I was recently revisiting old lecture notes on BJT amplifiers and am a little stuck. I have been a out of touch with analog electronics for some time so I'd really appreciate it if someone stepped up and rectified anything here that I may have understood incorrectly.
Assuming a NPN BJT under CE bias:
1. A given amount of base current, say Ib, would be able to drive a corresponding amount of collector current, Ic and no more (assuming the B-E junction is forward biased and C-B is reverse biased).
2. In effect, for a given base current the BJT would limit the collector current to an amount specified by its gain, thereby acting as a current regulator.
3. As we begin to drive more and more base current into the transistor and correspondingly increasing the collector current, the collector-base voltage keeps dropping (rate of drop depends on the load(?)).
4. After this collector-base voltage hits a certain minimum value, the transistor can source no more collector current and is said to be saturated.
5. To design a circuit to reliably saturate a BJT, we need to forward bias the collector-base diode. For this, we need to drive enough current into the transistor to make sure that the output current is always able to provide enough drop across a pull-up resistor to ensure collector voltage less than base volts.
6. Why is the collector-base voltage drop less than the emitter-base volt drop. I understand the BE junction behaves almost like a diode and has ~0.7V of drop (silicon) whereas the CB junction has something like 0.4V-0.5V.
(Clearly both cannot have equal voltage drops as that would mean the net drop across the BJT is zero, ideally?)
7. It says here
8. Saturated transistors are said to act like closed switches, in that the voltage across them is almost zero and the current is maximum. I know I am missing something here because the characteristic curves show the collector current to be more in the linear region for a given base current value. Reference.
This was a long and quite possibly a trivial question but I've been struggling with these basics for some time and some help would go a long way.
Thanks.
Assuming a NPN BJT under CE bias:
1. A given amount of base current, say Ib, would be able to drive a corresponding amount of collector current, Ic and no more (assuming the B-E junction is forward biased and C-B is reverse biased).
2. In effect, for a given base current the BJT would limit the collector current to an amount specified by its gain, thereby acting as a current regulator.
3. As we begin to drive more and more base current into the transistor and correspondingly increasing the collector current, the collector-base voltage keeps dropping (rate of drop depends on the load(?)).
4. After this collector-base voltage hits a certain minimum value, the transistor can source no more collector current and is said to be saturated.
5. To design a circuit to reliably saturate a BJT, we need to forward bias the collector-base diode. For this, we need to drive enough current into the transistor to make sure that the output current is always able to provide enough drop across a pull-up resistor to ensure collector voltage less than base volts.
6. Why is the collector-base voltage drop less than the emitter-base volt drop. I understand the BE junction behaves almost like a diode and has ~0.7V of drop (silicon) whereas the CB junction has something like 0.4V-0.5V.
(Clearly both cannot have equal voltage drops as that would mean the net drop across the BJT is zero, ideally?)
7. It says here
Aren't the voltage and current interdependent? As soon as the BE voltage rises above the 0.7 V threshold, isn't there base current to drive the transistor into ON state?Remember that bipolar transistors are current-controlled devices: they regulate collector current based on the existence of base-to-emitter current, not base-to-emitter voltage.
8. Saturated transistors are said to act like closed switches, in that the voltage across them is almost zero and the current is maximum. I know I am missing something here because the characteristic curves show the collector current to be more in the linear region for a given base current value. Reference.
This was a long and quite possibly a trivial question but I've been struggling with these basics for some time and some help would go a long way.
Thanks.