BJT Input AC Signal Amplitude

Question

This question is referring to npn BJT CE:

If we look at base voltage vs. collector current plot we can see that at approximately 0.7V on base to emitter junction, the BJT starts rapidly conducting.

In a circuit DC BJT is biased so that quiescent point is in the middle of the load line (for class A amplifier).

Questions:

• Does DC quiescent point represents reference point for the AC input signal?

• If we want to normally amplify input signal (without distortions), does the input signal have to vary from 600mV to 750mV (approximately)?

Answer 1

Does DC quiescent point represents reference point for the AC input signal?

Yes. If you have time try to analysis this simple CE amplifier.

Where Vcc = 12V; Vc = 6V (red plot); Ve = 2V (green plot) ; Vinput = 1V (blue plot).

And we have a ideal BJT (Vbe is fixed and equal to 0.7V, and Beta = 100)

DC quiescent point is Vb = 2.7V; Ve = 2V; Vc = 6V; Ic = 6mA; Ib = 60μA. No the AC input signal voltage will "modulate" our DC operating point the in the rhythm of the input signal.

The AC voltage at input will cause that the base voltage will change from 3.6V to 1.6V in "rhythm" of a AC input signal. Those changes will result that the emitter voltage will also change from 3V to 1V. The emitter resistor (Re) will "convert" this changes in Ve voltage into Ie current. This will result the changes in emitter current from 0.9mA (3V) to 0.3mA (1V). And now this changes in Ie and Ic current will once again be "convert" in to voltage in Rc resistor. And this change in collector current will cause change in VRc voltage, between 9V to 3V. We have a three times larger change in VRc voltage because Rc is three times larger then Re resistor. So the voltage gain is equal to Av = Rc/Re

Also when Vin is at peak Vin = +1V we have 3.7V on the base of the BJT, so the emitter current is equal Ie = 3V/330Ω = 9mA the collector current is also equal 9mA

And collector voltage is equal Vc = Vcc - Ic* Rc = 12V - 9V = 3V

The base current needed to provide 9mA of a collector current is equal:

Ib = 9mA / 100 = 90μA

The base current provided form Rb and Eb is:

I = (Eb - Vb)/Rb = (4.7V - 3.7V) / 33KΩ = 30uA

But the emitter current must be equal to 9mA, and that's requires the base current to be equal 90uA

So we have a situation: Eb delivers 30uA, base needed 90uA, so that extra current (60uA) will (must) be "deliver" by AC voltage source Vin.

So Vin will be deliver (sourcing) 60uA of a current.

For the negative peak swing at Vin (-1V) We have this situation Vb = 1.7V ; Ic = 3mA; Vc = 9V

The base current needed to ensure Ic = 3mA is

Ib = 3mA/100 = 30uA

The Eb delivers I_Rb = (4.7 - 1.7V)/33K = 90uA

But we only need 30uA for the base current to "deliver" 3mA at emitter.

But Eb delivers 90uA so that extra current (60uA) must flow to Vin (must be sink by Vin).

I hope this will help.

EDIT

The input impedance is:

Rin = RB||(beta+1)Re = 33k||101*330R = 33k||33k = 16.67k

Or Rin = Vin/Iin = 1V/60uA = 16.67k

As you can see everything falls into place.

If we want to normally amplify input signal (without distortions), does the input signal have to vary from 600mV to 750mV (approximately)?

No, the input signal don't have to vary from 600mV to 750mV. Because the input coupling capacitor provide necessary DC-offset (DC level is shift thanks to Cin capacitor). And if Vbe is increased by about 60mV the collector current will increase his value ten times. Also BJT is a highly nonlinear device, so distortion will always occur. And this is why almost no one uses a single CE stage anymore.

Answer 2

1: Yes. At least, the signal at the base; see 2. 2: No. Yes. Sort of.

To elaborate, a BJT must be biased into the appropriate mode of operation, and then your AC signal is typically capacitively coupled into it; this effectively just overlays your AC small signal on top of the DC that the bias network provides.

It should be noted, however, that amplifying 'without distortion' is not going to happen. There's going to be distortion no matter what; the question is how much.

Answer 3

Try this

^{simulate this circuit – Schematic created using CircuitLab}

The operating point has 1mA Collector current, because emitter current is ~ 1mA, because we have ~~ 2 volts across R5. We have 2 volts across R1.

To compute gain, we need value for 'reac', slope of the emitter diode. That slope is 1/gm. At 1mA, 'reac' is 26 ohms Gain is $$(R1 || Zload) / (R2 + reac) = 2,000 / 126 = 16$$

assuming a 10MegOhm||10pF scope probe is the load.

Because of the linearizing effect of 100_Ohm in emitter, at 1mA Ie, we can swing nearly +- 100mV before "serious distortion" occurs.

Look up the writings of Willy Sansen, on computing bipolar distortion and the IP2 and IP3 values. You can extend his math, for a linearizing resistor.

Assume the input distortion Intercept Points are 2 volts rms.

What is the input random noise floor? Ignore the large Cmiller, and just use the Collector F3dB; 2K ohm and 20pF = 40 nanosecond, 25Meg Radians, or 4Mhz. Assume our noise density is set by 'rbb' of 2N3904, assumed to be 1Kohm. Noise density is 4 nanoVolts/rtHz. Total input referred noise is 4nV * sqrt(4Mhz) = 4nV * 2,000 = 8 uV rms.

SPDR (spurious free dynamic range) is 2 volts rms / 8uV rms, or 250,000 or 108dB.

====================================================== edit Here is link to Sidney Darlington discussing the bipolar transistor

http://ethw.org/Oral-History:James_Early

And this next link gives you insight into bipolar distortion, including the Taylor Series expansion to describe the various coefficients (equation 12)

http://lapsyc.ingelec.uns.edu.ar/Guillermo/Second_and%20Third_order%20Distortion_Compensation_3.pdf