amplifiertransistors
Could somebody please check to see if what I have done is right? I posted a similar question a few days ago and someone said V(CE) is equal to V(CC)/2. Is it ok to do this?
So I got V(CEQ) = 6V and then I(C)=1mA
I tried to solve I(B) on the left side first but got a negative value for it? So that can't be right.
I don't need any help with part b)
Thanks
Yes. Give us the circuit and we will be able to spec a transistor for you.
AND a photo to go on with (and comment on thos below).
A copper side view of PCB as well would help muchly (for the terminally enthused to start in on).
If you have a working unit so much the better.
I you can tell us the supply voltage, power taken and frequency it would help greatly. That would be enough to hazard a good gusstimate. Even voltage would possibly be enough !!!
BUT the more you tell us the better.
For example, telling us that it was in a metal T03 case on a heatsink about xxx big and was made in 2008 or ... would help.
eg - like this? Yes? :-)
May be used elsewhere in system, too? Board level parts supplier here says
Item # Description Price
SS8G-PCA080 Generator Board, Acousti-pulse 6580 $91.91
SS8G-PCA120 Generator Board, Acousti-pulse 6580 $91.91
and... . Transistor is about certain to be the same given the peep of part number seen.
Item # Description Price
SS8G-PCA120X2 Generator/Driver Acousti-Pulse/Dual $165.44
You didn't tell us you knew board part numbers !!!! More photos
Because RL is not given, we have to make a suitable assumption. One common rule is to make VCE equal to the remaining DC voltage drops in the path. That means:
VCE=5V and IC*(RL+RE)=5V.
From this with IC*RE=2V we can derive a suitable RC value of RC=3V/1mA=3 kOhms.
For the remaining part it is recommended to calculate the resistors R1 and R2 based on the assumption that the current through R2 is app. 10 times larger than the base current.
Best Answer
Task 1 - find Ib.
We know that VBE is 0.7V (given).{quick check - is this possible with the given values? - without the base connected R2 would produce a drop of about 0.75V - this is a good result because we can always drop this voltage by taking current into the base}.
For VBE = 0.7 and R2 = 6.3k by Ohm's law I(R2) = 0.7/R2 = 0.11111 mA
The 11.3 volt drop (12 - 0.7) across R1 gives us a current of I(R1) = 11.3/93.7k = 0.120598mA
By Kirchoff's law the difference in current must be the base current 0.12 - 0.11 mA (2dp) = 0.01mA
Task 2 - the collector current
We know that current gain is 100 (given)
So Ic = 100Ib = 100 * 0.01mA = 1mA
Task 3 - find the voltage at the collector.
By Ohms law the voltage drop across Rc = Ic*Rc = 1(mA) * 6k = 6V
So the voltage at the collector (with respect to ground is 12 - 6 = 6V