Experimental design for small sig. amp

Thread Starter

hobbyist

Joined Aug 10, 2008
892
THANK YOU HOBBYIST! Your work is greatly appreciated! I've really wanted to understand transistors better and your efforts have made that possible.
Your welcome!

I'm not a professional, just a hobbiest, so just take my circuits as home brew projects, and learn from the experts on this board they have a wealth of info, to share.

Thanks..
 

Wendy

Joined Mar 24, 2008
23,421
Part of what millwood is saying (I think, and agree) is that you need to center your output voltage to the middle of the linear range. This can be hard sometimes, but the end result is clipping occurs on both peaks of the AC wave form when it is overdriven. Many times you need to manually balance a circuit out for just such a result, I helped a friend many orbits ago do somthing similar with a FET high power audio amp using my cheap sucky Oscope.

The idea is you can hear the overdrive, and then back it off.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
Part of what millwood is saying (I think, and agree) is that you need to center your output voltage to the middle of the linear range. This can be hard sometimes, but the end result is clipping occurs on both peaks of the AC wave form when it is overdriven. Many times you need to manually balance a circuit out for just such a result, I helped a friend many orbits ago do somthing similar with a FET high power audio amp using my cheap sucky Oscope.

The idea is you can hear the overdrive, and then back it off.
Thanks I agree.
In a sense than, didn't I manualy balance it when I adjusted the base bias watching the scope for a symetrical sinwave?

But it does make sense, I was biasing it for a standing current according to (1/2VCC / RE1) which did not take into acount that I need enough current to supply the load for pk. voltage.

So when designing for the current needed to drive a load in CC. amp, its best to have standing current at the least which is (Vpk / RLOAD) + (1/2VCC /RE1) which the best way was as he said to take the (Vpk. / RLOAD) = Iout and then (1/2VCC / Iout) to give the max. value for RE1, and then even try to go lower if possible. While adjusting base bias to keep the 1/2VCC across RE1.

But here is the big question, I had good signal output for stages Q1,Q2,Q3,
Q1,Q2,Q3.jpg

I had a good signal output with stage Q4, alone, but when I hooked Q4, stage to the rest of the amp, the oscope. just went fuzzy and noise, and maybe tremendous oscillations, the screen went on a blur, best way to put it, when I put the cap. C6 at the base it filtered the noise wonderfully and got excellent waveform output.

Also, This was before I put the AC bypass at Q4 stage.

What causes that to happen after the third stage, Q4 stage worked good alone. As well as first 3 stages worked good as one unit together.

Q1,Q2,Q3,Q4.jpg

Thanks.
 
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ELECTRONERD

Joined May 26, 2009
1,147
Okay folks, here is the CC amplifier that I designed. I'm just experimenting so I'm not sure I did this right; but it still worked. Okay, I wanted to have 4.5V at the base so I used two 10KΩ resistors. Now to find the emitter voltage, I had 4.5V-0.7V = 3.8V. I wanted to have around 2mA for the emitter current so I did 3.8V/2mA = 1.9KΩ but I just used 1.8KΩ (Close enough). I looked at hobbyist's diagrams and he just has the collector going to the supply voltage without a resistor so that's what I did. I have the speaker on the schematic which is actually the ceramic headphone amplifier. For the input to the base I just have a standard electret microphone. Let me know if I can improve anything ( to get more amplification), or letalone find out that I did it right. Thanks!

Hmmm......I can't attach the picture...wierd...
 

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Thread Starter

hobbyist

Joined Aug 10, 2008
892
Okay folks, here is the CC amplifier that I designed. I'm just experimenting so I'm not sure I did this right; but it still worked. Okay, I wanted to have 4.5V at the base so I used two 10KΩ resistors. Now to find the emitter voltage, I had 4.5V-0.7V = 3.8V. I wanted to have around 2mA for the emitter current so I did 3.8V/2mA = 1.9KΩ but I just used 1.8KΩ (Close enough). I looked at hobbyist's diagrams and he just has the collector going to the supply voltage without a resistor so that's what I did. I have the speaker on the schematic which is actually the ceramic headphone amplifier. For the input to the base I just have a standard electret microphone. Let me know if I can improve anything ( to get more amplification), or letalone find out that I did it right. Thanks!

Hmmm......I can't attach the picture...wierd...
The picture didn't show up as a pic. instead it's a doc. file.

do a goofle search called "circuitmaker" and the first hit that comes up shows a link to download a free student version of the program.

It is limited to 50 compinents.

Everything works good on it.

I use it mainly to draw schematics.

I draw a schematic with circuitmaker, then select and copy it.
Then open paint program, and paste it in, then save it as a JPG.

Yjen you can post it in the attachment. on this board.

Also remember, that a CC. amp, has no voltage gain, it is mainly used to match a low impedance load to a high impedance source, when it comes to amplifiers.

That's why I used it because by the time I would get done amplifying the signal from my generator, with 700 ohms output impedance, I knew I would need at least an input impedance to feed into for the final load stage of around 4K ohms considering the amount of Av. I was trying to achieve, thats why I used 2 CC amps, because the input impedance of Q1 stage was around 2K ohms,( min. beta 20) and the second stage Q2 brought the input impedance to around 6K ohms, high enough to connect a CE amp to.
 
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ELECTRONERD

Joined May 26, 2009
1,147
The picture didn't show up as a pic. instead it's a doc. file.

do a goofle search called "circuitmaker" and the first hit that comes up shows a link to download a free student version of the program.

It is limited to 50 compinents.

Everything works good on it.

I use it mainly to draw schematics.

I draw a schematic with circuitmaker, then select and copy it.
Then open paint program, and paste it in, then save it as a JPG.

Yjen you can post it in the attachment. on this board.

Also remember, that a CC. amp, has no voltage gain, it is mainly used to match a low impedance load to a high impedance source, when it comes to amplifiers.

That's why I used it because by the time I would get done amplifying the signal from my generator, with 700 ohms output impedance, I knew I would need at least an input impedance to feed into for the final load stage of around 4K ohms considering the amount of Av. I was trying to achieve, thats why I used 2 CC amps, because the input impedance of Q1 stage was around 2K ohms,( min. beta 20) and the second stage Q2 brought the input impedance to around 6K ohms, high enough to connect a CE amp to.
Thanks Hobbyist! How do you save it as a JPG file? Can you show me step through step?

The reason I chose the CC amplifier is because in my ARRL Amateur Radio Handbook they say that the CC amplifier has a high input impedance and low output impedance, eaxactly what you said. I realize that transistors don't have any voltage gain. Unlike tubes that amplify voltage, transistors amplify current.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
Thanks Hobbyist! How do you save it as a JPG file? Can you show me step through step?

.
1. I draw my circuit in circuit maker prog.

2. then in the edit menu I click on copy to clipboard.

3. open microsoft paint program, and in "edit" menu click paste.
when the circuit is pasted in the paint prog.

4. goto "file" menu and click on "save as" it will bring up a dialog box
atr the bottom, click on that box and it will open a list of all the formats to save it as,
one of them says JPG, click on it then give this file a name and click save. Make sure you know what file it saved to.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
Ok first rev.

I removed filter cap. C6. from Q3 base.


and made the filter for the front end. Q4 stage input. New C6 and R1.

Q1,Q2,Q3,Q4, rev. 1.JPG

results are:

Low freq. around 400HZ.
High freq. around 600KHZ.

So this method millwood showed did take care of the noise, and as well as getting rid of the base filter at Q3, restored the high freq. response.

good progress.

Low freq. response bad, is it due to the 100uf coupling caps?



Thanks guys.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
As you already know I am no expert at any of this, but I believe this to be true, you always want to have a low impedance output of any amp. with respect to the load it's driving.

I would venture to say you have the biasing pretty good, if your output at the emitter has the 1/2VCC or close to it, but the 8 ohm speaker will swamp the output voltage because while you have the proper voltage at the emitter, that;s because of the 1.8K ohm resistor, but when your signal comes in it won't see the 1.8K resistor, but the 8 ohm resistance of the load, and the voltage across the 8 ohm load would probably drop considerably.
 

ELECTRONERD

Joined May 26, 2009
1,147
As you already know I am no expert at any of this, but I believe this to be true, you always want to have a low impedance output of any amp. with respect to the load it's driving.

I would venture to say you have the biasing pretty good, if your output at the emitter has the 1/2VCC or close to it, but the 8 ohm speaker will swamp the output voltage because while you have the proper voltage at the emitter, that;s because of the 1.8K ohm resistor, but when your signal comes in it won't see the 1.8K resistor, but the 8 ohm resistance of the load, and the voltage across the 8 ohm load would probably drop considerably.
I see, well I didn't actually use a speaker just a ceramic headphone and it worked great. Although I did think of impedance, the way I saw it "schematicwise" was that I had the 1.8K and headphone impedance theoretically in parallel. So how could I get a lower emitter value resistor to acknowledge impedance? Will a lower base voltage help with this? I would assume that a higher emitter current would result in a lower impedance.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
If it worked good then you got a good circuit, design for that particular aplication,
which means is that the headphone has high enough impedance to be driven.
Yes ,your theory is right, both the resitor and the load is in parrallel, only at the signal voltage.
Good job.

To use a speaker or something as low as a speaker res. 8 ohms, would require a resistor low enough to hold a signal voltage across, but now you start getting into power amplification, low resistant loads like that require more power to reun properly, so you need to get a higher wattage power transistor, as well as resitors higher wattage ratings.

You may be able to scale down the entire voltage alltogether and build it, with 2n3904's.
 

ELECTRONERD

Joined May 26, 2009
1,147
If it worked good then you got a good circuit, design for that particular aplication,
which means is that the headphone has high enough impedance to be driven.
Yes ,your theory is right, both the resitor and the load is in parrallel, only at the signal voltage.
Good job.

To use a speaker or something as low as a speaker res. 8 ohms, would require a resistor low enough to hold a signal voltage across, but now you start getting into power amplification, low resistant loads like that require more power to reun properly, so you need to get a higher wattage power transistor, as well as resitors higher wattage ratings.

You may be able to scale down the entire voltage alltogether and build it, with 2n3904's.
Thanks Hobbyist! I was able to hear myself, not faintly but not all that loud either. I'm sure that the ceramic earphone has a low impedance so I will now try to scale down the emitter resistor. It will allow more current flow unfortunately, but I'll scale down the base voltage also. I'll let you know what the results are!
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
Good job.
Once you get that part made here is a simplified design note for a basic CE amp. to feed into it.
-------------------------------
-----------------------------

test amp.JPG


Design of the CE amp stage.
-----------------------------

Determine Zin CC. stage: simplified.

1. assume low value for beta = 20

2. (20 x (Rload x RE1) / (Rload + RE1)) = RoutQ1

3. ((RB1 x RB2) / (RB1 + RB2)) = RinQ1

4. ((RoutQ1 x RinQ1) / (RoutQ1 + RinQ1)) = ZinQ1
---------------------------------------------------

5. RC1 = < ZinQ1 best to make it around 10 times less, if possible!

6. choose voltage gain :Av: (RC1 / Av.) = RE2

7. (( 1/2 x VCC) / RC1) = ICQ2

8. (ICQ2 x RE2) = VRE2

9. (VRE2 + Vbe) = VBQ2

10. choose RB3 to be, 10 to 20 times RE2.

11. (VBQ2 / RB3) = IRB3

12. ((VCC - VBQ2) / IRB3) = RB4

Then test voltages at the collector and base, and adjust values as needed to get
close to the calculated voltages.
 
 
 
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millwood

Joined Dec 31, 1969
0
Zin is roughly the parallel of the divider network plus the amplified Re.

At high frequencies, the amplification goes away and the low impedance on the transistor leg dominates.
 

Thread Starter

hobbyist

Joined Aug 10, 2008
892
Millwood

I took your suggestion, about putting the filter at the input stage to get rid of noise.

Works great!

Removed the filter cap. from the base of Q3.
and put a filter at the input aod Q4, per drawing, and High freq. response is restored.

And noise is gone with good waveform again.

View attachment 10967

results are:

High freq. around 600KHZ.
 

ELECTRONERD

Joined May 26, 2009
1,147
Okay folks, i haven't had a whole lot of time to do this, but I was thinking about it on my way to swim team practice. I considered current and gain and imagined the picture in the attachment. You can either have high gain or high current. With higher gain and lower current, you will have a higher resistor value, thus, higher impedance. But with low gain and higher current, you will have a low impedance value, although the gain still suffers. Where does voltage come in place? Well I considered it and realized that with higher voltage comes higher impedance (R=E/I). So, I now knew that I wanted low voltage, and equal gain and current (or close to that). When I designed a CC amplifier with higher voltage, It wasn't as loud as it was with lower voltage CC amp I previously designed, proving my hypothesis. Does anyone know how to solve for the gain of the CC amplifier that i posted earlier? Thanks!
 

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millwood

Joined Dec 31, 1969
0
the (voltage) gain of that emitter follower would be slightly less than 1.0x - it being a follower.

but that works only with tiny voltage input (low Vpp), since it is idling at such a low current level.

emitter followers are CURRENT amplifiers - meaning they turn a source with high output impedance into a source with low output impedance.

so you amplify your signal, and then buffer it with an emitter follower before connecting it to a low impedance load (like a speaker).
 
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