor, but imagine a resistor, it would divide the voltage. I don't know much about inductors, don't know what it would do in this scenario.

Thank you all for the answers!

Don't think of the base-collector junction as a diode (in normal circuits). It is never forward biased enough to conduct as a diode would.

It is far more useful to think of saturation mode as being when there is so much base current that the transistor can't get the desired collector current (nominal current gain * base current) to flow even if it drops the collector-emitter voltage as low as it can. So the device goes from being essentially a constant current device (for a given base current) to a constant voltage device with a low Vce, making it look effectively like a closed switch.