Basic Transistors

Hello again,

I have a small question about BJT. I know that Vbe has to be at least 0.7 in most case. But I have a circuit here (juste npn connected to power supply ) with the base grounded. The emitter still put out a voltage. I though the base was acting as as gate that allowed the current from the collector to the emitter WITH A BIT OF CURRENT FROM THE BASE, ionizing the p region so the two n-region could conduct. I am right ? so why does it work even if there is no current from the base ???

ken

GERMANIUM TRANSISTOR has Vbe of .2 volts and SILICON TRANSISTOR has Vbe of .6 volts.

An easy memory aid to rember this is tell yourself G comes before S.

Sounds like you got it down. Regards CL. Maher Certified Electronics

"w4clm" wrote

A transistor turns on with any Vbe>0

Ic = Is * e^(Vbe/nkt)

It just doesn't turn on very much below 0.6v.

Transistors switching signals to static CMOS and other low-power circuits often have such low collector current that the transistor can be saturated with a Vbe of < 0.1V. Battery powered equipment often operates with Vbe's of ~0.3V

--
Nicholas O. Lindan, Cleveland, Ohio
Consulting Engineer:  Electronics; Informatics; Photonics.
Remove spaces etc. to reply: n o lindan at net com dot com
psst.. want to buy an f-stop timer? nolindan.com/da/fstop/

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Not exactly, still need to know about capacitor in that circuit.This is = a link,

if you scoll down to 2. = TYPICAL TRANSISTOR CIRCUIT you can see a small circuit. It is an amplifier circuit. I am just confused as how the circuit know = not to change the frequency. Does the input and output capacitor always = have to be the same ? could they both be 50uF ? how do we choose without = altering the frequency and just have a bigger amplitude ??

Ken

Not exactly, still need to know about capacitor in that circuit.This is a link,

if you scoll down to 2. TYPICAL TRANSISTOR CIRCUIT you can see a small circuit. It is an amplifier circuit. I am just confused as how the circuit know not to change the frequency. Does the input and output capacitor always have to be the same ? could they both be 50uF ? how do we choose without altering the frequency and just have a bigger amplitude ??

Ken

Ken, (1) No, the input and output capacitors do not always have to be the same values. (2) Yes, they could both be 50uF. The values of the capacitors are determined by calculations based upon the signal frequencies involved, and the impedances of the driving and driven circuits. (3) In an amplifier circuit, you do not change the frequency of the signal. The components in the circuit affect the frequency RESPONSE of the circuit. IOW, the amplification may be different at different frequencies. In the URL that you provided suggests that the reader do some additional reading at the links provided in the text. I admonish you to do that reading... it will explain a lot to you.

--
Dave M
MasonDG44 at comcast dot net  (Just subsitute the appropriate characters in 
the address)

Never take a laxative and a sleeping pill at the same time!!

"Tweetldee" wrote

A quick calculation of the gain of this class of circuit is

Rc / Re

A quick analysis of why this is so is:

Ignoring DC levels and looking at the AC signal: The emitter of the transistor will follow the base; The AC current in Re is Vb/Re; Ic ~= Ie; Vc = Ic * Rc; Vc = (Vb/Re) * Rc; Vout = Vc; Vin = Vb; Gain = Vout/Vin = Rc/Re.

As drawn the circuit has a gain of 1,000 / 480 ~= 2.1 However the in and out voltage graphs show a gain of 3.5/.325 ~= 8

You may want to take this website with a grain of salt.

--
Nicholas O. Lindan, Cleveland, Ohio
Consulting Engineer:  Electronics; Informatics; Photonics.
Remove spaces etc. to reply: n o lindan at net com dot com
psst.. want to buy an f-stop timer? nolindan.com/da/fstop/

The gain of 8 for an AC signal would appear to be in the ballpark for 1KHz and hFE 100 when C3 is not neglected, but it's been awhile since I did this.

In which I blithely ignored the presence of C3. And gave the wrong answer.

Now, though, without much analysis, it seems the input is cap. coupled directly to the base and the emitter to ground. This leads to the base current being set by the AC impedance of the input and reflected emitter capacitors. Gain is then (approx) directly proportional to frequency. Strange circuit.

But it seems the point of the circuit is to demonstrate the bias voltages of the C & E terminals of the transistor.

--
Nicholas O. Lindan, Cleveland, Ohio
Consulting Engineer:  Electronics; Informatics; Photonics.
Remove spaces etc. to reply: n o lindan at net com dot com
psst.. want to buy an f-stop timer? nolindan.com/da/fstop/