Altium Designer is used for PCB mainly but it also supports circuit simulation. Here we will see how to do this by designing and simulating amplifier circuit. The amplifier circuit uses a single transistor and is biased using VDB(Voltage Divider Biaising). We will also add coupling capacitor and show how to calculate these capacitor values. Finally we will simulate the circuit.

If you want this
+9V visible in the schematics, open the value parameter properties window and check the visible option as shown below.

Change the designation name to Vcc. Now the DC source VSRC will look like the one below-

Notice that when you first place a resistor it will be like R? with question mark. When we compile the schematic then altium designer will automatically designate labels such as R1, R2 etc.

But we can also manually rename them. For the purpose of discussion, we will rename the resistors as follows.

The resistors R1 and R2 forms the voltage divider resistors. Rc and Re are the collector resistor and emitter resistor.

Biasing the transistor means calculating the value of resistors so that we have the stable and required output voltage and current at the collector.

For VDB(Voltage Divider Biasing) we begin by making the emitter voltage drop of one tenth of the supply voltage-

\[V_E=\frac{V_{CC}}{10}=\frac{9V}{10}=0.9V\]

\[I_E \approx I_c =10mA\]

With this we can calculate the resistor at the emitter-

\[R_E =\frac{V_E}{I_E}=\frac{0.9V}{10mA}=90ohm\]

Then we get the collector resistor value-

\[R_C =4R_E=4*90ohm=360ohm\]

Using the formula below we can now calculate the value of R1:

\[R_1<\frac{\beta_{dc}*R_E}{10}\]

that is,

\[R_1<\frac{100*90ohm}{10}=900ohm\]

\[\frac{R_2}{R_1}=\frac{V_2}{V_1}\]

or,

\[R_2=R_1\frac{V_2}{V_1}\]

We have,

\[V_2=V_E+V_{BE}=0.9V+0.7V=1.6V\]

and,

\[V_1=V_{CC}-V_2=9V-1.6V=7.4V\]

Substituting the known values we get,

\[R_2=R_1\frac{V_2}{V_1}=900ohm\frac{1.6V}{7.4V}=194.59ohm\approx195ohm\]

Once the transistor is biased we add the signal source and load resistor.

We will be using assuming signal source of 10mV amplitude and 1kHz frequency. To set this double click on VSIN and edit its properties as shown below.

We assume the load resistor to have value of 10kohm.

The formula to calculate the coupling capacitors capacitance is,

\[X_c<\frac{Z}{10}\]

where, Xc is the reactance of the capacitor and Z is the load impedance

For the output coupling capacitor C2 we have,

\[X_{c2}<\frac{Z}{10}=\frac{R_{load}}{10}=\frac{10kohm}{10}=1kohm\]

Therefore,

\[\frac{1}{2\pi f C_2}<1kohm\]

or

\[C_2>\frac{1}{2\pi f*1k}=\frac{1}{2\pi 1k*1k}=0.159\mu F\]

at f=1kHz

We have calculated this capacitance at 1kHz. Thus it allows signal at 1kHz to pass through it, that is, short circuit at 1KHz. This tells us only what the value of capacitor must be to allow our 1kHz signal to pass through to the output. But we will need also to allow lower frequency than 1kHz to pass through. As you know the capacitor and frequency are inversely related. For lower value of frequency the capacitance should be increased. Lets take C2 as 0.8 uF for 20Hz. Remember that will still block the dc as required.

We will also take C1 as 0.8uF as approximate. One can go through the details to calculate the input coupling capacitor C1 but it is tedious and not beneficial to big extent in this case. But the process is the same as that of calculating for C2. The VDB circuit with transistor will be the load(parallel combination of R1, R2 and the Zin(base) of the transistor).

Update the input and output coupling capacitor with 0.8uF value.

A bypass capacitor is also needed parallel to the emitter resistor. The same formula used for the coupling capacitor above can be used.

\[X_c<\frac{Z}{10}\]

Here f=1kHz, and Z=90ohm are required to calculate value of this bypass capacitor C3.

\[C_3>17.69\mu F\]

But taking lower frequency into account say as before f=20Hz gives us-

\[C_3>0.088\mu F\]

taking approximate-

\[C_3>0.1\mu F\]

This bypass capacitor with its value is shown in the schematic below-

In altium designer, we need to attach labels to the wires in the circuit where we want to measure the signal. Here we are interested in the input signal and the output signal.

Right click on the schematic and choose Place> Net Label(or click the Net icon on the toolbar). Then add two net labels at the input and output and rename them as Vin and Vout as shown in the figure below-

Now compile the project by going to the Project >> Compile PCB Project

Next we have to specify what kind of simulation we are interested in. Click the Setup Mixed-Signal Simulation icon on the toolbar or if this is not visible you can use the Design>Stimulate>Mixed Sim.

See the figure below.

Now, in order to capture the input and output
waveform ensure that the collect data option has voltage node, supply
current, device current and power selected. Also set the simview setup
option to show active signals.

From the available signal lists select the two signals Vin and Vout and click on the '>' to move them to the active signal list. This is shown below.

Select the operating point analysis and the transient
analysis. To change the default transient settings, uncheck the Use
Transient Default. Set the Transient Stop Time to 10m and Transient Step
Time to 10u as shown below-

Now to run the simulation, click on the Run Mixed-Signal Simulation as shown-

This will generate the desired input(Vin) and output signal(Vout) waveform as shown below.

From the graph we can see that the magnitude of
the output waveform is approximately 23mV. The input Peak to peak voltage of input
is 20mV whereas the output peak to peak voltage of output waveform is
46mV. We also see that the input and output signal are 180 degree out of phase

### Creating the Schematic

Create a new project and make the following circuit. The circuit uses the 2N3904 transistor in voltage divider arrangement to bias the transistor.Fig: Voltage Divider Biasing |

#### Change the DC voltage to 9V

Double click on the VSRC, then edit. In the new pop up window select Parameters tab and enter +9V in the value box.Fig: Setting up Power Supply |

Fig: Setting up Power Supply |

Fig: Power Supply |

But we can also manually rename them. For the purpose of discussion, we will rename the resistors as follows.

Fig: Renaming Resistors |

The resistors R1 and R2 forms the voltage divider resistors. Rc and Re are the collector resistor and emitter resistor.

### Biasing the transistor

Biasing the transistor means calculating the value of resistors so that we have the stable and required output voltage and current at the collector.

For VDB(Voltage Divider Biasing) we begin by making the emitter voltage drop of one tenth of the supply voltage-

\[V_E=\frac{V_{CC}}{10}=\frac{9V}{10}=0.9V\]

#### Calculating R1 value

The collector and emitter current are approximately-\[I_E \approx I_c =10mA\]

With this we can calculate the resistor at the emitter-

\[R_E =\frac{V_E}{I_E}=\frac{0.9V}{10mA}=90ohm\]

Then we get the collector resistor value-

\[R_C =4R_E=4*90ohm=360ohm\]

Using the formula below we can now calculate the value of R1:

\[R_1<\frac{\beta_{dc}*R_E}{10}\]

that is,

\[R_1<\frac{100*90ohm}{10}=900ohm\]

#### Calculating R2 value

To calculate R2 we use the voltage divider rule:\[\frac{R_2}{R_1}=\frac{V_2}{V_1}\]

or,

\[R_2=R_1\frac{V_2}{V_1}\]

We have,

\[V_2=V_E+V_{BE}=0.9V+0.7V=1.6V\]

and,

\[V_1=V_{CC}-V_2=9V-1.6V=7.4V\]

Substituting the known values we get,

\[R_2=R_1\frac{V_2}{V_1}=900ohm\frac{1.6V}{7.4V}=194.59ohm\approx195ohm\]

#### Updating the schematic with resistor values

The following figure shows the schematics after entering the resistor values-Fig: Resistor Values |

### Adding Signal Source and Load Resistor

Once the transistor is biased we add the signal source and load resistor.

Fig: Adding signal source and load |

We will be using assuming signal source of 10mV amplitude and 1kHz frequency. To set this double click on VSIN and edit its properties as shown below.

Fig: Setting up Input Signal |

Fig:Setting up Load |

### Input and Output Coupling Capacitor

Add two coupling capacitor at the input(C1) and output(C2) of the biasing circuit as shown.Fig: Adding Coupling Capacitors |

\[X_c<\frac{Z}{10}\]

where, Xc is the reactance of the capacitor and Z is the load impedance

For the output coupling capacitor C2 we have,

\[X_{c2}<\frac{Z}{10}=\frac{R_{load}}{10}=\frac{10kohm}{10}=1kohm\]

Therefore,

\[\frac{1}{2\pi f C_2}<1kohm\]

or

\[C_2>\frac{1}{2\pi f*1k}=\frac{1}{2\pi 1k*1k}=0.159\mu F\]

at f=1kHz

We have calculated this capacitance at 1kHz. Thus it allows signal at 1kHz to pass through it, that is, short circuit at 1KHz. This tells us only what the value of capacitor must be to allow our 1kHz signal to pass through to the output. But we will need also to allow lower frequency than 1kHz to pass through. As you know the capacitor and frequency are inversely related. For lower value of frequency the capacitance should be increased. Lets take C2 as 0.8 uF for 20Hz. Remember that will still block the dc as required.

We will also take C1 as 0.8uF as approximate. One can go through the details to calculate the input coupling capacitor C1 but it is tedious and not beneficial to big extent in this case. But the process is the same as that of calculating for C2. The VDB circuit with transistor will be the load(parallel combination of R1, R2 and the Zin(base) of the transistor).

Update the input and output coupling capacitor with 0.8uF value.

Fig: Adding Coupling Capacitors value |

**Calculating the Bypass Capacitor**A bypass capacitor is also needed parallel to the emitter resistor. The same formula used for the coupling capacitor above can be used.

\[X_c<\frac{Z}{10}\]

Here f=1kHz, and Z=90ohm are required to calculate value of this bypass capacitor C3.

\[C_3>17.69\mu F\]

But taking lower frequency into account say as before f=20Hz gives us-

\[C_3>0.088\mu F\]

taking approximate-

\[C_3>0.1\mu F\]

This bypass capacitor with its value is shown in the schematic below-

Fig: Complete Amplifer Design schematic |

**Simulation**In altium designer, we need to attach labels to the wires in the circuit where we want to measure the signal. Here we are interested in the input signal and the output signal.

Right click on the schematic and choose Place> Net Label(or click the Net icon on the toolbar). Then add two net labels at the input and output and rename them as Vin and Vout as shown in the figure below-

Fig: Adding Net Labels |

Next we have to specify what kind of simulation we are interested in. Click the Setup Mixed-Signal Simulation icon on the toolbar or if this is not visible you can use the Design>Stimulate>Mixed Sim.

See the figure below.

Fig: Simulation Setting in Altium Designer |

From the available signal lists select the two signals Vin and Vout and click on the '>' to move them to the active signal list. This is shown below.

Fig: Simulation Setting in Altium Designer |

Fig: Simulation Setting in Altium Designer |

Fig: Running Simulation in Altium Designer |

Fig: Resulting Input and Output Waveform |

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