Ic flows into the collector for NPN and out of the collector for PNP

Ib flows into the base for NPN and out of the base on PNP

Ie flows out of the emitter for NPN and into the emitter for PNP

V3 creates a 50V offset on the left side of the circuit to simulate offset references on each side of the circuit.

Case 1 on:

Iled = 5/301 = 17mA

Ic * R2 = CTR * Iled * R2 = 17V

This is greater than V2, so the transistor is in saturation and Vce = 0.2

Case 2 off:

Ic = 0

Vr2 = Ic * R2 = 0

Vo = V2 – Vr2 = V2

Vb = Ve + 0.65 = 0.65

Case 1 on:

Ib = (Vin – Vb)/3.01k = (3.3-0.65)/3.01k = 0.88mA

Ic * Rl1 = Hfe * Ib * 300 = 100 * 0.88mA * 300 = 26V

This is greater than V2, so the transistor is in saturation and Vce = Vout = 0.2

Il1 = (12 - 0.2)/300 = 39mA

Case 2 off:

I b = 0, Ic = 0

Initially Il1 ≠ 0, which creates a “flyback” voltage

Eventually Il1 = 0, and Vout = V2 – Il1 * Rl1 = V2

This voltage will damage the transistor, and cause it to fail.

Vb = Ve + 0.65 = 0.65

Case 1 on:

Ib = (Vin – Vb)/3.01k = (3.3-0.65)/3.01k = 0.88mA

Ic * R2 = Hfe * Ib * 301 + 0.7 = 100 * 0.88mA * 301 + 0.7 = 27V

This is greater than V2, so the transistor is in saturation and Vce = Vout = 0.2

Ir2 = (3.3 - 0.2 – 0.7)/301 = 8mA

Case 2 off:

Ib = 0, Ic = Ir2 = 0

Id = 0, Vout = V2 – Ir2 * R2 – Id * Rd = V2 = 3.3V

Vb = Ve + 0.65 = 0.65

Case 1 on:

Ib = (V2 – Vb)/2k = (3.3-0.65)/2k = 1.7mA

Ic * R3 = Hfe * Ib * 300k = 10 * 1.7mA * 300k = 5.1kV

This is greater than V2, so the transistor is in saturation and Vce = Vout = 0.2

Ir3 = (300 - 0.2)/300k = 1mA

Case 2 off:

Ib = 0, Ic = 0

Ir3= 0, and Vout = V1 – Ir3 * R3 = V1 = 300V

Case 1 on:

Vin = 0

Vb = Ve + 0.65 = 0.65

Ib = (V3 – Vb)/2k = (3.3-0.65)/2k = 1.7mA

Ic * R3 = Hfe * Ib * 100k = 10 * 1.7mA * 100k = 1.7kV

This is greater than V1, so the transistor is in saturation and Vce = 0.2V

Vo = 0.2 + Vin = 0.2

Ir3 = (100 - 0.2)/100k = 1mA

Case 2 off:

Vb = 3.3 + 0.65 = 3.95V

Ib = 3.3 – 3.96 which is negative, the transistor is off, and Ib = Ic = 0

Ic = Ir3 = 0

Vout = V1 – Ir3 * R3 = 100V

R3 is the load.

If (Ie1 + Ie2) << Ir3,  then Ir1 = Ir3, the load current.

R1 << R3, so Ir3 = V1/R3 = 1A

Ir3 * R1 = 0.1V = V1 – Vr3

Vb2 = Vr3 – 0.65

Vb1 = Vb2

Ve1 = Vb1 + 0.65 = Vr3

Ir2 = (V1 – Vr3)/100 = 0.1V/100 = 1mA

Ic1 = Ie1 = Ir2 = 1mA

Vo = Ic1 * R4 = 1mA * 1k = 1V

R2 and R4 are the loads.

Q1 case 1, V2 = 0:

Vb = 5 – 0.65 = 4.35V

Ib = Ir1 = (4.35 – V2)/2k = 4.35/2k = 2mA

Ib * Hfe * R2 = 2mA * 100 * 100 = 20V

20V > V1, so the transistor is on (saturated)

Vo = V1 - 0.2 = 4.8V

Q1 case 2, V2 = 3.3:

Vb = 5 – 0.65 – 3.3 = 1.05V

Ib = Ir1 = (1.05 – V2)/2k = 1.05/2k = 0.5mA

Ib * Hfe * R2 = 0.5mA * 100 * 100 = 5V

5V ≥ V1, so the transistor is on (saturated)

Vo = V1 - 0.2 = 4.8V

Not a switch because it is stuck on, it never turns off

Lets fix this problem by adding Q3-

Q2 case 1, V2 > 0.65, Q3 is on, Vc3 = 0.2V:

Vb2 = 5V – 0.65 = 4.35V

Ib2 = (4.35 – 0.2)/1k = 4mA

Ib2 * Hfe * R4 = 4mA * 100 * 100 = 40,

40 > V1 so Q2 is saturated and Vo2 = V1 – 0.2 = 4.8V

Q2 case 2, V2 = 0, Q3 is off, Ib3 = Ic3 = 0:

Ib2 = Ic3 = 0, Ic2 = 0

The Q2 transistor is off, Ic2 = Ir4 = 0

vo2 = Ir4 * 100 = 0.

DC bias analysis:

Rb = Re * Hfe = 2k * 100 = 200k

200k >> 10k||7k so Vb = 3.3 * 10/(10+7) = 1.9V

Ve = 1.9 – 0.65 = 1.25V

Ie = Ic = 1.25/2k = 0.6mA

Vout = v2 – Ir2 * R2 = 3.3V – 0.6mA * 2k = 2.1V

AC analysis:

1/sC2 @ 1000Hz << Rb||10k||7k 

Vb = vin

Ve (ac) = Vb

R4 >> (1/sC1 + R5)

Ir5 = Ie

R5 >> 1/sC1

Ir5 = Ve/R5 = vin/R5

Ic = Ie

Vout = 0V – Ic * R2

Vout = -vin * R2/R5 = -.05 * 2k/200 = -0.5

The gain is R2/R5 = 10

DC:

Rb = Hfe * R4, Rb >> R3||R6

Vb = 9V * 2k/(2k+2k) = 4.5V

Ve = 4.5 – 0.65 = 3.85V

Ie = (4.5-0.65)/500 = 7.7mA

AC:

1/sC1 and 1/sC2 are <<50 at 10MHz

Rb = Hfe * R5, Rb >> R6||R3

Ve = Vb = Vin

re = 26mV/Ie [dc] = 26mV/7.7mA = 3.3 ohms

Ie = Vin/(50+Re) = 0.25/(50+3.3) = 4.6mA

The standard assumptions are valid.

DC:

Ie = (9 * 10/(10+1) – 0.65 – 0.65)/40 = 172mA

re = 26mV/172mA = 0.15

AC:

Vo = vin

Ir2 = vin/(R2+re) = 1/(8 + 0.15) = 122mA

R2 is the load.

Vb = Vd1 = 6.5V

Ve = vo = 6.5-0.65 = 5.9V

Ie = Ir2 = 5.9/10 = 590mA

AC:

Vb1 = 0V

Ve1 = 0V

Zin = 36 ohms to match the antenna

Ie = Ic = vin/36 ohms

Zfilter = (Zl || Zc || 30.1k || 100k) [tank band pass filter tuned to the carrier frequency]

Vo1 = Ic * Zfilter

Vb2 = vo1

Ve2 = vb2 = vo1

D1 R9 C5 R10 form a peak detector to track the envelope of vo2

Vo3 is the positive envelope of vo1

Vo3 = envelope [(vin / 36 ) * Zfilter]