**Input Impedance**, **Zin** or *Input Resistance* as it is also called, is an important parameter in the design of a transistor amplifier and as such allows amplifiers to be characterized according to their effective input and output impedances as well as their power and current ratings.

An amplifiers impedance value is particularly important for analysis especially when cascading individual amplifier stages together one after another to minimise distortion of the signal.

The *input impedance of an amplifier* is the input impedance “seen” by the source driving the input of the amplifier. If it is too low, it can have an adverse loading effect on the previous stage and possibly affecting the frequency response and output signal level of that stage. But in most applications, common emitter and common collector amplifier circuits generally have high input impedances.

Related Products: Amplifier IC Development Boards and Kits | RF Amplifier Chip

Some types of amplifier designs, such as the common collector amplifier circuit automatically have high input impedance and low output impedance by the very nature of their design. Amplifiers can have high input impedance, low output impedance, and virtually any arbitrary gain, but were an amplifiers input impedance is lower than desired, the output impedance of the previous stage can be adjusted to compensate or if this is not possible then buffer amplifier stages may be needed.

In addition to voltage amplification ( Av ), an amplifier circuit must also have current amplification ( Ai ). Power amplification ( Ap ) can also be expected from an amplifier circuit. But as well as having these three important characteristics, an amplifier circuit must also have other characteristics like high input impedance ( Zin ), low output impedance ( Zout ) and some degree of bandwidth, ( Bw ). Either way, the “perfect” amplifier will have infinite input impedance and zero output impedance.

In many ways, an amplifier can be thought of as a type of “black box” which has two input terminals and two output terminals as shown. This idea provides a simple *h-parameter model* of the transistor that we can use to find the DC set point and operating parameters of an amplifier. In reality one of the terminals is common between the input and output representing ground or zero volts.

When looking from the outside in, these terminals have an input impedance, Zin and an output impedance, Zout. The input and output impedance of an amplifier is the ratio of voltage to current flowing in or out of these terminals. The input impedance may depend upon the source supply feeding the amplifier while the output impedance may also vary according to the load impedance, R_{L} across the output terminals.

The input signals being amplified are usually alternating currents (AC) with the amplifier circuit representing a load Z to the source. The input impedance of an amplifier can be tens of ohms, (Ω’s) to a few thousand ohms, (kΩ’s) for bipolar based transistor circuits up to millions of ohms, (MΩ’s) for FET based transistor circuits.

When a signal source and load are connected to an amplifier, the corresponding electrical properties of the amplifier circuit can be modelled as shown.

Where, V_{S} is the signal voltage, R_{S} is the internal resistance of the signal source, and R_{L} is the load resistance connected across the output. We can expand this idea further by looking at how the amplifier is connected to the source and load.

When an amplifier is connected to a signal source, the source “sees” the input impedance, Zin of the amplifier as a load. Likewise, the input voltage, Vin is what the amplifier sees across the input impedance, Zin. Then the amplifiers input can be modelled as a simple voltage divider circuit as shown.

The same idea applies for the output impedance of the amplifier. When a load resistance, R_{L} is connected to the output of the amplifier, the amplifier becomes the source feeding the load. Therefore, the output voltage and impedance automatically becomes the source voltage and source impedance for the load as shown.

Then we can see that the input and output characteristics of an amplifier can both be modelled as a simple voltage divider network. The amplifier itself can be connected in *Common Emitter* (emitter grounded), *Common Collector* (emitter follower) or in *Common Base* configurations. In this tutorial we will look at the bipolar transistor connected in a common emitter configuration seen previously.

The so called classic common emitter configuration uses a potential divider network to bias the transistors Base. Power supply Vcc and the biasing resistors set the transistor operating point to conduct in the forward active mode. With no signal current flow into the Base, no Collector current flows, (transistor in cut-off) and the voltage on the Collector is the same as the supply voltage, Vcc. A signal current into the Base causes a current to flow in the Collector resistor, Rc generating a voltage drop across it which causes the Collector voltage to drop.

Then the direction of change of the Collector voltage is opposite to the direction of change on the Base, in other words, the polarity is reversed. Thus the common emitter configuration produces a large voltage amplification and a well defined DC voltage level by taking the output voltage from across the collector as shown with resistor R_{L} representing the load across the output.

Hopefully by now we are able to calculate the values of the resistors required for the transistor to operate in the middle of its linear active region, called the quiescent point or Q point, but a quick refresher will help us understand better how the amplifiers values were obtained so that we can use the above circuit to find the input impedance of the amplifier.

Firstly lets start by making a few simple assumptions about the single stage common emitter amplifier circuit above to define the operating point of the transistor. The voltage drop across the the Emitter resistor, V_{RE} = 1.5V, the quiescent current, I_{Q} = 1mA, the current gain (Beta) of the NPN transistor is 100 ( β = 100 ), and the corner or breakpoint frequency of the amplifier is given as: ƒ_{-3dB} = 40Hz.

As the quiescent current with no input signal flows through the Collector and Emitter of the transistor, then we can say that, I_{C} = I_{E} = I_{Q} = 1mA, so by using Ohms Law:

With the transistor switched fully-ON (saturation), the voltage drop across the Collector resistor, Rc will be half of Vcc – V_{RE} to allow for maximum output signal swing from peak-to-peak around the center point without clipping of the output signal.

Note that the DC no signal voltage gain of the amplifier can be found from -R_{C}/R_{E}. Also notice that the gain is negative in value as the output signal is inverted. i.e. 180^{o} out-of-phase with the input signal.

As the NPN transistor is forward biased, the Base-Emitter junction acts like a forward biased diode so the Base will be 0.7 volts more positive than the Emitter voltage ( Ve + 0.7V ), therefore the voltage across the Base resistor R2 will be:

If the two biasing resistors are already given, we can also use the following standard voltage divider formula to find the Base voltage Vb across R2.

The information given stated that the quiescent current is 1mA. Thus the transistor is biased with a Collector current of 1mA across the 12 volt supply, Vcc. This Collector current is proportional to the Base current as Ic = βIb. The DC current gain, Beta ( β ) of the transistor was given as 100, then the Base current flowing into the transistor will be:

The DC bias circuit formed by the voltage divider network of R1 and R2 sets the DC operating point. The Base voltage was previously calculated at 2.2 volts then we need to establish the proper ratio of R1 to R2 to produce this voltage value across the 12 volt supply, Vcc.

Generally, for a standard voltage divider DC biasing network of a common emitter amplifier circuit, the current flowing through the lower resistor, R2 is ten times greater than the DC current flowing into the Base. Then the value of resistor, R2 can be calculated as:

The voltage dropped across resistor R1 will be the supply voltage minus the Base bias voltage. Also if resistor R2 carries 10 times the Base current, upper resistor R1 of the series chain must pass the current of R2 plus the transistors actual Base current, Ib. In other words, 11 times the Base current as shown.

For a common emitter amplifier, the reactance Xc of the Emitter bypass capacitor is usually one tenth (1/10th) the value of the Emitter resistor, R_{E} at the cut-off frequency point. The amplifiers specifications gave a -3dB corner frequency of 40Hz, then the value of capacitor C_{E} is calculated as:

Now we have the values established for our common emitter amplifier circuit above, we can now look at calculating its input and output impedance of amplifier as well as the values of the coupling capacitors C1 and C2.

Related Products: RF Amplifier Module | SP Amplifier | Audio Amplifier | GPS Amplifier

The generalised formula for the input impedance of any circuit is Z_{IN} = V_{IN}/I_{IN}. The DC bias circuit sets the DC operating “Q” point of the transistor and as the input capacitor, C1 acts as an open circuit and blocks any DC voltage, at DC (0Hz) the input impedance (Z_{IN}) of the circuit will be extremely high. However when an AC signal is applied to the input, the characteristics of the circuit changes as capacitors act as short circuits at high frequencies and pass AC signals.

The generalised formula for the AC input impedance of an amplifier looking into the Base is given as Z_{IN} = R_{EQ}||β(R_{E}+ re). Where R_{EQ} is the equivalent resistance to ground (0v) of the biasing network across the Base, and re is the internal signal resistance of the forward biased Emitter layer. Then if we short out the 12 volt power supply, Vcc to ground because Vcc appears as a short to AC signals, we can redraw the common emitter circuit above as follows:

Then we can see that with the supply voltage shorted, there are a number of resistors connected in parallel across the transistor. By taking the input side of the transistor amplifier only and treating capacitor C1 as a short circuit to AC signals, we can redraw the above circuit to define the input impedance of the amplifier as:

We said in the previous Common Emitter Amplifier tutorial that the internal signal resistance of the Emitter layer was equal to the product of 25mV ÷ Ie with this 25mV value being the internal volt drop and I_{E} = I_{Q}. Then for our amplifier circuit above the equivalent AC resistance value re of the Emitter diode is given as:

Where re represents a small internal resistor in series with the Emitter. Since Ic/Ib = β, then the value of the transistors Base impedance will be equal to β x re. Note that if bypass capacitor C_{E} is not included within the amplifiers design, then the value becomes: β(R_{E}+ re) significantly increasing the input impedance of the amplifier.

In our example bypass capacitor, C_{E} is included, therefore the **input impedance, Z _{IN}** of the common Emitter amplifier is the input impedance “seen” by the AC source driving the amplifier and is calculated as:

This 2.2kΩ’s is the input impedance looking into the input terminal of the amplifier. If the impedance value of the source signal is known, and in our simple example above it is given as 1kΩ, then this value can be added or summed with Z_{IN} if required.

But lets assume for one minute that our circuit has no bypass capacitor, C_{E} connected. What would be the input impedance of the amplifier without it. The equation would still be the same except for the addition of R_{E} in the β(R_{E}+ re) part of the equation as the resistor will no longer be shorted at high frequencies. Then the unbypassed input impedance of our amplifier circuit without C_{E} will be:

Then we can see that the inclusion of the Emitter leg bypass capacitor makes a huge difference to the input impedance of the circuit as the impedance goes down from 15.8kΩ’s without it to 2.2kΩ’s with it in our example circuit. We will see later that the addition of this bypass capacitor, C_{E} also increases the amplifiers gain.

In our calculations to find the input impedance of the amplifier, we have assumed that the capacitors in the circuit have zero impedance (Xc = 0) for AC signal currents, as well as infinite impedance (Xc = ∞) for DC biasing currents. Now that we know the bypassed input impedance of the amplifier circuit, we can use this value of 2.2kΩ’s to find the value of the input coupling capacitor, C1 required at the specified cut-off frequency point which was given previously as 40Hz. Therefore:

Now that we have a value for the input impedance of our single stage common Emitter amplifier circuit above, we can also obtain an expression for the output impedance of the amplifier in a similar fashion.

The **Output Impedance** of an amplifier can be thought of as being the impedance (or resistance) that the load sees “looking back” into the amplifier when the input is zero. Working on the same principle as we did for the input impedance, the generalised formula for the output impedance can be given as: Z_{OUT} = V_{CE}/I_{C}.

But the signal current flowing in the Collector resistor, R_{C} also flows in the load resistor, R_{L} as the two are connected in series across Vcc. Then again, by taking the output side of the transistor amplifier only and treating the output coupling capacitor C2 as a short circuit to AC signals, we can redraw the above circuit to define the output impedance of the amplifier as:

Then we can see that the output signal resistance is equal to R_{C} in parallel with R_{L} giving us an output resistance of:

Note that this value of 833Ω’s results from the fact that the load resistance is connected across the transistor. If R_{L} is omitted, then the output impedance of the amplifier would be equal to the Collector resistor, R_{C} only.

Now that we have a value for the output impedance of our amplifier circuit above, we can calculate the value of the output coupling capacitor, C2 as before at the 40Hz cut-off frequency point.

Again the value of coupling capacitor C2 can be calculated either with or without the inclusion of load resistor R_{L}.

The voltage gain of a common emitter circuit is given as Av = R_{OUT}/R_{EMITTER} where R_{OUT} represents the output impedance as seen in the Collector leg and R_{EMITTER} is equal the the equivalent resistance in the Emitter leg either with or without the bypass capacitor connected.

Without the bypass capacitor C_{E} connected, (R_{E}+ re).

and with the bypass capacitor C_{E} connected, (re) only.

Then we can see that the inclusion of the bypass capacitor within the amplifier design makes a dramatic change to the voltage gain, Av of our common emitter circuit from 0.5 to 33. It also shows that the common emitter gain does not go to infinity when the external emitter resistor is shorted by the bypass capacitor at high frequencies but instead the gain goes to the finite value of R_{OUT}/re.

We have also seen that as the gain goes up the input impedance goes down from 15.8kΩ’s without it to 2.2kΩ’s with it. The increase in voltage gain can be considered an advantage in most amplifier circuits at the expense of a lower input impedance.

In this tutorial we have seen that the input impedance of a common emitter amplifier can be found by shorting out the supply voltage and treating the voltage divider biasing circuit as resistors in parallel. The impedance “seen” looking into the divider network (R1||R2) is generally much less that the impedance looking directly into the transistors Base, β(R_{E}+ re) as the AC input signal changes the bias on the Base of the transistor controlling the current flow through the transistor.

There are many ways to bias the transistor. Thus, there are many practical single transistor amplifier circuits each with their own input impedance equations and values. If you require the input impedance of the whole stage plus source impedance, then you will need to consider Rs in series with the base bias resistors as well, (Rs + R1||R2).

The output impedance of a common emitter stage is just equal to the collector resistor in parallel with the load resistor (R_{C}||R_{L}) if connected otherwise its just R_{C}. The voltage gain, Av of the amplifier is dependant upon R_{C}/R_{E}.

The emitter bypass capacitor, C_{E} can provide an AC ground path for the Emitter, shorting out the emitter resistor, R_{E} leaving only the signal Emitter resistance, re in the Emitter leg. The effect of this is an increase in the gain of the amplifier (from 0.5 to 33) at high frequencies but also a decrease in the amplifiers input impedance value, (from 18.5kΩ to 2.2kΩ).

With this bypass capacitor removed, the amplifiers voltage gain, Av decreases and Z_{IN} increases. One way to maintain a fixed amount of gain and input impedance is to include an additional resistor in series with C_{E} to create what is called a “split-emitter” amplifier circuit that is a trade-off between an unbypassed and a fully bypassed amplifier circuit. Note that the addition or removal of this bypass capacitor has no effect on the amplifiers output impedance.

Then we can see that the input and output impedances of an amplifier can play an important role in defining the transfer characteristics of an amplifier with regards to the relationship between the output current, Ic and the input current, Ib. Knowing an amplifiers input impedance can help to graphically construct a set of output characteristics curves for the amplifier.

Error! Please fill all fields.

I am a study of electronic engineering if you can to help me to understand this topic I will very happy.

Thanks

Muntasser Mosleh

Can someone please tell me how to find low cut-off frequency and high cutoff frequency of any given BJT or FET amplifier?

“Since Ic/Ib = β, then the value of the transistors Base impedance will be equal to β x re.”

Why is that true?

Hi there! It was not really clear to me either, till found this excellent video :”Common Emitter Amplifier, from David Williams” https://www.youtube.com/watch?v=DVLO4YBURSo

If you do not want to watch it (i do not trust any links either):

Zin=R1 II R2 II Zinq, where Zinq is the “Base impedance”, Wayne was talking about. Zinq=Vin/Iin, which equals to re*IE/Ib, that is re*B*ib/ib. But Ib is cancelled out and bingo! Zinq=re*B (when RE resistor is shorted by CE capacitor, if not Zinq=B*(re+RE))

Rather watch the video!

best info ever seen .is there any tutorial for biasing of common collector amplifier for impedance matching with power amplifier

Best tutorial.Anyone can teach to the electronics students after study this tutorial… Thank You!

Hi Wayne,

Thanks for replying.

Consider an open circuit output (RL = infinity). Now if RL is in parallel with RC as far as the capacitor is concerned (as you claim) then the resistance seen by the capacitor would be just about RC and the capacitor would charge and discharge with one side open circuit. Actually RL is in series with RC as far as the output coupling capacitor is concerned and therefore, with one side open circuit the resistance seen by the capacitor is just about infinity which is why the capacitor, in practice, can’t charge or discharge with one side open circuit.

Great tutorials

Cheers

James

The output resistance always equals Rc.

RL is the load resistance, quite often the input resistance of a following amplifier stage.

To calculate the size of the output coupling capacitor use C = 1/(2*pi*f*R) but R equals the sum of Rc and RL not the parallel combination as you have used.

Use the parallel combination of Rc and RL to calculate the gain as you have used.

Av = (Rc//RL)/(RE+Re).

The AC analysis of the output resistance, Ro shows that Rc is grounded through the supply (as all DC sources are set to zero, and all capacitors are replaced with short circuits) and therefore in parallel with RL. Then the equivalent resistance seen by C at the required frequency is RC||RL

You stated: With no signal current flow into the Base, no Collector current flows, (transistor in cut-off) and the voltage on the Collector is the same as the supply voltage, Vcc. A signal current into the Base causes a current to flow in the Collector resistor, Rc generating a voltage drop across it which causes the Collector voltage to drop.

This is true, but incomplete and needs to be clarified.

The bias circuit ensures that a quiescent (“standing”) collector current (Iq) flows for Class A operation, and low-level voltage amplifiers (such as your example) almost universally use Class A. This is the essential condition for non-distorting amplification in all Class A circuits, and especially so for signals whose ampliude is less than the forward bias voltage.

A silicon transistor, with *no* bias, would not respond to signals of less than about 0.6 V.

The input signal current *adds to*, or *subtracts from*, the bias current, increasing or decreasing Iq to form the signal current component of the collector current.

With the bias at zero (that is, in Class B operation) , the circuit could only respond to *positive* signal excursions (for an NPN device), thus the collector current would only ever show half-cycles of the amplified signal. (remembering that a signal exceeding some 0.6 V p-p would be needed for *any* output).

The resulting output signal would be amplified, negative-going half-cycles of the input signal.

Please also note that output impedance *does* change with emitter bypassing. See, for example “Transistors, Theory and Applications”, Coblenz and Owen, McGraw-Hill, 1955, p. 132.

why to fill?

“Note that the addition or removal of this bypass capacitor has no effect on the amplifiers output impedance.”

This doesn’t quite make sense to me. Isn’t any unbypassed emitter resistance in the output circuit as well as the input circuit?

Hello Arisian, As shown above, the emitter resistor Re, forms part of the input circuit through the Base-Emitter forward-biased junction but is effectively isolated from the output circuit by the two reverse-biased Collector-Emitter junctions. Then any bypass capacitor connected across Re has no effect on the amplifiers output impedance, only its input impedance, Zin.