Simple emitter followers

I'm reading this (fascinating reading):
http://www.dself.dsl.pipex.com/ampins/discrete/ef.htm

and I noticed the capacitor. What value is this, typically? Should
it be electrolytic, ceramic, ???

I also noticed one end is black, and the other white. Is the black
supposed to be the (-) terminal of the cap?

Thanks,

Michael

 

[John Popelish]

mrdarrett@gmail.com wrote:

I'm reading this (fascinating reading):
http://www.dself.dsl.pipex.com/ampins/discrete/ef.htm

and I noticed the capacitor. What value is this, typically? Should
it be electrolytic, ceramic, ???

I also noticed one end is black, and the other white. Is the black
supposed to be the (-) terminal of the cap?

Since both input source and output load are referenced to
ground, and the NPN follower produces an output voltage one
diode drop more negative than the input voltage, I think the
negative side of the electrolytic capacitor goes to the
emitter. The value of the capacitor is set by the lowest
frequency that you want amplified and the load resistance.
By Xc=1/(2*pi*f*C), you want to solve for C that produces a
Xc lower than the load resistance at minimum frequency, f.

For example, if the load resistance is 32 ohms and you want
little voltage drop across the capacitor down to 20 Hz you
might shoot for a Xc of something like 10 ohms (less than
1/4 of the total signal voltage dropped across the capacitor
at 20 Hz).

So rearranging,

C=1/(2*pi*f*Xc)=1/(2*pi*20*10)=0.0008F=800uF.

I would probably use something between 470uF or a 1000uF.

For audio and low resistance loads, you can see why
electrolytics are so practical. A very low voltage
capacitor can be used because there is only about 0.6 volts
across it, in this circuit.

--
Regards,

John Popelish

 

On Oct 15, 2:50 pm, John Popelish <jpopel...@rica.net> wrote:

mrdarr...@gmail.com wrote:
I'm reading this (fascinating reading):
http://www.dself.dsl.pipex.com/ampins/discrete/ef.htm

and I noticed the capacitor. What value is this, typically? Should
it be electrolytic, ceramic, ???

I also noticed one end is black, and the other white. Is the black
supposed to be the (-) terminal of the cap?

Since both input source and output load are referenced to
ground, and the NPN follower produces an output voltage one
diode drop more negative than the input voltage, I think the
negative side of the electrolytic capacitor goes to the
emitter. The value of the capacitor is set by the lowest
frequency that you want amplified and the load resistance.
By Xc=1/(2*pi*f*C), you want to solve for C that produces a
Xc lower than the load resistance at minimum frequency, f.

For example, if the load resistance is 32 ohms and you want
little voltage drop across the capacitor down to 20 Hz you
might shoot for a Xc of something like 10 ohms (less than
1/4 of the total signal voltage dropped across the capacitor
at 20 Hz).

So rearranging,

C=1/(2*pi*f*Xc)=1/(2*pi*20*10)=0.0008F=800uF.

I would probably use something between 470uF or a 1000uF.

For audio and low resistance loads, you can see why
electrolytics are so practical. A very low voltage
capacitor can be used because there is only about 0.6 volts
across it, in this circuit.

Ah, ok. Low voltage...

--
Regards,

John Popelish

Thanks for the clarification

Michael