Lab 14 RS

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The Common Emitter Amplifier

Circuit

Construct the common emitter amplifier circuit below according to your type of emitter.

Calculate all the R and C values to use in the circuit such that

a. Try $R_B \approx 220 \Omega$ and $I_C \approx 100 \mu A$

b. $I_C \gt 0.5$ mA DC with no input signal

c. $V_{CE} \approx V_{CC}/2 \gt 2$ V

d. $V_{CC} \lt V_{CE}(max)$ to prevent burnout

e. $V_{BE} \approx 0.6 V$

f. $I_D \approx 10 I_B \lt 1$ mA

Let's $V_{CC} = 11\ V$, $R_E = 0.2\ k\Omega$ and $R_C = 2.0\ k\Omega$.

The load line equation becomes:

[math]I_C = \frac{V_{CC}-V_{CE}}{R_E+R_C} = \frac{(11 - V_{CE})\ V}{2.2\ k\Omega} [/math]

About $R_1$, $R_2$ and $R_E$, $C_1$, $C_2$ and $C_E$ see my calculations below.

Draw a load line using the $I_{C}$ -vs- $I_{EC}$ from the previous lab 13. Record the value of $h_{FE}$ or $\beta$.

On the plot below I overlay the output transistor lines (from the previous lab report #13) and the Load Line calculated above.

My reported values of $\beta$ in lab report #13 was $\beta \approx 150$. But this because we have used the approximation $V_E \approx 0$ because small resistor value $R_E = 100\ \Omega$ was used.

If we will do the accurate calculation of $\beta$ based on my lab report #13 measurements using exact formula:

[math]I_B = \frac{(V_{bb}-V_{BE}-V_E)}{R_B}[/math]

we will end up with value of $\beta$:

 [math]\beta \approx 170[/math]

Set a DC operating point $I^{\prime}_C$ so it will amplify the input pulse given to you. Some of you will have sinusoidal pulses others will have positive or negative only pulses.

I will set up my operating point in the middle of the load line:

[math]I_C = 2.5\ mA[/math], [math]V_{EC} = 5.5\ V[/math].

Let's calculate all bias voltage needed to set up this operating point. Because the knowing of $V_{BE}$ and $\beta$ is very important for this calculation I did the preliminary set up to measure this quantities. They are the only parameters which depends from transistor. I was able to find:

[math]V_{BE} = 0.68\ V[/math]

[math]\beta = 173[/math]

Now

[math]V_E = I_E \cdot R_E = 2.5\ mA \cdot 0.2\ k\Omega = 0.5\ V[/math]
[math]V_B = V_E + V_{BE} = (0.50 + 0.68)\ V = 1.18\ V[/math]

To set up the above operating point we need to set up $V_{B} = 1.18\ V$.

We have:

[math]I_B = \frac{I_C}{\beta} = \frac{2.5\ mA}{173} = 14.4\ uA[/math].

To get operating point independent of the transistor base current we want $I_D = I_{R1} \gg\ I_B$

Let's $I_{R1} = 590\ uA \gg\ I_B = 14.4\ uA$

So

[math]R_1 = \frac{V_B}{I_1} = \frac{1.18\ V}{590\ uA} = 2\ k\Omega[/math]

And $R_2$ we can find from Kirchhoff Voltage Low:

[math]V_{CC} = I_2 \cdot R_2 + V_B[/math].

and Kirchhoff Current Low:

[math]I_2 = I_1 + I_B[/math]

So

 [math]R_2 = \frac{V_{CC}-V_B}{I_1+I_B} = \frac{(11-1.18)\ V}{(590 + 14.4)\ mA} = 16.25\ k\Omega[/math]

I tried to adjust my calculation by varying the fee parameters $V_{cc}$ and $I_1$ to end up with all my resistor values I can easily set up.

Now when we know all resistor values we are ready to calculate capacitor values.

1) To prevent decreasing in gain we need to make capacitance $C_E$ large enough so the $X_C$ will become small enough compared to $R_E$ to make AC signal grounded. Also note that the DC bias is not affected because $C_E$ will not pass the direct current.

[math]X_C = \frac{1}{\omega C_E} \ll R_E[/math]

So

[math]C_E \gg \frac{1}{\omega_{min}R_E} = \frac{1}{(2\pi)(20\ Hz)(200\ \Omega)} = 40 \mu F[/math]

I choose the electrolytic capacitor:

[math]C_E = 100\ \mu F[/math].

2) Now we need to block the amplifier from input and output signal to avoid change the DC bias. We can do it by placing the "blocking" capacitor, in series in-front and in-back of the amplifier. They both form the high-pass filter. And if we want to amplify the frequency more than $20\ Hz$ we need to choose the breakpoint frequency to be:

[math]\omega_B = \frac{1}{RC} \ll (2\pi\ 20)\ Hz[/math] 

For the output high-pass filter from the ac equivalent circuit we have $R = R_L$ where the resistance $R_L$ is the resistance of the output device. Let's $R_L = 10\ k\Omega$

So

[math]C_2 \gg \frac{1}{(2\pi\ 20)\ R_L} = \frac{1}{(2\pi\ 20)\ 10\ k\Omega} = 0.8\times10^{-6} F[/math]

For the input high-pass filter from the ac equivalent circuit we have

[math]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_{BE}} =  \frac{1}{2\ k\Omega} + \frac{1}{16.25\ k\Omega} + \frac{1}{2.6\ k\Omega} = 0.94\ k\Omega [/math]

So

[math]R = 1.06\ k\Omega[/math]

And

[math]C_2 \gg \frac{1}{(2\pi\ 20)\ R_L} = \frac{1}{(2\pi\ 20)\ 1.06\ k\Omega} = 7.5\times10^{-6} F[/math]

To be safe I have choose the following values for $C_1$ and $C_2$

[math]C_1 = C_2 = 100\ \mu F[/math]

Measure all DC voltages in the circuit and compare with the predicted values.(10 pnts)

My predicted DC voltages are: (from the calculation above):

[math]V_{EC} = 5.50\ V[/math] 
[math]V_{BE} = 0.68\ V [/math]

[math]V_E = 0.50\ V[/math]
[math]V_B = 1.18\ V[/math]
[math]V_C = V_E + V_{EC} = (0.50 + 5.50)\ V = 6.00\ V[/math]

My measured DC voltages are:

Here is very important to set up all resistor values as close as possible to my assumed values.

After many tries and errors I was able to end up with the following values of my resistors:

[math]R_E = (200.0 \pm 0.1)\ \Omega[/math]
[math]R_C = (2.002 \pm 0.001)\ k\Omega[/math]
[math]R_1 = (2.004 \pm 0.001)\ k\Omega[/math]
[math]R_2 = (16.26 \pm 0.01)\ k\Omega[/math]

And my measurements of DC voltages looks like:

[math]V_{cc} = (11.00 \pm 0.01)\ V[/math]

[math]V_E = (0.500 \pm 0.001)\ V[/math]
[math]V_B = (1.183 \pm 0.001)\ V [/math]
[math]V_C = (6.03 \pm 0.01)\ V [/math]

[math]V_{BE} = (0.683 \pm 0.001)\ V[/math]
[math]V_{EC} = (5.53 \pm 0.01)\ V[/math]

[math]V_{R_2} = (9.82 \pm 0.01)\ V[/math]
[math]V_{R_C} = (4.97 \pm 0.01)\ V[/math]

All my measurements are in agreement with each other within experimental errors.

I mean here that $V_B = V_E+V_{BE}$, $V_C = V_E+V_{EC}$ and $V_{cc} = V_B+V_{R_2}$, $V_{cc} = V_C+V_{R_C}$

Also all my predicted values are in agreement with my measured DC voltage values except of the fact that my measured $V_{BE} = 0.683\ V$ instead of $V_{BE} = 0.680\ V$ as I initially assumed. That gives me the correspondent corrections to $V_B$. But if I will consider only the one significant sign all my predicted and measured DC voltage values are in total agreement.

Below are my current measurements which I did using millivoltmeter and which are also in agreement with each other and with all my previous calculation:

[math]I_E = (2.48 \pm 0.01)\ mA[/math] 
[math]I_C = (2.49 \pm 0.01)\ mA[/math] 
[math]I_B = (14.4 \pm 0.1)\ uA[/math] 

[math]I_1 = (588 \pm 1)\ uA[/math] 
[math]I_1 = (603 \pm 1)\ uA[/math]

Measure the voltage gain $A_{\nu}$ as a function of frequency and compare to the theoretical value.(10 pnts)

In the table below are my input and output voltage measurements and voltage gain calculation. Here the $V_{in}$ is the peak-to-peak value of the input voltage, the $V_{out}$ is the peak-to-peak value of the output voltage, and $A_{\nu} = \frac{V_{out}}{V_{in}}$ is the voltage gaine.

My theoretical voltage gain are:

[math]A_{\nu} = \frac{\beta R_C}{R_{BE}}[/math].

Here I do not really know the values of $R_{BE}$. But typically $R_C$ and $R_{BE}$ are the same order of magnitude so the approximate voltage gain are:

[math]A_{\nu} \approx \beta = 173[/math].

As we can see from the table above my theoretical voltage gain are in good agreement with the measured ones in wide range of frequencies from the low frequency $f = 300\ Hz$ up to high frequency $f = 200\ kHz$

The decreasing of measured gain in low frequencies can be explained from the fact that capacitors $C_1$ and $C_2$ actually form the high-pass filter for input and output AC signals. And because I have measured $V_{in}$ before $C_1$ and $V_{out}$ after $C_2$ that will drop down my measured gain at low frequencies.

The decreasing of measured gain at high frequency can be explain from the fact that the transistor have some nonzero capacitance value $C_{BE}$. So the transistor by itself will behave as low-pass filter. Because this $C_{BE}$ capacitance should be small enough the breaking point frequency $\omega_B = \frac{1}{RC}$ should be high enough. As the result we have the attenuation of measured voltage gain at high frequencies.

Measure $R_{in}$ and $R_{out}$ at about 1 kHz and compare to the theoretical value.(10 pnts)

How do you do this? Add resistor in front of $C_1$ which you vary to determine $R_{in}$ and then do a similar thing for $R_{out}$ except the variable reistor goes from $C_2$ to ground.

Input impedance $R_{in}$ measurements and comparing with theory

To measure the input impedance of my amplifier I have set up the circulant below. Here I have attached the resistor $R_L$ in-front of capacitor $C_1$ and I replaced all my input internal resistor of amplifier by equivalent resistor $R_{inp}$. Here I have draw the capacitor $C_1$ only for clearness and it does not included in calculation below. My input signal frequency is $1\ kHz$

Using Ohm's Law:

[math]I = \frac{V_A-V_B}{R_L}[/math]

and

[math]R_{inp} = \frac{V_B}{I} = R_L\ \frac{V_B}{V_A-V_B}[/math]

Below is the table with my measurements and $R_{inp}$ calculation. I did that calculation for several values of $R_L$. Here $V_A$ and $V_B$ are the RMS values.

The minimum error for $R_{inp}$ I have for the case $R_L=4030\ \Omega$.

So my best estimate for $R_{inp}$ is:

[math]R_{inp} = (1727.1 \pm 35.7)\ \Omega[/math]

The theoretical values of input impedance can be described using the so called $h$ parameter as:

[math]R_{in} \approx (h_{ie}||R_1||R_2)[/math]

where

[math] h_{ie} \cong \frac{\beta kT}{e I_C}[/math]

or using $\beta \approx 173$, $T = 300\ K$ and $I_C = 2.5\ mA$

[math] h_{ie} = 1.79\ k\Omega[/math]

So,

[math](R_{in})_{theory} \approx (1.79\ k\Omega||2.00\ k\Omega||16.26\ k\Omega) \approx 0.89\ k\Omega[/math]

This result is in not really well agreement with my measured values of

[math](R_{inp})_{exp} = (1727.1 \pm 35.7)\ \Omega[/math]

The discrepancy probably can be explained by approximate nature of the theoretical formula and maybe assumed values of temperature $T = 300\ K$ which I do not really measured. Maybe my method of measuring input impedance was not good. Anyway this big discrepancy need more detailed study.

Output impedance $R_{out}$ measurements and comparing with theory

To measure the output impedance of my amplifier I have set up the circulant below. Here I have attached the resistor $R_L$ after of capacitor $C_2$ and I replaced all my output internal resistor of amplifier by equivalent resistor $R_{out}$. Here I have draw the capacitor $C_2$ only for clearness and it does not included in calculation below.

Here to find out the output impedance we can use the battery method from lab#1. By Kirchoff voltage law:

[math]V_B=V_A - I\cdot R_{out}[/math]

and

[math]I = \frac{V_B}{R_L}[/math]

So by graphing the current on the x-axis and the measured voltage $V_B$ on the y-axis for several values of the resistance $R_L$ we can find the output internal impedance of our amplifier as the slope of the line $V_B=V_A - I\cdot R_{out}$

Below is the table with my measurements and current calculation. Here $V_B$ is the RMS values.

And below is my plot of output voltage $V_B$ as function of current $I$:

The line equation is $V_B[mV]=p_0[mV] + p_1\cdot I[\mu A]$ The slope of this line is $p_0=(1.971\pm 0.004)\ \frac{mV}{\mu A} = (1.971\pm 0.004)\ k\Omega$.

So my measured output impedance is:

[math] R_{out} = (1.971\pm 0.004)\ k\Omega[/math]

The theoretical values of output impedance can be described as:

[math]R_{out} \approx \left(\frac{h_{ie}}{\Delta_e h}\right)||R_C[/math]

I do not really know here the $\Delta_e h$ but assuming the values of $\Delta_e h \approx 2\cdot 10^{-2}$ and using $h_{ie} = 1.79\ k\Omega$ calculated above

[math]R_{out} \approx (89.5\ k\Omega)||R_C[/math]

Using the actual values of $R_C = 2\ k\Omega$ we get finally for the theoretical values of output impedance:

[math]\left(R_{out}\right)_{theory} \approx (89.5\ k\Omega||2.0\ k\Omega) = 1.96 k\Omega[/math]

This theoretical values is in very well agreement with my measured values

[math] \left(R_{out}\right)_{exp} = (1.971\pm 0.004)\ k\Omega[/math]

that tell me that the method I have used to measure the output impedance is pretty good and accurate.

Measure $A_v$ and $R_{in}$ as a function of frequency with $C_E$ removed.(10 pnts)

To measure $A_v$ and $R_{in}$ as a function of frequency with $C_E$ removed I have set up the following circuit:

Here the box represents the equivalent circuit for my amplifier (with removed $C_E$). Here I have draw the capacitor $C_1$ and $C_2$ only for clearness and it does not included in calculation below.

The input impedance $R_{in}$ is as before:

[math]R_{inp} =  R_{load}\ \frac{V_b}{V_a - V_b}[/math]

and the voltage gain $A_{\nu}$ is as before

[math]A_{\nu} =\frac{V_{out}}{V_{inp}}[/math]

In the table below are presented my measurements and $R_{in}$ and voltage gain $A_v$ calculation. Here $V_A$, $V_B$ and $V_C$ are the RMS values. I have measured peak-to-peak values by oscilloscope and than convert them to RMS. It's why I have fractional values for voltage errors.

As we can observe from the table above the measured voltage gain for the case with removed $C_E$ is much less than the measured voltage gain values for the case with $C_E$. It's in completely agreement with the main function of emitter conductance as $C_E$ to avoid decreasing the gain of amplifier.

Questions

Why does a flat load line produce a high voltage gain and a steep load line a high current gain? (10 pnts)

If we have a flat load line then by changing a little the input signal (so we will change the $I_B$ and will move up or down in load line) we will change a lot of $V_{CE}$. As the result we have high voltage gain.

If we have a steep load line then by changing a little the input signal (so we will change the $I_B$ and will move up or down in load line) we will change a lot of $I_C$. As the result we have high current gain.

What would be a good operating point an an $npn$ common emitter amplifier used to amplify negative pulses?(10 pnts)

The operating point should be as higher as possible in the load line so we will have the freedom of moving down in the load line to amplify the negative pulses. But it still has to be in saturation region of transistor.

What will the values of $V_C$, $V_E$ , and $I_C$ be if the transistor burns out resulting in infinite resistance. Check with measurement.(10 pnts)

With burned transistor (removed) I will end up with the DC following circuit:

As we can easily see because we have the open loop there are no current in collector and emitter and all voltage $V_{bb}$ will drop between collector and emitter. As the result:

[math]V_C = V_{bb} = 11\ V[/math]

[math]V_E = V_C - V_{BE} = (11 - 11)\ V[/math]

[math]I_C = 0\ A[/math]

My measured values are:

[math]V_C = (11.00 \pm 0.01)\ V[/math]

[math]V_E = (0.0 \pm 0.0)\ V[/math]

[math]I_C = (0.0 \pm 0.0)\ A[/math]

that is in total agreement with my predicted values

What will the values of $V_C$, $V_E$ , and $I_C$ be if the transistor burns out resulting in near ZERO resistance (ie short). Check with measurement.(10 pnts)

With burned transistor (with zero resistance) I will end up with the DC following circuit:

Now we have the perfect voltage divider:

[math]V_C = V_E = \frac{(R_E||R_1)}{(R_E||R_1) + (R_C||R_2)}\cdot V_{bb}[/math]

where

[math]R_E||R_1 = \frac{R_E R_1}{R_E + R_1} [/math]

[math]R_C||R_2 = \frac{R_C R_2}{R_C + R_2} [/math]

and using my actual values of resistors

[math]R_E = (200.0 \pm 1.0)\ \Omega[/math]
[math]R_C = (2.002 \pm 0.001)\ k\Omega[/math]
[math]R_1 = (2.004 \pm 0.001)\ k\Omega[/math]
[math]R_2 = (16.26 \pm 0.01)\ k\Omega[/math]

I end up with

[math]R_E||R_1 = (0.1818 \pm 0.0009)\ k\Omega  [/math]

[math]R_C||R_2 = (1.7825 \pm 0.0002)\ k\Omega [/math]

[math]V_C = V_E = \frac{(0.1818 \pm 0.0009)}{(0.1818 \pm 0.0009) + (1.7825 \pm 0.0002)}\cdot (11.00 \pm 0.01)\ V  = (1.018 \pm 0.005)\ V [/math]

[math]I_C = \frac{V_{bb}-V_E}{R_C} = \frac{(11.00\pm 0.01)V-(1.018 \pm 0.005)V}{(2.002 \pm 0.001)k\Omega} = (4.986 \pm 0.009)\ mA[/math]

My measured values are:

[math]V_C = V_E = (1.021 \pm 0.001)\ V[/math]

[math]I_C = (5.00 \pm 0.01)\ mA[/math]

that is in total agreement with my predicted values

Predict the change in the value of $R_{in}$ if $I_D$ is increased from 10 $I_B$ to 50 $I_B$(10 pnts)

Taking my biased base current value calculated above $I_B = 14.4\ uA$ my corresponded $I_D$ are:

[math](I_D)_1 = 10 \cdot 14.4\ uA= 144\ uA[/math]
[math](I_D)_2 = 50 \cdot 14.4\ uA = 720\ uA[/math]

My change in $V_B$ becomes:

[math](V_B)_1 = (I_D)_1 \cdot R_1 = (144\ uA \cdot 2\ k\Omega) = 0.288\ V[/math]
[math](V_B)_2 = (I_D)_1 \cdot R_1 = (720\ uA \cdot 2\ k\Omega) = 1.440\ V[/math]

and corresponding change in $V_E$ becomes:

[math](V_E)_1 = (V_B)_1 - V_{BE} = (0.288 - 0.68) = -0.392\ V[/math]
[math](V_E)_2 = (V_B)_2 - V_{BE} = (1.440 - 0.68) = 0.76\ V[/math]

Here I ended up with the negative $(V_B)_1$ value that is impossible for my case. The point here that my $R_1$ was calculated to be able to conduct current $I_D = 590\ uA \gg 14.4\ uA$. So the case of increasing $I_D$ from 10 $I_B$ to 50 $I_B$ is just impossible for my circuit just because I can get so low $I_D = 10\ I_B$ values.

To make some reasonable estimate I will pick up the change in $I_D$ current from 40 $I_B$ to 200 $I_B$, so my relative change is 5 times as before. Now I have:

[math](I_D)_1 = 40 \cdot 14.4\ uA = 576\ uA[/math]
[math](I_D)_2 = 200 \cdot 14.4\ uA = 2880\ uA[/math]

My change in $V_B$ becomes:

[math](V_B)_1 = (I_D)_1 \cdot R_1 = (576\ uA \cdot 2\ k\Omega) = 1.152\ V[/math]
[math](V_B)_2 = (I_D)_1 \cdot R_1 = (2880\ uA \cdot 2\ k\Omega) = 5.760\ V[/math]

and corresponding change in $V_E$ becomes:

[math](V_E)_1 = (V_B)_1 - V_{BE} = (1.152 - 0.68) = 0.472\ V[/math]
[math](V_E)_2 = (V_B)_2 - V_{BE} = (5.760 - 0.68) = 5.08\ V[/math]

and the $I_E$ current change:

[math](I_E)_1 = \frac{(V_E)_1}{R_E} = \frac{0.472\ V}{200\ \Omega} = 2.36\ mA[/math]
[math](I_E)_2 = \frac{(V_E)_1}{R_E} = \frac{5.08\ V}{200\ \Omega} = 25.4\ mA[/math]

Now I can use formula for transistor input impedance:

[math] h_{ie} \cong \frac{\beta kT}{e I_C}[/math]

and because $I_E \approx I_C$

[math] (h_{ie})_1 \cong \frac{\beta kT}{e (I_C)_1} = 1.89\ k\Omega[/math]
[math] (h_{ie})_2 \cong \frac{\beta kT}{e (I_C)_2} = 0.18\ k\Omega[/math]

Now I can use formula for the amplifier input impedance:

[math]R_{in} \approx (h_{ie}||R_1||R_2)[/math]

and using corresponding values of resistors I will end up with:

[math](R_{in})_1 \approx (1.89||2.0||16.26)\ k\Omega = 0.92\ k\Omega[/math]
[math](R_{in})_2 \approx (0.18||2.0||16.26)\ k\Omega = 0.16\ k\Omega[/math]

So if I increase my $I_D$ current from 40 $I_B$ to 200 $I_B$ I will decrease my amplifier input impedance from $0.92\ k\Omega$ to $0.16\ k\Omega$

Sketch the AC equivalent circuit of the common emitter amplifier.(10 pnts)

The point here is to remove the resistors $R_E$ which is shorted out by capacitor $C_E$ and remove all capacitors $C_1$, $C_2$ and $C_E$ which are have zero resistance for ac circuit. Also because $V_{bb}$ terminal of power supply is also ac grounded we can combine the $R_1$ and $R_2$ into one resistor $(R1||R2)$ and we can connect the resistor $R_C$ to the ground. So the ac equivalent circuit of the common emitter amplifier becomes:

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