# What's the rationale behind this choice of resistor for a BJT common collector amplifer?

As part of a design lab, we're using a common collector amplifier, and the instructor's choice for the emitter resistor makes no sense to me.

The power rail is at V+=15 volts, and the load is 8 ohms. The design specifications give us a power across the load of 5mW. We assume capacitors are large enough to ignore for the AC signal.

We want maximum voltage swing, so we say the emitter DC voltage is approximately 7.5V, half of our positive rail... ok, makes sense for a rough design. From the given power and resistance, we know the current through the load has an amplitude 35mA, and the voltage across the load has an amplitude of 0.28V, from ohm's law and the usual power equations. Then we say the total emitter voltage is AC + DC, and swings from 7.22V to 7.78V.

Here's the part that confuses me: the instructor claims that the emitter resistance should follow the relation

$$R_{E}\geq \frac{7.22V}{35mA} \approx 200\Omega$$

Where does this relation come from? What's the rationale behind it?

By the problem description, the amplifier is to source a 35mA sinusoidal current through the load.

Note that during the positive half of the cycle, the output current is from left to right through the capacitor but, during the negative half, the current is from right to left.

Note also that the BJT current is (assuming the transistor is active) always out of the emitter.

Thus, during the negative half cycle, the current through the emitter resistor is the sum of the load current and BJT emitter current.

If follows that the current 'down' through the emitter resistor at the negative peak is at least $35ma$. This would be the case if the BJT were just 'turning off' at the negative peak.

Since it is already given that the emitter voltage should be $7.22V$ at the negative peak, and we know the resistor current is at least $35mA$ then, Ohm's Law gives

$$R_E \le \frac{7.22V}{35mA}$$

In other words, we have established an upper bound on the value of $R_E$.

We want maximum voltage swing, so we say the emitter DC voltage is approximately 7.5V, half of our positive rail...

For sinusoidal operation and assuming the capacitor voltage is constant (the capacitor is an AC short circuit) and equal to the DC emitter voltage $V_E$, we have:

$$v_{O+} = V_+ - V_E$$

$$v_{O-} = -V_E \frac{R_L}{R_E + R_L}$$

Typically, one wants these to be equal in magnitude, i.e., we want the output voltage swing to be symmetric.

Setting $|v_{O+}| = |v_{O-}|$ yields the following condition for symmetric load voltage swing:

$$V_E = V_+ \frac{R_E + R_L}{R_E + 2R_L}$$

• I appreciate that not only did you answer the question in an easily understandable way, you gave a better method for determining VE for maximum voltage swing. – David Feb 17 '14 at 21:39

Basically you want $V_E$ to be nominally half the supply rail AND, you want to be able to deliver a peak current to the load of 35mA. When the negative peak current is being delivered, the emitter voltage will reduce to 7.22V (7.5V - $\sqrt2\cdot 20mV^1$).

It follows that $R_E$ should be 200 ohms. Making it less than 200 ohms will serve to be able to drive bigger voltages to the 8 ohm load but as you only need about 60mVp-p, 200 ohms will do fine.

If you wanted 2Vp-p into the load, the peak current would be 125mA and $R_E$ would be calculated from: -

Re = $\dfrac{7.5V - 1Vpk}{0.125}$ = 52 ohms i.e. significantly smaller.

A little bit of an explanation is here

$^1$The 20mV comes from the power requirements of 5mW into 8 ohms of course.

• I understand that Ve reduces from 7.5V to 7.22V when the current through the load is at its peak. To me, it seems like we assumed that the current through Re should be the same as the peak current through the load resistor. Where did that assumption come from? In the link you gave, for example, they just took the current to be 2mA. – David Feb 17 '14 at 10:33
• @David The emitter resistor supplies current to the load on negative half cycles of the input waveform into the base - the BJT and the emitter resistor, at all times are acting as a potential divider with the top half being the active element only. Therefore the current through the load is the same thru the BJT current (positive half cycles) and it is the same in the Re resistor (on negative half cycles). Re can be smaller without detriment to the signal but quiescent power consumption will be greater (no advantage). Re can also be a bit bigger but this might start to cause distortion. – Andy aka Feb 17 '14 at 10:47