Electronic – Class A Amplifier Design: Emitter Resistance Voltage Drop

amplifieraudioclass-acommon-emitter

I am trying to understand the design principles of Class A amplifiers, primarily for audio purposes.

I have read several times that "the voltage drop across the emitter resistor should be in the region of 1 to 2 volts"…

To me this seems like a backwards way to design an amplifier, and it confused me. Why would one specify this voltage drop before anything else? Surely the voltage drop should be calculated from the current flow through the amplifier? (This is what I was thinking until today.)

Then I had a thought – I believe that the transistor will have a parasitic resistance which (for some reason that I do not yet understand) appears as if it were a small resistor connected in series at the emitter.

This resistor will create some voltage drop – so my thoughts are that perhaps this voltage drop (for some reason that I also do not yet understand) is always much less than 1 to 2 volts? At least this is my guess, I have no idea if I am along the right lines here.

To make this concrete with an example, please see the following webpage:

https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

Under Single Stage Common Emitter Amplifier the author states,

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, VRE = 1.5V, the quiescent current, IQ = 1mA

My question here is, why was this value, of 1.5V for VRE, chosen first, and why was it chosen to be this value?

Further to this, the author states that IQ = 1mA. Does this mean that the output current from this type of amplifier must be between 0 and a maximum of 2mA? (My understanding is that the quiescent current is half the maximum output current. Is that correct?)

Best Answer

The main reason that the 1.5V emitter voltage design point is chosen is to reduce the effect of temperature variations and device variations on the bias current.

The voltage across the base-emitter junction will change by about 2mV per degree C. Since the voltage at the base is assumed constant the voltage across the emitter resistor will increase by 2mV/deg resulting in a current increase. Over a range of say 50deg this could result in 100mV change. By selecting a voltage across the resistor that is large relative to the 100mV the current change will be minimized.

Notice that the 1.5k emitter resistor has a 27uF capacitor across it. This bypassing means that the presence of the resistor does not need to be included in the calculation of AC gain as at 1kHz it has an impedance of ~6ohms.

The AC gain can be calculated by multiplying the gm and the effective output impedance which is 5k//1k = 833 ohms.

The effective gm is 1/re. The effective emitter resistance 1ma/40 = 25 ohm giving a gm of 40ma/V. The AC gain is therefore 40/1000 * 833 = 33.

Personally, I think that 1.5V across the emitter resistor is being excessively conservative and for normal temperatures, something less would be acceptable.

Also, this circuit arrangement is rarely used these days even when designing discrete circuits and one that uses two or more transistors with DC feedback is more common.

This type of arrangement was used in the early days of transistors when they were much more expensive with very inferior characteristics compared to modern types (modern being within the last 40 years!). It is, however, a good example circuit for demonstrating the principles.