Why does a common base amplifier give a non inverting output?

Question

I am unable to understand why a common base amplifier gives a non-inverting output.

My reasoning so far:

Let's take a look at the common collector (emitter follower) amplifier. Here \$V_e = V_b -0.7\$. So on differentiating we get \$ d{V_e} = d{V_b}\$ which clearly implies that output is non-inverting.

Similarly in the case of a common emitter amplifier \$ V_{out} = V_{cc} - I_cR_c \$. which implies that \$d{V_{out}} = -d{{I_c}R_c}= \beta d{I_b}R_c \$. The negative sign implying that output is inverted.

In the case of a common base amplifier, I can't figure it out with similar reasoning. When the emitter voltage increases then \$I_e\$ increases which in turn increases the \$I_c\$ and the situation again becomes like that of a CE amplifier. Please provide an explanation of why this is the case and where am I wrong.

Answer 1

No. When \$V_E\$ increases, \$I_E\$ decreases, because \$V_{BE}\$ decreases. Any extra current that flows through \$R_E\$ is supplied by the signal source, not the transistor.

Answer 2

With this incredible story, I want to congratulate @Tony Stewart EE75 - the virtuoso of circuits and Falstad simulations. Tony, welcome back to us!

Looking for similarity

It is a great idea to see the connection between an electronic circuit today and an electrical circuit from two centuries ago. It is even greater to make the connection between several electronic circuits with a basic electrical circuit. In this way, we will understand electronic circuits better than "This is common-emitter and this is common-base... It consists of... and so on..."

Looking at the schematics of the three basic transistor amplifier configurations so neatly arranged by OP, we can see that there is something very basic and common between them... and it is known since the 19th century.

Balanced bridge circuit

It is about the famous Wheatstone bridge circuit. It can be represented (in the schematic below) as two voltage dividers (half-bridges) R1-R2 and R3-R4 supplied by a common voltage source Vcc. When they have the same voltage ratios, their output voltages are the same and the difference between them (measured by a sensitive voltmeter or the null indicator NI) is zero; as they say, the bridge circuit is balanced. In this condition, the voltage drops across opposite resistors are equal (eg, VR4 = VR2), and the voltage drops across resistors on the same side (eg, VR3 and VR4) are connected through the common current in the relation VR3/VR4 = R3/R4.

^{simulate this circuit – Schematic created using CircuitLab}

Transistor "bridge circuit"

An emitter follower, driven by a voltage divider R1-R2 can be considered as such a "balanced bridge circuit" where the collector-emitter junction of the transistor acts as R3 and the load as R4. The transistor directly monitors the voltage difference by its base-emitter junction (no need for a zero indicator) and changes its collector-emitter "resistance" to zero the difference.

^{simulate this circuit}

Adjustable bridge circuit

To balance the bridge circuit and take the output, the upper resistor is divided into two parts - variable Rvar and constant R3. The procedure (aka "negative feedback") is simple - we change Rvar (in the right direction) until the indicator is zeroed.

^{simulate this circuit}

"Ideal" transistor bridge circuit

As incredible as it sounds, the 21st century transistor bridge circuit is, because of this "annoying" Vbe, more imperfect than the 19th century 4-resistor Wheatstone bridge circuit. This is simply intolerable and we must do something to destroy this Vbe and make the circuit (almost) perfect. The simplest thing we can think of, as soon as a voltage drop of 623 mV is lost, is to add a voltage of 623 mV (there are more sophisticated op-amp solutions). So, let's connect a compensating voltage source before the base and adjust its voltage equal to Vbe read from the CircuitLab parameters window.

^{simulate this circuit}

Great! Our bridge circuit became perfect (almost like the one from two centuries ago:-) As a result, VRc = VRe = VR2 = 2.5 V and, as they say, the circuit is properly (DC) biased.

Common-emitter stage

Now we need to find a point (input) at which to apply an input signal. The first idea that comes to mind is the base of the transistor...

Electrical implementation

... but let's stick to the sequence of exposition and make it purely electrical first - as they would have done in the 19th century... if they had only wanted to. I will present the operation of this "transistor-man circuit" in three frames - at zero, positive and negative input voltage.

Zero input voltage 2.5 V (0 V referenced to 2.5 V): The "zero" input level of the circuit is 2.5 V. So we need to apply exactly that much input voltage Vin = 2.5 V. We see that nothing changes.

^{simulate this circuit}

Positive input voltage 3.5 V (1 V referenced to 2.5 V):

We need to add (bias) the input voltage to the zero level. In this conceptual diagram, we can simply increase it by 1 V and apply 3.5 V to the input.

^{simulate this circuit}

Negative input voltage 1.5 V (-1 V referenced to 2.5 V): Now we can decrease it by 1 V and apply 1.5 V to the input.

^{simulate this circuit}

Transistor implementation

In the practical transistor circuit, we divide the input source into two sources...

DC follower and amplifier: ... 1 V AC input source and 2.5 V DC bias source. The emitter voltage Vout1 across Re is a copy of the input base volage. The voltage across Rc is also a copy since Rc = Re... but if we change the ratio Rc/Re we can obtain attenuator (Rc < Re), follower (Rc = Re) or amplifier (Rc > Re). With this feature the circuit resembles an op-amp inverting amplifier.

^{simulate this circuit}

AC amplifier: To make the circuit a real AC amplifier with a high gain, we fave to fix the emitter voltage. For this purpuse, let's connect a big capacitor Ce in parallel to Re. As a result, when the input voltage varies, Ve does not vary and the input voltage variations are directly applied to the base-emitter junction.

^{simulate this circuit}

We see that the gain is more than 100. This has forced us to reduce the amplitude of the input signal to 10 mV in order not to saturate the transistor.

Common-base stage

With the same success we can drive the transistor from the emitter side keeping the base voltage constant.

Electrical implementation

In the "19th century circuit", this means setting Vref = 2.5 V on the left and varying Vin on the right relative to 2.5V. Also imagine that Rvar is a voltage-controlled resistor. So when Vin varies, the difference voltage, current and Vout vary as well.

^{simulate this circuit}

Transistor implementation

First we have to fix the base voltage by connecting a big capacitor Cref to the divider's output. Then, we "raise" the input voltage using a charged capacitor Ce to the 2.5 V level of the emitter and then begin changing Vin. Because of the fixed base voltage, the input voltage variations are directly applied to the base-emitter junction (like in the common-emitter stage above).

^{simulate this circuit}

As a result, an amplified copy of Vin appears across Rc and OUT.

How does the input part operate?

Let's take a closer look at what is going on with the currents in the input part of the circuit.

Real circuit. When the power of this conceptual circuit is turned on...

^{simulate this circuit}

... Ce quickly (exponentially) charges up to Vref (5 V).

After the voltage across the capacitor is established, we see that the current IRe through the emitter resistor is relatively constant, while the current IQ through the transistor and the input source varies widely.

Equivalent circuit. The best way to investigate this is to manually vary the input voltage. To do this, let's replace the capacitor and the AC input voltage source with a varying DC voltage source. Also, let's connect ammeters in the three branches to monitor the currents.

Vin = 5 V: No current flows through the input source as if it is not there. The 1 mA quiescent current flows through the transistor and the emitter resistor.

^{simulate this circuit}

Vin = 5.1 V: The transistor is almost off. The current through the emitter resistor (almost the same 1 mA) is provided entirely by the input voltage source.

^{simulate this circuit}

Vin = 4.9 V: The collector current vigorously increases and passes through the input voltage source.

^{simulate this circuit}

Why is CB circuit non-inverting?

The voltage Vbe is a difference between two voltages - Vb which increases the collector current (common-emitter stage) and Ve which decreases the collector current (common-base stage). If we use the voltage drop across Rc as the output, the common-emitter circuit would be non-inverting and the common-base inverting. But since we are not using the voltage VRc itself, but its complement to Vcc, the circuit operation is inverted and the common-base circuit is non-inverting.

For more explanations, see my answer to a related question.