BJT question_2 - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: September 22, 2022, 06:51:04 pm
content: In the book Electronic Devices and Circuit Theory by Robert Boylestad, I can not understand few lines.Q1) Here, to fully understand common base amplifiers why we required two sets of characteristics? what is the importance of the second characteristic?Q2) Why output set consider Output current(Ic) to an output voltage(VCB) for various levels of input currents(IE)? What is the problem if I wanted to consider Output current(Ic) to an output voltage(VCB) for various levels of input voltage(VBE?

author: TimFox
author_rank: Super Contributor
timestamp: September 22, 2022, 07:14:50 pm
content: Short answer:An amplifier comprises the input network, active device, and output network.The two networks are external to the active device.Specifically, with the load impedance connected to the output, that means that the output voltage depends on the output current as well as the input voltage and current.

author: khatus
author_rank: Regular Contributor
timestamp: September 23, 2022, 08:13:59 am
content: Why output set consider Output current(Ic) to an output voltage(VCB) for various levels of input currents(IE)? What is the problem if I wanted to consider Output current(Ic) to an output voltage(VCB) for various levels of input voltage(VBE)?

author: Vovk_Z
author_rank: Super Contributor
timestamp: September 23, 2022, 08:45:32 am
content: We consider that BJT output (collector) current is driven by base current. There is a relation between Vbe and Ic too, of cause, but it is more complicated.

author: MrAl
author_rank: Super Contributor
timestamp: September 23, 2022, 12:15:14 pm
content: Quote from: khatus on September 22, 2022, 06:51:04 pmHello there,This is what happens when you get these side track books that try to explain things in a way that makes it hard to understand and what they show you is often just a single way to view that type of circuit so when you go to do another circuit you need a whole 'nother chapter just for that.The best way to learn is to go to the root of the problem in the most fundamental way, while at the same time keeping the reference model simple.This this end, a bipolar transistor can be modeled as a current controlled current source with a particular gain we call the Beta, and it's usually between 10 and 150, but could be a bit lower or a bit higher. In any case, you choose one Beta and go with that. Since typical values are 10, 50, and 100, you could start out with 50 and go from there. This way you get a really simple circuit that will actually show you many of the characteristics of any circuit used for amplification, and even for switching but that would be an overkill for this model.With the CCCS (current controlled current source) you use one sense input as the base and the other connected to the emitter which is also one of the outputs of the CCCS. Other output terminal of the CCCS is the collector. For an NPN, the current arrow would point downward (to start with) so the more base current you provide the more current flows down through the collector into the emitter mixing with the base current and that makes up the emitter current.This concept is extremely simple. You just have to know how a CCCS works, and if you dont, you can find out easily by asking here or looking on the web.What i can promise you is you will not have a problem with bipolar transistors after you analyze a couple circuits with the CCCS in it. The basic operation will stand out and you can apply that directly to many amplifier circuits without modification.One modification that is common though is to add a diode in series with the base to represent the base emitter diode. That in turn can be modeled as a voltage source, but dont worry about that until you've worked with the stand alone CCCS as the transistor.If you need help with this i would be happy to help or you could ask more about this in this forum. It just so happens that i am currently working with another student on this very same topic and using this very model.

author: khatus
author_rank: Regular Contributor
timestamp: September 23, 2022, 02:36:46 pm
content: Thank youQuotefor sure

author: the_cake_is_a_lie
author_rank: Contributor
timestamp: September 23, 2022, 06:12:03 pm
content: It's poorly worded and I had to think for a minute of what it's actually trying to say. Don't overthink here.The Q Point of Ic and Vcb you can set with resistor biasing. This is the driver. Maybe with that you calculate an expected output of 5x the input voltage but in practice it's 2x. Why? You have to consider the input and output impedance of the transistor, the power supply, as well as the load. These are the input and output referred to. If your collector's supply is +9V, can't exactly output more than +9V. A load of, say, 50 ohms against a circuit with a medium to high output impedance can't obtain the full 5x.Same idea as two resistors in series forming a voltage divider in DC. If the same resistance, the second resistor gets half the input voltage but maximum power transfer. Sometimes that's what you want. In 5V and 3.3V digital logic, you'd probably rather have the maximum voltage transfer at the cost of power. In this case you want a small resistor as the input and one 10x or greater as the output. 10/(10+1) is a voltage transfer of 91%.Here we probably can't change the load. One way of solving the low gain problem is with another transistor in between the first transistor and the load as a common collector to serve as a buffer. This common collector has no voltage gain but a high input impedance and a low output impedance. One transistor for voltage and/or current gain and another as a buffer to enable maximum voltage transfer to the load. You have to consider the transistor setup, as well as the input and the output to which it's driving.

author: MrAl
author_rank: Super Contributor
timestamp: September 24, 2022, 03:34:02 am
content: Quote from: khatus on September 23, 2022, 02:36:46 pmOk, so here is an image you can look at. The top circuit is the actual transistor circuit, the bottom circuit is the same circuit with a CCCS put in place of the transistor. The would be the starting point.Does this make sense to you, and if not, what does not make sense.