Home      Projects     Experiments     Circuits     Theory     BLOG     PIC Tutorials     Time for Science

24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

Maximum Unclipped Oscillation

Again, there are situations in which the signal clipping is eligible, for example in B and C class amplifiers. But in many application the signal must be amplified without any distortion or clipping whatsoever. Therefore, we must be able to design amplifiers with maximum unclipped amplification gain.

There are 2 steps to design an amplifier with maximum unclipped output. The first step is to determine the maximum output peak to peak voltage (Vp-p Max), and the second is to set the operation point at the half of the max Vp-p Max.

Let's first see the case that the load resistance is very high or no load is connected. In that case, the AC load line is almost the same as the DC load line, so we can safely work only with the DC load line. The maximum oscillation output can be from 0.5V to VCE(cut). We avoid operating the transistor near the saturation area because in that area the output signal is highly distorted, therefore we use the number 0.5 for our calculations. So, to set the Q point, all we have to do is to find the middle. There is a simple formula which does exactly this:

VCEQ = (0.5 + VCE(cut)) / 2

Let's see an example. Suppose that we calculated the IC(sat)=4mA and VCE(cut)=12V. From these points we draw the DC load line as shown bellow. To set the optimum Q point, we need to know only the VCE(cut) which is 12V:

VCEQ = (0.5 + 12) / 2 = 12.5 / 2 = 6.25V

This way we achieve maximum oscillation within the complete linear area of the transistor, without exceeding the VCE(cut) value (12V).

Suppose now that the load resistor is not that big, and the AC load line has different slope than the DC load line. First of all, let's make something clear: Since the load is connected in parallel (AC analysis) to the collector or emitter resistor, this means that the resulting AC resistor (rc or re) can only be smaller than the collector or emitter DC resistor (RC or RE). Therefore, it is easy to understand that the Ic(sat) current of the AC load line can only be higher than the IC(sat) current of the DC load line. And since the AC and DC load lines intersect at the Q point, it is absolutely certain that the Vce(cut)-AC voltage of the AC load line can only be less than the VCE(cut)-DC voltage of the DC load line.

The previous statement makes clear that, if the AC load line is not the same as the DC load line, the output oscillation without clipping becomes smaller. As a matter of fact, the new oscillation range will be 0,5V up to Vce(cut-off)-ac. To calculate the Q point we use the same formula as before, but we replace the VCE(cut)-DC term with the Vce(cut)-AC:

VCEQ = (0.5 + Vce(cut)-ac) / 2

This formula tells us that, in order to achieve the maximum unclipped output, we need to set the Q point in the middle (approximately) of the AC load line. Here is an example:

How to set the Optimum Q Point

There are many ways to change the Q point, since any change on the DC bias will also change the Q point. The designer may choose to go with the trial and error method, or by solving the mathematics equations. But no matter which method is used, the designer must be able to locate the biasing part that needs to be changed, so that this change will have big effect on the Q point and small or no effect on the rest of the circuit and its characteristics.

Changing the Q point in Common Emitter connection

As always, we suppose that the transistor is biased with a voltage divider, and the emitter has also a small feedback resistor. Let's remember how VCE is calculated:

VCC = IC x RC + VCE + IC x RE = > VCE = VCC - IC x RC - IC x RE

So, by changing either RC or RE, we can change the VCE, thus we change the VCEQ of the Q point. But which one to choose? The answer is simple. The capacitor CE acts as a bridge in AC signal, which means that that the resistor RE does not have any affect on the AC signal, and therefore has no affect on the AC load line. Therefore, we prefer to change the emitter resistor RE, since it affects only the DC load line. By increasing the IE (=IC) current the VCEQ point shifts rights. If IE is decreased the VCEQ point shifts left.

Changing the Q point in Common Collector connection

Here is an example of a common collector connection:

It is obvious that if RE is changed, it will have an affect on both AC and DC load lines, since there is no bypassing capacitor across this resistor. As we know, this type of connection is also called "emitter follower", because the emitter voltage follows the base voltage:

VE = VB - VBE (1)

Moreover, from the schematic we can calculate the VCE:

VCC = VCE + IE x RE = VCE + VE => VCE = VCC - VE (2)

We replace the term VE in the second formula (2) from the first formula (1):

VCE = VCC - (VB - VBE) => VCE = VCC - VB + VBE

This equation tells us that we can change the VCE and thus the VCEQ of the Q point by changing the base voltage VB. So, we can simply change the RB1 or RB2 resistor to achieve the optimum Q point. There is something that we need to take into account here. First, let's see the AC equivalent of the circuit:

As you see, RB1 and RB2 are still active components in the AC equivalent. These components have an affect at the input signal, since RB1 and RB2 will eventually define the input stage impedance. A large change on either resistor may require to re-design the circuit.

Relative pages
• Basic transistor circuits
• 555 theory of operation
• Learn how to interface ICs
• Learn how dimmers work
• The TRIAC theory
• Learn about the most popular PC Cooling methods
• Dr.Calculus: Checking transistor functionality
• Op-Amp IC Pinouts