One of the nice things about the 4-resistor bias approach you now have is that you can stop worrying about Beta (for the gain calculation). The gain becomes approximately Rc/(Re+RE), where Re is the intrinsic emitter resistance (which changes with current), and RE is the emitter resistor.I am still not getting good current measurements at transistor pins. With this configuration, Ic = .62mA, Ib = .1uA, and Ie = 0. I have no other explanation, other than that I am probably measuring the current incorrectly. To measure, I break the circuit and insert the test leads on each end of the break. There is a picture attached showing how I measured.
Thanks for the excellent video! It looks like that breadboard has seen some use! What is your Youtube channel name, so I can subscribe?Hi,
I made a short video, using your circuit design values, for the resistors, to verify biasing currents and voltages, then I give it the dynamic test on my scope, and try to explain in my own opinion, what I think may be some of the problems your running into, and how to possibly fix it.
Your collector voltage is running a little on the low side of the load line, but not enough to cause any kind of heavy distortion you were getting, in my video I demonstrate the voltage gain you should be getting, using my scope and signal generator.
This is the video :
Now all of my opinions is on a hobby experience only.
I have more circuit demonstrations on my youtube channel. enjoy...
Hope this helps...
Didn't realize you could simply calculate gain from the resistor values. That will make the calculations very simple from now on. I took out the capacitor and it solved my distortion problem.One of the nice things about the 4-resistor bias approach you now have is that you can stop worrying about Beta (for the gain calculation). The gain becomes approximately Rc/(Re+RE), where Re is the intrinsic emitter resistance (which changes with current), and RE is the emitter resistor.
You're seeing distortion because you have completely bypassed the emitter resistor, which means your gain is Rc/Re. Re in your case is around 26 ohms (=26mV/Ie), so your gain is 3000/26=115. With 0.1V peak input signal, that would give 11.5V output, but your supply is only 8.73V, so it distorts.
Are you shooting for a particular value of gain? You can take out the emitter capacitor and get a value of 3k/(26+432) = 6.55. Or you can split the emitter resistor so that you're only bypassing part of it. If you split it in half so you have two 216 ohm resistors, your gain is 3k/(26+216) = 12.4.
Is there a way I can determine the impedance of the generator?A function generator will be low impedance, but in reality the impedance will be higher and everyone has been running around in circuits and failing to ask the most important question, that should have been determined before anything else.
The emitter bypass capacitor can be useful if you want higher gain for the same DC bias current. But is has the added drawback of creating a high-pass filter pole with the un-bypassed emitter resistance. For example, with no bypass capacitor your gain is 16.4dB and the high-pass frequency is 6.6Hz. If you bypassed half the emitter resistance like this:I followed your advice and removed the bypass capacitor, and I was able to get a clean, amplified output all the way up to Vin = 1.5V. So, when would I ever need to use a bypass capacitor? Many, many circuits I have seen use it. Does it depend on the frequency range I am working with? I am building this for an audio application, so is it unnecessary for low frequencies?
It's an approximation based on the fact that the collector current and emitter currents are pretty much equal. So if your collector resistor is larger than the emitter resistor, more voltage will be dropped across it, so you get voltage gain.Didn't realize you could simply calculate gain from the resistor values. That will make the calculations very simple from now on. I took out the capacitor and it solved my distortion problem.
Don't worry about current level from the guitar, just think about it's output impedance relative to the input impedance of your amplifier. I've seen guitar pickups quoted at 5-20kΩ output impedance, which means you need higher input impedance than what your circuit has now or you will lose level. Check out some other guitar-related BJT circuits, and you'll see resistors in the 250k-2Meg range on the input stage.I am honestly not sure how much gain, and how many amplifier stages I need... What I am doing is amplifying output from an electric guitar to drive a 4 ohm, 15 Watt speaker. So, I guess I need to figure out the current from the guitar output, the current needed to drive the speaker, and calculate gain needed. Then I can figure out how many stages, given the gain per stage.
The working point of that circuit is highly dependent on the hFE of the transistor used.Can anyone find a mistake in my calculations or circuit build?
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by Jake Hertz
by Jake Hertz