## The Common-Emitter Amplifier

### SPICE Simulations

#### Operating point Analysis

Here, we are calculating DC voltages (bias voltages) at every node under no input source.

i.e. DC analysis with Vin = 0V (capacitors opened, inductors shorted).

Relevant source code lines:

``````OP
print all
print (n1-n2)       ; vbe bias point.
``````

Results: node DC voltages.

``````Node       Voltage
----       -------
n_pos           12
n_in             0
n1         2.07461
n_out      7.00346
n2         1.29397
(n1-n2)    .780638
``````

Note a DC operating point analysis is always performed before transient or AC analyses, thus we did not really need to perform this analysis excplicitly, it was done for instructional/learning purposes.

#### Transient Analysis

The transient analysis was performed over a time-length of 2mS (tstop) with steps (sample points) every 10nS (tsep).

An AC voltage source was used as the input signal: simple sinusoid with 0V offset, 1Vpeak amplitude and 1KHz frequency.

We are plotting the output voltage at “net out” (n_out), the input voltage at “net in” (n_in), and the voltage at “net1” (n1) after AC coupling.

Relevant source code lines:

``````V2   n_in 0  dc 0 ac 1.0 sin(0 1 1KHz)

TRAN 10nS 2mS

gnuplot \$filename n_out n_in n1 title \$title xlabel \$xlabel ylabel \$ylabel
``````

#### DC Analysis

In our DC analysis, we are measuring the DC output voltage dependence over power supply variations (over a DC sweep).

We are applying a DC sweep to our supply voltage (V1) from 0 to 12V in 0.1V increments.

We are plotting the DC output voltage at “net out” (n_out) vs supply voltage (V1).

Relevant source code lines:

``````DC V1 0 12 0.1

gnuplot \$filename n_out title \$title xlabel \$xlabel ylabel \$ylabel
``````

#### AC Analysis

In our AC analysis, we are performing a logarithmic sweep with decade scaling (decade variation) with 10 points per decade. The frequency sweep goes from 0.01Hz to 1KHz.

We are computing the transfer function for the circuit, to calculate the gain of the amplifier, and phase shift at different frequencies.

We are plotting:

• The transfer function magnitude in a linear scale: i.e. the gain of the circuit in V/V.

• The transfer function magnitude in a logarithmic (dB) scale: i.e. the gain of the circuit in dB.

• The transfer function phase in degrees: i.e. the phase shift of the output voltage waveform with respect to the input.

Relevant source code lines:

``````AC dec 10 0.01Hz 1KHz  ; Parametters: dec(log sweep), 10 pts/dec, 0.01->100Hz

gnuplot \$filename mag(n_out/n_in) title \$title xlabel \$xlabel ylabel \$ylabel

gnuplot \$filename db(n_out/n_in) title \$title xlabel \$xlabel ylabel \$ylabel

gnuplot \$filename phase(n_out/n_in)*180/pi title \$title xlabel \$xlabel ylabel \$ylabel
``````

### Results (mostly a sanity check)

Let’s start with a quick qualitative analysis. From a Common-emitter Amplifier circuit topology we should expect:

• We have an AC coupled input to the CE amplifier, thus we should expect a signal shifted “up” from 0V to the voltage at n_1 (our bias voltage for Q1, more on this later).

Check.

• A CE amplifier is an inverting amplifier, thus we should expect an amplified and inverted output waveform (and accordingly a 180 degrees phase shift at the output w.r.t the input).

Makes sense: we can see this both from the transient analysis plot and the AC analysis phase transfer function plot, check.

• As we sweep our supply voltage from 0 to 12V. Initially our transistor is OFF, there should be no current flow (our transistor is an “open” at this point) and we should expect to see our output voltage tracking our supply voltage (at least until our transistor turns ON, more on this later).

OK, we can see this in our DC sweep plot: a 1 to 1 slope initially, check.

• In our AC analysis, we should see no gain initially (input capacitor is open at DC), and an increase in gain as we raise the frequency until our gain figure settles into a constant factor (at high frequency our capacitor is a short), furthermore we should see a phase shift of -180 degrees as this is an inverting amplifier (also mentioned above).

Check and check.

OK, From a very quick qualitative analysis, our intuition agrees with our simulation results, but let’s calculate our node voltages at Vin = 0V just to make sure our operating point results are correct.

Our base voltage (i.e. the voltage at n_1) is just a voltage divider (not quite, minus a little bit due to our base current, but more on this later)

``````Vn_1 = 12V * (R2 / (R1 + R2))
= 12V * (24KR / 124KR)
= 12V * 0.193
= 2.32V
``````

Our emitter voltage (i.e. n_2), should be our base voltage minus about ~0.65V, i.e:

``````Vn_2 = Vn_1 - 0.65V
= 2.32V - 0.65V
= 1.67V
``````

Now given our emitter resistor R4, we can calculate the current through our transistor (collector to emitter) and R3, mainly:

``````I_qc = Vn_2 / R4
= 1.67V / 1KR
= 1.67mA
``````

Now, let’s talk about the base current. Given a generic discrete NPN BJT we can assume it will have a (beta) current gain (Ic/Ib) of about 100. Thus we can calculate how much bias base current will go into our transistor:

``````I_qb = I_qc / beta
= I_qc / 100
= 1.67mA / 100
= 16.73uA
``````

Now this will reduce our bias voltage at n_1 a little bit. Given our branch current through R1 and R2 of:

``````
I_r2 = 12V / (R1 + R2)
= 12V / 124KR
= 96.77uA
``````

Minus our base current, we will have an actual current through R2 of:

``````I_r2 = 96.77uA - 16.73uA
= 80.05uA
``````

And a bias voltage at the base (i.e. at n_1):

``````Vn_1 = I_r2 * R2
= 80.05uA * 24KR
= 1.92V
``````

Now, we can repeat the process to calculate our emitter voltage (at n_2), and our collector current:

``````Vn_2 = Vn_1 - 0.65V
= 1.92V - 0.65V
= 1.27V
``````
``````I_qc = Vn_2 / R4
= 1.27V / 1KR
= 1.27mA
``````

And given our collector current we can calculate the voltage drop through R3 and our output voltage Vn_out:

``````Vn_out = 12V - (I_qc * R3)
= 12V - (1.27mA * 3.9KR)
= 7.04V
``````

Thus our DC analysis (at Vin = 0V) i.e. our hand calculated operating point figures are as follows:

``````Vn_1   = 1.92V
Vn_out = 7.04V
Vn_2   = 1.27V
``````

And these figures are rather close for our purposes.

Thus our analyses of what the circuit should do and what the circuit actually does in our simulation are in agreement – that’s a good thing.