@rayma Agreed that OpAmp solution is easy option but its still interesting to do this with discrete components using a few 2N3904. In fact if I were doing this I would be tempted to do two versions one with an OpAmp and one with a few 2N3904 and compare performance of both designs. The comparison could be done between physical built designs and using the simulator.
I think the output voltage maximum peak to peak voltage swing mentioned of 3 volts can easily be achieved by a 2N3904 design.
I think the output voltage maximum peak to peak voltage swing mentioned of 3 volts can easily be achieved by a 2N3904 design.
@rectifryer "My main technical hangup is deriving the internal transistor resistance, sometimes referred to as Ree. Wouldn't it just be the Voltage drop divided by the base current? Vforward/ib?"
OK Ree no idea about that but it is mentioned, have a look at this link voltage divider bias Further down it explains "voltage divider bias" and gives the equations for calculating R1 R2 RL and Re
OK Ree no idea about that but it is mentioned, have a look at this link voltage divider bias Further down it explains "voltage divider bias" and gives the equations for calculating R1 R2 RL and Re
As for the AC gain Av - for a transistor operating with large hfe Av = Rc/Re as long as Av is low.
To understand this for an ideal transistor think of this "voltage divider bias" circuit as a Emitter-Follower where the Emitter AC voltage follows the input AC voltage. Now its clear that
ie = Ve/Re and since ie flows through Rc then Vc/Rc = ie so Vc = VeRc/Re so Vc/Ve = Rc/Re since Ve follows Vin and Vc is Vout then Vout/Vin for AC signal equals Rc/Re and we can write Av = Rc/Re
For low distortion the key is to set an operating point (using bias resistors) to work at maximum hfe, also keep Av low for a stage. Assuming three stages in series, for the first two Rc needs to be low compared to the next stage R1//R2 for the output stage Rc can be higher because it feeds into a high impedance load.
So hfe >> Av needs to be true. Typically Stage 1 might have an Av of 10 stage 2 might have an Av of 10 stage 3 should have an Av of 3 - Total Av AC input to AC output then is 300 and will remain stable.
Any distortion will be because hfe is finite.
To understand this for an ideal transistor think of this "voltage divider bias" circuit as a Emitter-Follower where the Emitter AC voltage follows the input AC voltage. Now its clear that
ie = Ve/Re and since ie flows through Rc then Vc/Rc = ie so Vc = VeRc/Re so Vc/Ve = Rc/Re since Ve follows Vin and Vc is Vout then Vout/Vin for AC signal equals Rc/Re and we can write Av = Rc/Re
For low distortion the key is to set an operating point (using bias resistors) to work at maximum hfe, also keep Av low for a stage. Assuming three stages in series, for the first two Rc needs to be low compared to the next stage R1//R2 for the output stage Rc can be higher because it feeds into a high impedance load.
So hfe >> Av needs to be true. Typically Stage 1 might have an Av of 10 stage 2 might have an Av of 10 stage 3 should have an Av of 3 - Total Av AC input to AC output then is 300 and will remain stable.
Any distortion will be because hfe is finite.
That is because you need to use the old formulas instead of the inverse conductance derivative formulas they teach in place of old formulas.
REE? i Imagine that is what I called the 'base to emitter trans-resistance' That is derive from temperature coefficient at operating collector current. That people shortcut with an Isat formula that sometimes work (because its more prevalent).
I have to doge out for a bit, but I'll be back and post some math work for you to study.
@rectifryer "The values chosen aren't completely arbitrary. I chose around 20-50mA collector current based off the rise time performance at that collector level. That could be in error, but I worked back from there until I got the voltage gain I needed"
You are right to choose collector current based on something like rise time performance but in this instance rise time is way faster than what you need for audio frequencies. Better to choose collector current for things like maximum hfe and noise performance. After another look at the 2N3904 datasheet it looks to me like a collector current in the 1mA to 10mA range would be good.
I have posted more than enough on this subject so will leave others with more experience than I to advise you further. Good luck and I hope you get a good understanding of how transistors work.
You are right to choose collector current based on something like rise time performance but in this instance rise time is way faster than what you need for audio frequencies. Better to choose collector current for things like maximum hfe and noise performance. After another look at the 2N3904 datasheet it looks to me like a collector current in the 1mA to 10mA range would be good.
I have posted more than enough on this subject so will leave others with more experience than I to advise you further. Good luck and I hope you get a good understanding of how transistors work.
`yea, no OK, I'll show you how I would do this transistor, and screw you guys up Q-bias parameters, since you are not going to use their parameter in the datasheet to do your shortcut formulas that don't work for all circuits.
Datasheet says 200mA max continuous. With safety margin of 80%, that would be 160ma @IcSat RrL=5V/106mA= Resistance before temperature compensation = 31.25 RL= (Rrl+(RrL/2)) =46.875 So RL should be 47 ohms QbiasSat=Vcc/RL= 106.4mA
Class A Qbias = 1/2 Vcc so 2.5V @ the base input impedance is 1Kohms so voltage divider should be two 2K resistors (in AC Vcc is gnd so both resistors are in parallel in AC)
someone just walked in with a repair...will continue later
Datasheet says 200mA max continuous. With safety margin of 80%, that would be 160ma @IcSat RrL=5V/106mA= Resistance before temperature compensation = 31.25 RL= (Rrl+(RrL/2)) =46.875 So RL should be 47 ohms QbiasSat=Vcc/RL= 106.4mA
Class A Qbias = 1/2 Vcc so 2.5V @ the base input impedance is 1Kohms so voltage divider should be two 2K resistors (in AC Vcc is gnd so both resistors are in parallel in AC)
someone just walked in with a repair...will continue later
That would kill the bias for the transistor. You'll probably also want a capacitor on the output of the amp.
Usually in a single-ended design like that one designs for around 1/3 of the supply voltage at the base of the transistor and 1/2 to 2/3 of the supply voltage on the collector. Keep in mind that the output impedance of the amp will be high-ish, so your 1 kΩ load might be too heavy for the amp to drive.
Tom
That is what I was looking for...
but the real world hfe is 100 instead of 300 and that would even vary with current
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The average HFE at 5V is 40 with 2n3904 the ones I tested here.
So the only thing that might work is a common base to emitter follower, but for that voltage with that transistor, its not going to be a linear class A.
Raspberry Pi are engineered more for mosfets being their build out devices. especially on that 3.3V i/o That is why those people have to do so many things to make a transistors work on its i/o pins