Is this biasing a good idea?

Consider a simple common emitter amplifier

Here I will set r1 = r2. Now I think this is what you call a class A amplifier, with the base voltage being half of vcc . I have a feeling this is not a good idea as it cannot be this simple.

The idea is that if I keep the base voltage at half of vcc, the voltage source will be able to pull half way and push half way if it ever needs to.

• if I keep the base voltage at half of vcc, the voltage source will be able to pull half way and push half way if it ever needs to. That is correct but is that what limits the performance of the amplifier ? If it has a voltage gain of 5, does the base need to be able to swing almost Vcc/2 up and Vcc/2 down ? What happens at the collector ? – Bimpelrekkie Oct 20 '16 at 13:03
• @FakeMoustache having a voltage gain of 5 will make it clip i guess? I completely forgot about voltage gain, now how do i fix this? – Vrisk Oct 20 '16 at 13:30
• Exactly, it will clip but at the output. The voltage at the base in combination with Re sets the biasing current. What will set the gain ? – Bimpelrekkie Oct 20 '16 at 13:39
• Rc yes, beta: no but also: gm which is dependent on the collector biasing current Ic. Beta is a large signal parameter, it mainly comes into play when you want to calculate how much biasing current will flow into the base once you know the collector current, so: beta = Ic / Ib and that is only used for DC currents. – Bimpelrekkie Oct 20 '16 at 14:03
• There are many "common emitter" explanations to be found by Google, some very theoretical, many too vague (to learn the thing properly). But this one might be what you need: wiki.analog.com/university/courses/electronics/text/chapter-9 Yes it starts with the "real" common emitter (no Re and Ce) but that's good, you need to understand that one first then add Re and Ce. – Bimpelrekkie Oct 20 '16 at 14:56

Rule of thumb is that the voltage across the emitter resistor (as set by the base bias voltage) should be between 0.5 volts and 1 volt. This is just a hand-wavy rule of thumb so, if you decide on 0.5 volts then the base needs to be set at approximately 0.7 volts higher i.e. 1.2 volts.

The whole point here is that you usually want maximum swing of signal on the collector and any standing dc voltage across the emitter resistor will eat into this requirement. Here's the general idea: -

I've chosen bias resistors that set the collector at about 4.5 volts DC on a 9V supply. The voltage gain is approximately 10 (Rc/Re) and, if you look closely at the bottom of the blue trace (collector voltage), it's starting to crash into the voltage on the emitter resistor. This is starting to clip the bottom peaks because the BJT is entering saturation (collector-emitter voltage less than about 0.1 volts).

When saturating the hFE of the BJT reduces rapidly and you get distortion.

Hence you want to keep the emitter voltage reasonably low to maximize the undistorted collector voltage swing.

• Wait, if the voltage gain is 10, shouldn't vc = 1.09 * 10? – Vrisk Oct 20 '16 at 16:12
• And you may want to keep $V_e \gg \frac{k T}{q}$ for several important reasons. – jonk Oct 20 '16 at 16:35
• The ac voltage gain is ten not the dc voltage gain as implied by the emitter dc voltage. – Andy aka Oct 20 '16 at 18:12
• I didn't want to ask more info because I'm a very big noob and didn't want to waste more of your time but aright - isn't the emitter voltage always base voltage - .7? How do I fix the emitter voltage then? – Vrisk Oct 21 '16 at 11:17
• The emitter voltage in an NPN circuit operating as this one can be between 0.4 volts and maybe 1 volts lower than the base voltage. We generally use 0.7 volts as a rule of thumb but, as you can see above, it's more like 0.6 volts. Base-emitter is a forward biased diode in effect. You fix the emitter voltage by setting the base voltage. – Andy aka Oct 21 '16 at 11:28