- #1
Saladsamurai
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Hello all I am a mechanical engineering student taking an introductory circuits course. I find the things we are doing in lab quite interesting, but I really lack a good understanding of what we are doing. We are currently building logic gates using BJTs. I have some basic questions regarding these. I get the basic idea (I think): We set up a circuit that has some constant source voltage connected to it. Then, if we give it an input voltage of proper magnitude, a *switch* is thrown allowing current to flow from our source voltage to ground. The output is a measured voltage that is either high or low (1 or 0).
I know it really isn't a switch in the mechanical sense of the word, but I don't really want to know about all of the intricacies of doping, so I will just think of it as a switch. Here is the schematic I am looking at of a BJT in such a circuit:
Figure 1
Now I really need to clear up what I think is going on here in this circuit (from my reading). The +6V is my constant input voltage connected to the collector C of the transistor. Since the circuit is "open" from C to E, no current can flow and hence the voltage drop from E to ground is zero. Now if we apply the appropriate voltage at B, that *switch* I spoke of is thrown and shorts C to E so that current can now flow. The output is what is labeled here in Figure 1 as Vout, which is measured with respect to ground and should be 6V if the BJT offers no resistance.
Now here is one of my questions that I need to have answered before I continue on to logic gates. The 4.7 kΩ resistor is referred to in my handout as the pull-down resistor.
1) Why is called the "pull-down" resistor?
2) What function(s) does it serve? Why can't we just have a wire from E to ground instead? Is it so the current doesn't go to "infinite"? I am picturing the circuit without that 4.7kΩ resistor as just a voltage source with a wire connecting (+) directly to (-).
I will post logic gate questions once I feel comfortable with this stuff here. Any thoughts are appreciated.
I know it really isn't a switch in the mechanical sense of the word, but I don't really want to know about all of the intricacies of doping, so I will just think of it as a switch. Here is the schematic I am looking at of a BJT in such a circuit:
Figure 1
Now I really need to clear up what I think is going on here in this circuit (from my reading). The +6V is my constant input voltage connected to the collector C of the transistor. Since the circuit is "open" from C to E, no current can flow and hence the voltage drop from E to ground is zero. Now if we apply the appropriate voltage at B, that *switch* I spoke of is thrown and shorts C to E so that current can now flow. The output is what is labeled here in Figure 1 as Vout, which is measured with respect to ground and should be 6V if the BJT offers no resistance.
Now here is one of my questions that I need to have answered before I continue on to logic gates. The 4.7 kΩ resistor is referred to in my handout as the pull-down resistor.
1) Why is called the "pull-down" resistor?
2) What function(s) does it serve? Why can't we just have a wire from E to ground instead? Is it so the current doesn't go to "infinite"? I am picturing the circuit without that 4.7kΩ resistor as just a voltage source with a wire connecting (+) directly to (-).
I will post logic gate questions once I feel comfortable with this stuff here. Any thoughts are appreciated.