I just want to thank everyone for the support! You've all helped me out a LOT and I hope that in the future I can achieve my goal with transistors! That is, look at any transistors specs and be able to saturate it and make a class-A amp out of it.
Well, you misunderstand Audioguru.You've said that with the 2N3904, the base current should be 1/10th that of the collector current in other posts as well. But, I thought that it was for saturation. I recall reading in your other posts that you needed this to saturate the transistor, is it supposed to help it be a Class-A amp?
And this mean, that current that is flow through voltage divider should be at least ten times larger the the base current.Its base current is about 1/10th the voltage divider's current so the different hFE of each transistor has a small effect on the divider's voltage.
Also, you said "The graph" and I would like to know which one you're referring to. The current gain over collector current one?
Typical Vce(sat) is equal 0.2V so 1V is very close to saturation.What should Vce be then if It shouldn't be 1V What should Vce be then if It shouldn't be 1V
No, because in saturation Ib=Hfe*Ic don't hold any more.Also, you mentioned that hFE isn't important in saturation, where I think that it could be. Since transistors act like a switch in saturation, couldn't you use hFE to calculate how much current gain you get for your transistor switch?
I thought I did.Well, you misunderstand Audioguru.
I thought that was the graph as well.
Interesting, you cleared that up nicely. So actually the "On Characteristics" represent saturation of the transistor? According to the specs, how would I find a good value for Vce if I want it as a Class-A amp? In addition, what would be a good typical value, aside from that of the specs Vce?Typical Vce(sat) is equal 0.2V so 1V is very close to saturation.
When you set the bias point at Vce=1V then for positive swing you get form Vce=1V to Vcc. And for negative swing of a output voltage, you get from Vce=1V to 0V (Vce(sat)).
And if you set bias point at Vcc/2 then positive swing is equal negative swing.
Positive swing form Vce=4.5V to 9V and negative swing from Vce=4.5V to 0V
Ahh...ok. Thanks for clearing that up too.No, because in saturation Ib=Hfe*Ic don't hold any more.
Ok, I'll keep that about Vce in mind when I do more designs.The spec's for a 2N3904 transistor show the hFE when the Vce is 1V and 5V so the transistor is not saturated.
They show the max saturation voltage drop when there is plenty of base current that is 1/10th the collector current.
You want the Vce of a class-A amplifier transistor to be half the supply voltage so that the collector can swing equally in a positive direction and in a negative direction then its output is max with low distortion. You must allow for the DC voltage across the emitter resistor that effectively reduces the output swing.
hFE is beta which is current gain, not voltage gain.
current that is flow through voltage divider should be at least ten times larger the the base current
I think that there is some mixing up of saturated with non-saturated designs in this thread.So with the voltage divider, The Electrician said that I don't need to calculate the current output of the voltage divider. How am I supposed to know if the base current is 1/10th of the collector current? I suppose it's just like power supplies, where a power supply can supply 1A, but the load only needs 500mA (example). Being the case, the voltage divider can supply so much current but the base will take what it needs. Am I right?
Thanks, The Electrician.I think that there is some mixing up of saturated with non-saturated designs in this thread.
For a non-saturated design, the base current isn't going to be just 1/10 of the collector current.
The base current is related to the collector current by the β of the transistor.
First determine the collector current, Ic, you need based on the impedance of the load you need to drive. Then the base current is Ib = Ic/β; use the minimum β at your operating point. Then select the bias divider resistors so that the current in the divider is 10 times Ib.
How do we set the current in the divider?
If Vcc is known, the the current in the divider is Vcc/(R1+R2), where R1 and R2 are the divider resistors. That is, the total resistance of the divider is R1+R2, so (R1+R2) = Vcc/(10*Ib). Then you have to select R1 and R2 individually such that their ratio (while keeping their sum constant) gives the desired output voltage, ignoring the effect of Ib; in other words, assume Ib = 0 at this point. The error due to that assumption will be small enough to ignore.
It's just a little algebra.
I guess I'm not doing so bad for a kid my age...I'm pretty young.It's just a little algebra.
by Jake Hertz
by Duane Benson
by Duane Benson