I have done some research and found a transistor biasing textbook type PDF and it cleared some things up except this. People tell me that a resistor in series with a voltage source will cause a voltage drop, but they don't tell me how to calculate this voltage drop they just tell me to use Ohm's Law to calculate the current through the resistor which I have known for quite a while. What happens to the voltage? If there is a change in current there must be a change in voltage because energy cannot be created nor destroyed! Any help? I couldn't find anything on this anywhere and this was probably my downfall! Please... 
Ohm's law works either way, but obviously you need to know two of the parameters to calculate the third.
In the case of the voltage drop along the resistor, you have to know the current drawn; multiply x resistor and you get thems = voltage drop.
For example, assume an ideal (absolutely constant) 20V source and a 5 Ohm resistor. You're drawing 1 amp from it. Voltage across the resistor is 1 amp * 5 Ohms = 5 Volts, so after the resistor you only get 15 volts.
Every real-life voltage source will have some amount of internal resistance, so that its voltage drops like in our example as you're drawing current.
Then of course there's the next case where you don't know the current you're drawing from the source plus resistor, but you know the resistance you are connecting.
In that case, calculate the effective resistance as seen from the voltage source as the sum of resistors. In our example, if you have the 20 Volts source and the 5 Ohms resistor, and you connect a load resistor of let's say 10 Ohms, the source sees (5 Ohms + 10 Ohms) = 15 Ohms. Current is therefore 20 Volts / 15 Ohms = 1.33 Amps. Voltage drop on the 5 Ohms is therefore 1.33 Amps * 5 Ohms = 6.67 Volts. 20 Volts - 6.67 Volts leaves 13.3 Volts on the load resistor.
Next lesson waiting for next question ;-)
In the case of the voltage drop along the resistor, you have to know the current drawn; multiply x resistor and you get thems = voltage drop.
For example, assume an ideal (absolutely constant) 20V source and a 5 Ohm resistor. You're drawing 1 amp from it. Voltage across the resistor is 1 amp * 5 Ohms = 5 Volts, so after the resistor you only get 15 volts.
Every real-life voltage source will have some amount of internal resistance, so that its voltage drops like in our example as you're drawing current.
Then of course there's the next case where you don't know the current you're drawing from the source plus resistor, but you know the resistance you are connecting.
In that case, calculate the effective resistance as seen from the voltage source as the sum of resistors. In our example, if you have the 20 Volts source and the 5 Ohms resistor, and you connect a load resistor of let's say 10 Ohms, the source sees (5 Ohms + 10 Ohms) = 15 Ohms. Current is therefore 20 Volts / 15 Ohms = 1.33 Amps. Voltage drop on the 5 Ohms is therefore 1.33 Amps * 5 Ohms = 6.67 Volts. 20 Volts - 6.67 Volts leaves 13.3 Volts on the load resistor.
Next lesson waiting for next question ;-)
So if I wanted to bias Q1 at .7V and 10mA I would divide 120V by 10mA and make this value "A". And then A*?=.7V so -.7V/A="B". And then subtract B from A and make the result=R2 and make A=R1. Will this work? In case I made a mistake I have made an equation to represent this process:
(Vs/Ib)=A, (Vb/(Vs/Ib))=B, ((Vs/Ib)-(Vb/( Vs/Ib )))=R2, A=R1
This may not be right as to how I was doing lots of stuff other than math over the summer and I'm kindof rusty.
This is assuming that Q1 is NPN with the same biasing arrangement.
I loooooove making equations. every time I find a problem that requires more than one ohm's law equation or some strange resonent circuit thing, I make an equation. This is probably good for me, I think. So if you need an equation send me all the info and I'll make one!
(Vs/Ib)=A, (Vb/(Vs/Ib))=B, ((Vs/Ib)-(Vb/( Vs/Ib )))=R2, A=R1
This may not be right as to how I was doing lots of stuff other than math over the summer and I'm kindof rusty.
This is assuming that Q1 is NPN with the same biasing arrangement.
I loooooove making equations. every time I find a problem that requires more than one ohm's law equation or some strange resonent circuit thing, I make an equation. This is probably good for me, I think. So if you need an equation send me all the info and I'll make one!
hmm. I have just noticed that this design won't work over a range of currents because the biasing has to be so precise... I will have to count the voltage fluctuation range and that will be the varying voltage through Q1 And I will have to take this into account when biasing because this will be the voltage source of the bias...Am I right? This will be a bit hard, but I will try. Any help is appreciated!
but I will try. Any help is appreciated!- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.