If you look at the scope capture for the common emitter amp that exhibited acoustic feedback, you will see that the oscillation occurs at about 500 Hz, which is the speaker resonance. A power transformer works just fine there, and over the band of frequencies needed for an intercom it would be adequate.

This isn't intended to be high fidelity, after all!

 

This isn't intended to be high fidelity, after all!

Sorry.
In my career I customized, installed and programmed some huge and expensive high fidelity intercom systems for bank head offices. I even made a bass-boost circuit for the background music.

Then the government wrote a spec for a high fidelity PA system for airports so I souped up the good intercom system so it met the FM radio station broadcast quality spec's (audio up to 15khz and noise very low).

The good intercom system was made in Norway by Stentofon. It used pulse-amplitude multiplexing (PAMEX).

 

Sorry its taking me a long time to actually physically do my experiment, I have many other things going on. Please read what I wrote below carefully because I explain what I actually did step by step.

Anyway, before trying the Darlington I did one more try with the circuit Bill posted in post # 8 of this thread. Just to get some values. This below. Left one. I don't have a carbon mic.

I connected my super accurate multimeter and started from the rheostat R1 at 100KΩ. I measured the dc values of ic and ib at this time and got

ib =.036mA

ic = 9.6mA

This gives a β of 266.

Then I yelled into my microphone and set the multimeter to measure AC current. As I yelled into the microphone the AC current at the collector was 1mA. This is the current that drives the speaker coil. Too tiny to cause an audible sound of large enough magnitude.

Then I lowered the resistance of the rheostat and measured the AC part of ic again. This time it gave me 0mA. (i.e. the current is smaller than that, meter cannot pick it up.)

As I decreased the resistance of the rheostat the AC of ic was still 0. When I made the rheostat 10Ω (its smallest resistance), I could feel the lead on the collector get warm as it should since ic increased to 100mA+. But still when I measured the AC of ic it gave me 0.

Then to check whether my original 1mA AC was an error of the meter I went back to 100kΩ on the rheostat and again it gave me

i collector = 1mA AC So we can probably assume that is the real value (or in the neighborhood of).

When the DC part of ic is itself tiny the AC setting on the multimeter shows me 1mA through the speaker when I yell into the mic. (not if I speak softly, i.e. the circuit is sorta working even though we cant hear the result)

Is this because when the DC part of ic becomes huge the multimeter can no longer pick up the AC part? What I would give for an oscilloscope.


Here is a question to you guys, Bill I noticed that your circuit (shown above) does not have a bios to pass AC like was taught in All about circuits. I know the magnitude of the input voltage (sound) is so tiny its almost insignificant, but does that mean we don't need some sort of bios anyway?

Next I will try Bill's Darlington circuits. Once against sorry I'm slow at doing this, but got lots of other stuff going on that I need to do as well.

 

In the left schematic, you will get much better results if you connect a 100μF electrolytic capacitor from the junction of the "mic" and the 470Ω resistor to the emitter.

If you do that, and adjust the bias so that your collector current is about 50 mA, you should be able to get acoustic feedback if you place the "mic" close to the output speaker.

 

In the left schematic, you will get much better results if you connect a 100μF electrolytic capacitor from the junction of the "mic" and the 470Ω resistor to the emitter.

If you do that, and adjust the bias so that your collector current is about 50 mA, you should be able to get acoustic feedback if you place the "mic" close to the output speaker.

Will it be like a high pass filter only allowing AC through?

Doesn't electrolytic not allow AC by design because its polarized and sort of acts like a diode?

 

The polarized cap is more like a battery, but with the RC charge curve of a cap. Interestingly, you can make a non polarized cap by putting two polarized caps back to back, facing the opposite directions. In no way does a cap, polarized or not, ever act like a diode. Point a polarized cap the wrong way and you have a stink bomb.

It is like a high pass filter, but for the entire audio spectrum. I don't remember where the numbers are exactly, but it is around 10Hz to 20Khz. You want the DC values to be used to put the transistor in a useful area of its bias characteristics.

I haven't given up on this set of experiments, but time is always my most limited resource nowdays.

 

Be sure that the negative terminal of the electrolytic is connected to the emitter.

Without the electrolytic, the signal from the "mic" has to pass through a relatively high resistance (the bias resistor), and this attenuates the signal that is applied to the base. The capacitor will provide a low impedance path for the signal from the "mic" so that more signal gets to the base.

 

Count_Volta,
Its great to hear these results, keep it up!

Guys,
Learning about amplifiers, i read in my book that they recommend VCE to be in the midway of VCC (4.5V in this case) and VEE (0V in this case).
Why is it recommended, if Count has received such an high DC beta when IC was only 9mAdc - meaning when VCE was 4.428Vdc (VCC - 9mAdc * 8Ω = 4.428Vdc).
4.428Vdc is no where near the midway of VCC and VEE - which is 2.25Vdc.

Thanks.

 

It's all about voltage swing

 

Thanks alot Jony.

In the case of UE = 0.5Uzas (the left circuit),
Is it really possible for the AC signal to swing from VCC to VEE, like you showed?

I mean, that will cause the BJT to swing from Saturation to Cut-off, respectively.
And since the amplification exists only when the the BJT is forward biased, then such large swing will create no linearity between the input signal and the output signal.

-- Addition.
After some more thoughts about that,
I assume that in the left circuit, the max swing isnt between VCC and VEE but between (VCC-VCEsat) and VEE+ (VEE+ is just a little above VEE, enough to stay away from cut-off region).
Right?

 

If the peak of the sine wave is barely in saturation and the other peak is barely in cutoff, this is referred to as clipping, and is a normal condition of an over driven amplifier. The sound will sound distorted, but recognizable. Back the signal strength on the input a little and it will probably be OK.

This is the ideal design condition, but is rarely achieved. Most times one or the other is reached first, and sounds a lot uglier. Overdriving an amp can be done to almost every design.

 

In the case of UE = 0.5Uzas (the left circuit),
Is it really possible for the AC signal to swing from VCC to VEE, like you showed?

In real circuit of course this is impossible, you always have some voltage drop on a saturate transistor.
This figure was to show you why it is recommended to set Vc=0.5VCC

But this is not always the case.
For example in this circuit

 

Thanks again Jony and Bill.

Jony,
Are you saying that in the last example, we would want to bias VC at (12V + 2V) / 2 = 7V, instead of 6V?

Thank you.

 

Are you saying that in the last example, we would want to bias VC at (12V + 2V) / 2 = 7V, instead of 6V?

Yes,
theoretical the optimal Vce:

\(Vce_{opt}\approx Vcc*\frac{Rc+Re }{2(Rc+Re)}\approx 0.5*Vcc\)
And
\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2(Rc+Re)}\)

And if we add a Ce capacitor we get:
\(Vce_{opt}\approx Vcc*\frac{Rc}{2Rc+Re}\)
And
\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2Rc+Re}\)

 

Lots of thanks Jony!
Before i read your messages, i've just arrived to voltage swings in my book and didnt manage to understand this.
Now that i read your several posts about it, it became very clear to me.
So I learnd a lot from you.

 

Yes,
theoretical the optimal Vce:

\(Vce_{opt}=Vcc*\frac{Rc }{2Rc+Re}+Vce(sat)*\frac{Rc+Re }{2Rc+Re}\approx Vcc*\frac{Rc }{2Rc+Re}\)
And

\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2Rc+Re}\)

I think this should be:

\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2(Rc+Re)}\)

and with this change the expression for Vce(opt) will be different.

 

Allright thanks to my good friend Alphacat I understand the purpose of the capacitor. It only makes sense if you do small signal analysis. I should go back and read over that section of sedra and smith again.

So I will try the capacitor and see what happens. If it fails, Darlington here we come.

 

I think this should be:

\(Ic_{opt}=\frac{Vcc-Vce(sat)}{2(Rc+Re)}\)

and with this change the expression for Vce(opt) will be different.

Of course you're right, my equation apply for circuit whit Ce capacitor.

 

Of course you're right, my equation apply for circuit whit Ce capacitor.

Since you were responding to alphacat about the circuit in post #172, perhaps you should edit post #174 and give the equations for the case without a capacitor Ce, which would apply to the circuit in post #172.

That way, if someone follows this thread later, they won't be confused by the equations you originally gave, which aren't applicable to the circuit in post #172.

As an additional bonus, you could retain the equations you originally gave, but identify them as applying to the case with a capacitor Ce.

 

The Electrician thanks for your suggestion of the capacitor. I tried it. Below I give my results. I connected it like you said and got ic= 50mA DC like you said to do. Alphacat was there and can confirm what I did.

I hooked up my multimeter to measure AC current through the collector. When I spoke into the microphone I got 2 microamps AC. When I yelled into it like crazy I got 25 microamps AC. I could not get acoustic feedback.

This was a major improvement over the 2 microamps I got when yelling without the capacitor. Even though its a major improvement the values are still in microamps, i.e. too tiny for audible sound.

Bottom line is the gain of one BJT is too small, I need to try Bill's darlington circuits next.