Common Emitter Amplifier Design Help

Thread Starter

just_gaze

Joined Jan 3, 2016
14
Thanks for all of the tips earlier. I redesigned the circuit, using a feedback resistor at emitter, and voltage divider at base. I calculated the values according to this [ http://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/biasing-calculations/ ] guide, and my calculations are in the attachments. With Vin = .1V, 20Hz, Vout = 6.96 V (peak to peak). You can see this is in attachment "scope 1" (look at the blue wave, ignore the purple one). The waveform is nearly undistorted, but the bottom of the peak starts to flatten. I also tried with Vin = 0.1V, 100Hz, and the waveform is very flat on both top and bottom peaks (attachment "scope 2"). With Vin = 1V, 20Hz, there is also distortion (attachment "scope 3"). I included both the breadboard circuit and a picture of the simulation with voltage and current probe values for comparison. The values I used on the breadboard are those shown in the simulation. I had to tweak from the calculations because I didn't have the right resistors.

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.

So, I am getting an amplified waveform, as intended, but with distortion at any input higher than 0.1V. But the current measurements aren't matching up to what is expected. Am I simply measuring incorrectly, or is something else going on here? How can I tweak my biasing to get a clean, amplified waveform at higher voltage and frequency input?
 

Attachments

hobbyist

Joined Aug 10, 2008
889
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...
 
Last edited:

Veracohr

Joined Jan 3, 2011
740
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.
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.
 

Thread Starter

just_gaze

Joined Jan 3, 2016
14
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...
Thanks for the excellent video! It looks like that breadboard has seen some use! What is your Youtube channel name, so I can subscribe?

I am amazed I didn't use your method to measure current earlier, since it's just a simple application of Ohm's Law. I was just so fixated on getting a raw current measurement with the ammeter. It's weird, to me, that my ammeter is not giving me reliable values. It makes the thing unusable. I bought mine at Radioshack, so maybe their meters are just bad? I was able to measure the current how you did. I placed a 100 ohm resistor before the base, so I could get base current, and my DC current gain came out to Beta = 232, which seems pretty damn high! Oh, and I had the wrong value in the simulation for collector resistor. I was actually using 2.3k ohms, not 3...

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?
 

Thread Starter

just_gaze

Joined Jan 3, 2016
14
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.
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.

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.
 

Colin55

Joined Aug 27, 2015
519
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.
 

Thread Starter

just_gaze

Joined Jan 3, 2016
14
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.
Is there a way I can determine the impedance of the generator?
 

Veracohr

Joined Jan 3, 2011
740
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?
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:

Screen Shot 2016-01-10 at 1.09.52 PM.png

Your gain is now 22.1dB but the high-pass frequency is 10.7Hz. The emitter DC current is the same in both cases.

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.
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.

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.
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.

4Ω at 15W means your speaker can handle up to √(15*4)=7.4Vrms maximum. You won't get that with a 9V supply, so just get the most gain you can, then worry about a current amplifier to drive the speaker. The speaker can take √(15/4)=1.93Arms maximum.

Function generators are often 50Ω output impedance, but the manual will say. Much lower than a guitar pickup.
 

Colin55

Joined Aug 27, 2015
519
"Don't worry about current level from the guitar"
That's the very thing you have to worry about.
That's why none of you have the slightest idea how to design a per-amplifier circuit.
The current from a magnetic pick-up is minuscule (microamps) and needs a very high impedance per-amplifier stage to prevent the signal being attenuated to nothing.
The input and output capacitors are far too high and the H-bridge set-up is an absurd design for a pre-amplifier.
 

dannyf

Joined Sep 13, 2015
2,197
Can anyone find a mistake in my calculations or circuit build?
The working point of that circuit is highly dependent on the hFE of the transistor used.

Two ways to alleviate that:

1) tie the base resistor to the collector: the negative feedback would help stablize the working point;
2) use a small emitter resistor: it also helps define the gain of the amplifier.
 
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