Thanks so much, for all your replys, the equations are a real big help, I now understand how I arrive at the values I need to design my circut now.. I will post results, thxHere are equations you can study to bias your given circuit.
Voltage at the collector terminal ref. to ground, is called (VC).
Volt. at the base term. ref. to ground is called (VB).
Volt at the emitter term. ref. to ground is called (VE).
Current flow from the supply into the collector is called (IC).
Current flow through the base into the emitter is called (IB).
Current flow from the emitter to ground is called (IE).
Supply voltage is called VCC.
REdesignating your schematic:
R2 will be called (RC).
(RE) will be chosen, ..(it originally was R4.)
R1 and R3 will be calculated.
Equations:
1). VC = (VCC / 2)
2). IC = (VC / RC)
3). RE = (2 to 20) times less than RC. (choose a value)
4). VE = (IC x RE), where you made a choice for RE in above step.
5). VB = (VE + 0.65v)
6). R3 = (10 x RE)
7). ID = (VB / R3)
8). R1 = [(VCC - VB) / ID]
I wonder if this has done you so much of a favour. Someone has given you a tool-kit of formulae (or at any rate rules-of-thumb) which may let you make a working circuit. Do you actually understand why these rules are as stated?Thanks so much, for all your replys, the equations are a real big help, I now understand how I arrive at the values I need to design my circut now.. I will post results, thx
Hi,I wonder if this has done you so much of a favour. Someone has given you a tool-kit of formulae (or at any rate rules-of-thumb) which may let you make a working circuit. Do you actually understand why these rules are as stated?
One could also argue over whether these given rules are really good practice, for instance in the arbitrary choice of RE, but maybe it's best not to confuse matters further.
Thx hobbyist,Hi,
Electronics can be very confusing learning, when you are just new to the field,
sometimes the best encouragement to stick with it comes by hands'on, training, actually getting a circuit to work. (That's why I'll show pics. or a video of it working).
That's my experience with these forums are about, since I'm not a professional as most are on these boards,
I can help the newcomers get a chance to be encouraged, by designing a circuit that works, enough for them to want to continue on in there learning on the subject.
Since I am not a professional in this, I can relate to the students on here as I am a circuit designer on the most BASIC level of electronics, learning curve.
That's why my circuits are so simplistic in design, it all comes from working at the beginners level in this field.
Then as these students progress, they come into you guys' level, for more professional detail in a design.
@ stinky : I must apologise to you - I had assumed that you were one of the many young students (typically from certain eastern countries) who come seeking answers to assignments, seemingly without making much effort of their own. In some cases the aim appears to be to get answers for on-line assessments, with no aim other than to get better grades. In my opinion these students often have little real understanding of their subject and do not seem to wish to acquire any. The imperative is to get answers so as to avoid failing their course. This forum attracts a large number of enquiries of that kind, which may lead to some of us developing rather negative attitudes on the subject.Thx hobbyist,
Just to confirm to all I am very much a newcomer to electronics, (Its obvious I guess), I have just done a complete 180 career path change.
I have only started studying again this year to become an electrician...
After being out of school for over 20 years & the oldest in my class never that great at maths I'm struggling with the theory, but am top of my class at all practical assessments.
I realize that some of my questions may seem like no brainers for most, I can assure u all that they are genuine & I appreciate all help given as I'm trying my best to absorb all I can, to pass my cert & progress with my new career.
Then hopefully one day be on the electronics level as most posters on this forum..
Leaving aside whether the values you start with were the right ones, the correct result of 2.25V/200uA is 11 250Ω or 11.25kΩ.R2=(VR/IR2)=(2.25v/200uA) R2=11.25m ohms
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