How big can the gain of grounded emitter amplifier be? - Page 1
Forum section: electronics / beginners

author: ZongZi
author_rank: Newbie
timestamp: October 14, 2017, 11:32:38 am
content: I was reading The Art Of Electronics 3rd Edition and I came across this:D. Comments on biasing and gainOne important point about grounded emitter amplifier stages: you might think that the voltage gain can be raised by increasing the quiescent current, since the intrinsic emitter resistance re drops with rising current. Although re does decrease with increasing collector current, the smaller collector resistor you need to obtain the same quiescent collector voltage just cancels the advantage. In fact, you can show that the small-signal voltage gain of a grounded-emitter amplifier biased to 0.5VCC is given by G = 20VCC (in volts), independent of quiescent current.Exercise 2.15. Show that the preceding statement is true.--Paul Horowitz & Winfield Hill, The Art of Electronics 3rd Edition, page 98I know that re=25/I so with increased Ic, re get smaller, also, the small signal gain is Rc/re so when Ic increase the gain can get bigger. However, Since Rc seems to have its value fixed when we apply a signal to Vb to change Ic, the gain seems to vary and is related with Ic. Also, I wonder how to get that when Vb is baised to 0.5 VCC, the gain is 20VCC.Please help me figure it out THANK YOU.

author: bd139
author_rank: Super Contributor
timestamp: October 14, 2017, 12:35:53 pm
content: I really don’t like the explanation in that book myself. Watch the following video - far more elegantly explained. Forum member w2aew is the originator of the video:https://youtu.be/NizrzRKQqIIIt makes more sense as transconductance at a certain collector current than the ratio of Rc/re.

author: CD4007UB
author_rank: Contributor
timestamp: October 14, 2017, 01:09:47 pm
content: Gain = Rc/re.Collector current Ic = Vcc/(2Rc) (with Ic in amps),.With Ic in milliamps, Rc = 1000*Vcc/(2Ic) = 500Vcc/Ic.Since re= 25/Ic (with Ic in milliamps), we get Rc/re = (500Vcc/Ic)/(25/Ic) = 20Vcc.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 14, 2017, 08:35:27 pm
content: Note that you can keep increasing Vcc while keeping the same Vce and Ic. Rc goes up, and Av goes up proportionally, until Early effect ultimately limits gain*.This breaks the Vcc/2 assumption, which is how the original question can have such a well-defined answer.This approach has been used in audio amplifiers before, using an excessively high supply voltage to enhance gain, as well as provide more dynamic range. A transistor biased at 30V and supplied from 100V has triple the gain and twice the dynamic range of a transistor supplied from +/-15V. Dynamic range being the available voltage swing before clipping occurs.*Which can be extended further by keeping collector voltage constant: if you use a cascode configuration, the effect is to isolate the output voltage (which is driven by the top transistor) and current (which is set by the bottom transistor), giving a more ideal CCS output. You need to do the same for the load, of course, so instead of a very large resistor, you'd use a current source, which either uses a fairly large emitter resistor (in a sense, cascoding the resistor with a transistor, to make it a better current source -- same idea), or two transistors in cascode proper. After these steps, the DC gain is now extremely high (easily over 10k), but very weak (if you connect any resistance to it, the gain drops like a stone), and very slow (because the capacitive reactance of the transistors themselves dominates, even in the audio frequency range).Tim

author: David Hess
author_rank: Super Contributor
timestamp: October 14, 2017, 09:16:47 pm
content: I agree with bd139, the book's explanation leaves a lot out and T3sl4co1l pretty much covered my response. There are ways to support the collector current without lowering value of the collector load resistance so the gain may be increased including using an active or inductive load or raising the supply voltage. Then other factors like the Early effect, Miller effect, and output capacitance limit the gain.

author: Audioguru
author_rank: Super Contributor
timestamp: October 15, 2017, 12:35:44 am
content: Since the transistor amplifier has no negative feedback then its distortion is high. I see the top half of the sinewave squashed a little resulting in about 5% distortion even though the p-p output is less than half the supply voltage. When the output level is higher then the distortion becomes extreme even when the output level is lower than clipping. The top half of the waveform is where the transistor current is low resulting in reduced gm and a squashed output.

author: bd139
author_rank: Super Contributor
timestamp: October 15, 2017, 09:44:06 am
content: Distortion there is because the collector resistor is too big or the supply voltage is too low or the input signal is too large. You don’t need negative feedback with a grounded emitter to get good distortion characteristics. What you do need to do is keep it well away from saturation and cut off.