Current series feedback amplifier theory of everything movie
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Electrical Engineering
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts.
It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. I am experimenting with moderately high frequency circuits, so decided to analyze, build and test an amplifier. This is the circuit:. Q1 is a CE amplifier, Q2 is Emitter follower to buffer the load. R2-R4 bias the transistors and R4 provides negative feedback. Due to negative feedback, the input impedance of this amplifier is quite low and a strong function of gain.
Nonetheless, the 50ohm source is terminated by a 50ohm R6. RLOAD is set to The goal is to build an amplifier that has a reasonably flat response from 1 MHz to about 40 MHz.
I calculated the -3dB point and that came to about MHz, due to the Miller Impedance of the feedback network, in parallel with R1. SPICE confirms this calculation also. At 1 MHz, the gain I measure is about 25, and a theoretical gain of about The trouble is the frequency response is very unexpected. The transistors are not suitable: The transistors are standard BCs with a transition frequency of about MHz.
The resistors I have used are not suited for high frequency operation: I am not sure about this one, I am using carbon film resistors which should have reasonable low inductances for reliable sub MHz operation?
Both input and output measurements are taken with an x10 probe with the standard ground clip. The output is terminated at 50 ohms right at the output. I tried a direct BNC cable to measure the output instead of a probe, but that made things worse in terms of gain loss and distortion. I tried the springy ground tip for the probes, which made no difference. EDIT: The physical circuit does not show a capacitor C1, this is because i was trying to eliminate a possible bad output capacitor.
EDIT 2: After a lot of thought provoking input, i believe i am alot closer to understanding why i am "missing" gain. Ali Chen aptly pointed out the presence of power lead inductance, and after adding that to the sim model, it has bought a smile to my face. Consider the 2 frequency response of gain. The blue trace is with lead inductance, the green without. Most recently i measured a gain of 5 at 40Mhz and the model predicts a gain of about 8 when lead inductance is taken into account! The problem with this prototype is in the absence of bypass capacitance on power rail.
If a model for power lead of 30cm in length nH inductance is added to LTspice simulations, the output does match observations - -3dB fall-off at 15 MHz.
There's the main problem. At MHz, the current gain is unity i. This will not produce the voltage gain you require and will be quite a bit below your expectations for a gain of 30 at 40 MHz. I can't tell you why your sim didn't show this up - maybe try looking at the BC model it uses. Here is another way to look at Cbc. Looking from the base, this has the effect of making the capacitor look much larger.
This is the Miller effect. The way this is usually taken into account is to put and equivalent capacitor in parallel with Cbe. From this crude calculation it sees to explain the 16MHz.
Cbe is added to the new equivalent capacitor here. The common solution to this problem is to use a cascode amplifier. This means following the common emitter stage with a common base stage.
The cascode stage has no voltage change at the common emmitter stage collector so the capacitance is not multiplied. If you are stuck with 5V a cascode may be difficult to bias. The usual solution to this problem is to use a folded cascode.
A folded cascode implements the common base stage with a PNP. You would still follow this with a common collector stage. Sign up to join this community. The best answers are voted up and rise to the top.
Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Why is my amplifiers frequency response unexpected? Ask Question. Asked 5 years ago. Active 5 years ago. Viewed 1k times. Here is the small signal circuit I think applies here and my calculations: I calculated a gain of about 30 with the Load present and SPICE backs me up. Here is the circuit all built up: Here is how I probed everything, the probes are x10, and the scope is a MHz Rigol.
Possible explanations I have are: The transistors are not suitable: The transistors are standard BCs with a transition frequency of about MHz. Adil Malik. Adil Malik Adil Malik 1, 10 10 silver badges 24 24 bronze badges. But I don't see it in your physical circuit. That's going to change the bias point of Q2 quite a bit. The results outlined above were with a capacitor C1.
Ill edit that in. So much for my gut feeling. Show 8 more comments. Active Oldest Votes. However, changing it to 50 Ohms does not change the fundamental issue. Since we don't know the length of your wire leads, nH was a guess. If inductance is bigger, drop in gain is also bigger. This is with bypass cap C5 at 0. But you did show the picture from LTspice, why don't you experiment with your model?
What kind of caps did you add? Show 2 more comments. Here's a graph of what I mean but for a much faster transistor : - This will not produce the voltage gain you require and will be quite a bit below your expectations for a gain of 30 at 40 MHz. Andy aka Andy aka k 21 21 gold badges silver badges bronze badges. If i understand correctly, the presence of fT is modeled by Cbe and Cbc? Don't get me wrong, everyone falls into the sim trap now and then. Also, fT is measured with the collector tied to a voltage source so that miller capacitor effects are not allowed to disturb things.
The reason i checked with sim after was because my circuit wasnt performing as per the analysis. I still do not understand where exactly in my SS model the effect of fT been ignored?
The gain of 30 is voltage gain so it's a little more complex than that. Add a comment. The gain in this case is not gmRl because of feedback. This sets the pole at about Mhz i believe? I don't get 25 for this. What you are calling is Av if the close to the forward gain.
Would you agree that rbe2 and R2 included in Av calculation should be time beta? I dont quite understand why rbe2 and R2 should be times beta? Are you saying that because of the effect of the emitter follower? The way i look at it is: Due to the emitter follower, R2 and any RLoad appears very large beta times larger to be exact , so i ignore it completely from the gain calculation.
This resistance appears in parallel with R1, to form the Rl in your gmRl. This is what ive done in the SS model i drew. I believe this is correct? When a emitter follower is called a buffer, it is because it tends isolate the load from the previous stage. The feedback makes the input look very low impedance. Show 5 more comments. Sign up or log in Sign up using Google.
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Non-inverting op-amp
Operational amplifiers op-amps are some of the most important, widely used, and versatile circuits in use today. The first op-amp used vacuum tubes and was released in by Bell Labs. The ubiquitous ua was released in and is considered by many to be the standard upon which others are based. It is still in production today from various manufacturers. Designed to amplify a small signal up to something useful, op-amps are applicable in an extremely wide range of projects, everything from audio circuits, to data acquisition, to signal processing. My goal is to simplify the op-amp into something easy and fun to use, highlighting the important stuff and keeping it simple. If you could really care less about the theory behind op-amps or just don't want to read right now, skip this step. There won't be any heavy math involved, just some summarizing.
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Electrical Engineering (EL ENG)
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RC circuits work as filters high-pass or low-pass filters , integrators and differentiators. Here we explain how, and give sound files examples of RC filters in action. For an introduction to AC circuits, resistors and capacitors, see AC circuits. Low pass filter High pass filter Filter applications and demonstrations Integrator Differentiator. From the phasor diagram for this filter, we see that the output lags the input in phase. Now a reduction in power of a factor of two means a reduction by 3 dB see What is a decibel?
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