Example 4.9 Inverting Feedback Amplifier - Page 1
Forum section: electronics / beginners

author: Shturval
author_rank: Newbie
timestamp: June 23, 2024, 01:06:20 pm
content: Hello everyone, please tell me how the author of this book found G1 and G2. Thanks in advance. If possible, please write in detail.

author: m k
author_rank: Super Contributor
timestamp: June 30, 2024, 03:41:01 pm
content: A bit higher level.I guess RF, Microwave, Ham Radio section would be more suitable.Must be from Blackman's formula.One part being without a feedback.

author: jwet
author_rank: Frequent Contributor
timestamp: July 02, 2024, 05:39:38 pm
content: There are a number of typos between this spice figure and the hand analysis shown.Examples-RE1 is 7.5K in the spice circuit and netlist but the author uses 5.57K in his hand analysis.- See if there is an addendum published for the book.The calculation is laid out pretty succinctly. Because the circuit has feedback, he has to break the loop and do some calcuation and then close the loop for the final gain. G1 is the through gain. G2 is the gain (loss) of the loading effect of the RF.He first calculates the input impedance (open loop) for the follower and the CE amp. This is reflected emitter load of Q1 in parallel with Q2's load. He grinds this down to 180K (994K||220k). The .9879 is the fraction of the input that gets divided by the 2.2k and Rin. He calls this little alpha. This is what you'd expect from a follower. He doesn't use this further- just uses 1.The G1 calc has the wrong value for RE1 but is the gain of the first stage. The second product term is the gain of G2 with collector load being RL2 in parallel with RF. G1 comes out to -105.7. Slightly wrong because of RE1 error- but very little.G2 is the loading effect of RF on the 15K output Z- it is negligible. So the total gain G is the sum -105.6. H is the feedback factor, = Rb/(RF+rb), about 1%.1+GH is denominator of the feedback circuit which is 2.0433. This is loop gain that the open loop gain gets reduced by- to give a gain of -51.The last couple of steps recalculate rin and rout based on the actual loop gain at zero input.I don't understand the H calc completely- he's referencing Blackman's definitions for a very old Bell Labs paper.The final spice results might actually agree if the 5.57k and 7.5k were fixed?Good luck on your quest- messy derivation. I think this designed as a learning lesson which is made harder by a typo.Take Care

author: MrAl
author_rank: Super Contributor
timestamp: July 02, 2024, 06:55:17 pm
content: Quote from: jwet on July 02, 2024, 05:39:38 pmHi there,I am not sure how you concluded that he used 5.57k in place of 7.5k for RE1 in:"Examples-RE1 is 7.5K in the spice circuit and netlist but the author uses 5.57K in his hand analysis."My analysis tells me he actually did use 7.5k as follows.Following his first formula from the paper for ri (not mine BTW) I get:ri=(B1+1)*(((B2+1)*RE1*(RE2+re2))/((B2+1)*(RE2+re2)+RE1)+re1)and since he is using the same Beta for both transistors we can make B2=B1 which simplifies it a little bit:ri=(B1+1)*(((B1+1)*RE1*(RE2+re2))/((B1+1)*(RE2+re2)+RE1)+re1)Since B1=170 and re1=244 and re2=26.6 then:ri=171*((171*RE1*(RE2+26.6))/(171*(RE2+26.6)+RE1)+244)and this corresponds to the second line.If we divide by 171 and then subtract 244 we get a resistance:Rx=(171*RE1*(RE2+26.6))/(171*(RE2+26.6)+RE1)Now with RE1=7500 and RE2=100 we end up with:Rx=5570.233218748 Ohms.This matches the paper which states "5.5700k" close enough.Then we can match the paper again:ri=(171)*(244+5570.233218748)=994233.880405908which seems close enough to 994.2k.I did not go over the whole thing like how he developed the forumlas in that paper, I just started with what was written first and took it from there.Next I'll go over the whole formula from scratch.

author: MrAl
author_rank: Super Contributor
timestamp: July 02, 2024, 08:51:51 pm
content: Hello again,I took a second quick look at this and it seems there may be a problem.First a minor problem, it looks like the author added 7.5k+244 Ohms to add RE1 and re1, but that would not be correct at least not exactly. That's because in the first stage re1 is not directly in series with RE1, instead it forms a voltage divider where re1 would be above RE1 and the center tap of those two would feed the base of the next stage. That's different than having the emitter of the first transistor feed the base of the next stage directly. It is drawn that way, but since re1 is internal to the first transistor adding the two is not really the correct way to do it.The difference should be small though because 7.5k is so much larger than 244 Ohms, and besides that any 're' is usually taken to be an estimate anyway not an exact value.The second stage with the RE2+26.6 Ohms is ok because the output is not taken from the emitter.That's a minor issue. However, a more serious issue seems to be in the calculation of the input impedance, or input resistance ri in this case. The entire circuit is an inverting amplifier. That means it is using negative feedback. Since the input resistor is 2.2k if we wanted to get that up to nearly 1 Megohm as the paper suggests, we'd normally use positive feedback. Positive feedback artificially raises the input voltage at the base so that the input voltage source does not have to work as hard in order to raise the input at the base and this reduced current means the input impedance looks larger than the input resistor. Thus it looks like this needs more looking into. A full analysis or at least a good simulation should be performed.In general, there seems to be a lot of mistakes in amplifier analysis in the literature. This may be one of them if this is really what is happening.