Activity for LvW
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #291454 |
Post edited: |
— | 6 months ago |
| Edit | Post #291454 |
Post edited: |
— | 6 months ago |
| Edit | Post #291454 | Initial revision | — | 6 months ago |
| Answer | — |
A: Calculating base voltage of NPN transistor For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected), Let me try: Vb1-Vx=I5R5 I5=IC4+Vb1/R3 IC4=Ue4/R4 Ue4=VB2-0.7 These 4 equations (simple insertion procedure) result in: Vb1-Vx=R5[(Vb2-0.7... (more) |
— | 6 months ago |
| Comment | Post #287728 |
No input signal? Therefore, does the circuit produce this output? Oscillation? (more) |
— | almost 2 years ago |
| Comment | Post #287435 |
Using the keyword "Slew Rate" you will find a lot of information in the internet. (more) |
— | almost 2 years ago |
| Comment | Post #287689 |
I agree to everything Olin L. has stated.
Moreover, all commonly used formulas are (and must be) "inaccurate".
Typical example: The often used closed-loop gain expressions for opamps (-R2/R1 or (1+R2/R1)) are only approximations because many secondary effects are neglected (input/output impedances... (more) |
— | almost 2 years ago |
| Comment | Post #287689 |
@Olin Lathrop: Of course, I know that a phrase like "as clean as possible" says nothing. On the other hand, a "hobbyist of electronics" has certainly not a good feeling for "what is possible". Therefore, one should explain to him (and that`s what I have always done in similar situations) why the osci... (more) |
— | almost 2 years ago |
| Comment | Post #287689 |
I think, when a "hobbyist of electronics" uses a wording like "pure stable sinewave" he is referring to a sinusoidal output "as clean as possible" - in contrast to other oscillators (which, I think, should better be called "generators") producing squarewaves or other signal forms.
At least, this w... (more) |
— | almost 2 years ago |
| Comment | Post #287678 |
Quote: "Until now all the topologies that I have found if the frequency is changed, a change in the circuit component is mandatory here is some topologies: :
Courtesy of TEXAS INSTRUMENTS".
How do you want to tune the frequency if not by "changing" one or more circuit element(s)? More than that, ... (more) |
— | almost 2 years ago |
| Comment | Post #287435 |
Please note that the "10% to 90%" criterion is valid for small-signal responses only. The slew rate is a large signal parameter (valid for opamps with heavy feedback) which requires that the first stage is saturated for a very short time until feedback brings it back to linear operation. (more) |
— | almost 2 years ago |
| Comment | Post #287366 |
I am not sure if this comment is an appropriate answer to an unexperienced forum member...("deliberately disrespectful"..).
(more) |
— | about 2 years ago |
| Comment | Post #287211 |
Sorry, but I have problems to agree.
To me the quantity F is the feeedback factor - which is identical to "the fraction of the output fed back to the negative input" (your definition) .
That means: F is always smaller than unity.
Example: Feedback resistor is R2 and resistor R1 is from the inv. ... (more) |
— | about 2 years ago |
| Comment | Post #287211 |
You wrote: Gain = F / (1 + F/G).
When G is the open-loop gain and F the feedback factor , the closed-loop gain is (according to Blacks famous formula):
Gain Acl=G/(1+F*G) (more) |
— | about 2 years ago |
| Comment | Post #286689 |
You wrote: "A single pole low pass filter isn't totally flat up to the rolloff frequency...".
From this, one could derive that there is a multi-pole filter which is "totally flat" up to the roll-off frequency - but this is not possible, (more) |
— | about 2 years ago |
| Edit | Post #286931 |
Post edited: |
— | about 2 years ago |
| Edit | Post #286931 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Design high -pass filter with 2 points of the bode plot Miss Mulan - I am a bit late with my answer. Nevertheless - there is formula which gives you the necessary filter order "n" - as a function of two frequencies and both associated damping figures. If applied to the most simple case (Butterworth approximation) the result is n>1.2 Hence, a 2nd-... (more) |
— | about 2 years ago |
| Comment | Post #286793 |
Just to avoid misunderstandings - the feedback network does not contain any lowpass filters.
The oscillators working principle is as follows:
The feedback network of a Hartley oscillator consists of a 3rd-order HIGHPASS filter in classical lader topology: Ro-L1-C1-L2.
This highpass introduces... (more) |
— | over 2 years ago |
| Comment | Post #286784 |
@Concerned Citizen: I must admit that I cannot agree to your conclusion that "their value is chosen such that it doesn't influence the overall oscillator".
While it is true that the finite input resistance of the transistor stage has only a minor influence on the pole frequency of the R-L-C-L hig... (more) |
— | over 2 years ago |
| Comment | Post #286784 |
It seems that the given expression for the pole frequency contains a small error :
For L1=L2 the pole frequency is w²=1/2LC.
With your formula it is 1/4LC.
However, the given condition for oscillation is correct (based on the mentioned assumptions and simplifications).
Nevertheless - one additio... (more) |
— | over 2 years ago |
| Edit | Post #286783 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286783 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Results of analysis of Hartley oscillator dont make sense MissMulan - at first, you should show which nodes are connected (crossing points). Secondly, where are the voltages Vo and Vin in your diagram? Please keep in mind, that it is essential to OPEN the feedback loop at a suitable point for injecting a test signal. Then, you can/must analyze the LOOP G... (more) |
— | over 2 years ago |
| Comment | Post #286738 |
There is a contradiction in the title of your contribution.
A small-signal analysis is possible (makes sense) only in case of a linear (in practice: quasi-linear) region of the transfer function.
Saturation mode is a heavy non-linear mode and is defined for the case that BOTH pn-junctions are forwa... (more) |
— | over 2 years ago |
| Comment | Post #286700 |
@ MissMulan You were asking "why is that true" that the transfer function is defined for sinusoidal signals only. I think, definitions cannot be "true" or "false" - definitions are arbitrary determinations which are logical and make sense.
The transfer function T(s) of a system is derived from its f... (more) |
— | over 2 years ago |
| Comment | Post #286700 |
Quote: "The transfer function is a description of what happens with any input signal."
What is the meaning of "any input signal"?
To avoid misunderstandings it should be clear that the transfer function is defined for sinusoidal signals only! (more) |
— | over 2 years ago |
| Comment | Post #286679 |
Let me try to "translate" your requirements:
Lowpass (no order specified) with -3dB at fo=100 Hz
A frequency increase beyond this frequency fo by another 100Hz (fx1=200 Hz) must correspond to a damping of -10dB (because of 0.1dB/Hz).
A first-order lowpass (fo=100 Hz) has a damping of -10 dB ... (more) |
— | over 2 years ago |
| Comment | Post #286589 |
So I ask myself: What is the purpose of a comments section? Not to clarify misunderstandings/misconceptions? (more) |
— | over 2 years ago |
| Comment | Post #286589 |
Why not? I am really surprised. For my understanding it directly concerns the question (from Miss Mular). He/she was asking for the role of re=1/gm where gm is a quantity which connects the input VOLTAGE with the output CURRENT. And the transconductance gm is the key parameter for the explanation of... (more) |
— | over 2 years ago |
| Comment | Post #286589 |
continued: To understand your position it would be great if you would mention to me one single application (circuit) where the current-controlled view has advantages during design of the circuit and/or for a better understanding. (more) |
— | over 2 years ago |
| Comment | Post #286589 |
Quote: "...most of the electrons that leave the emitter get swept away to the collector before they have a chance to come out the base. That view is definitely current-dominated."
My response: I cannot share such a view. What means "dominated" in this context? When most of the electrons leaving th... (more) |
— | over 2 years ago |
| Edit | Post #286589 |
Post edited: |
— | over 2 years ago |
| Edit | Post #286589 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Re model of transistor is the same in 2 different configuration I never would recommend to use the "re model" for a BJT. The reason is as follows: The BJT is a voltage-controlled device (Ic is controlled by Vbe and the base current is an unwanted by-product). Hence, the characteristic transfer parameter is the transconductance gm=d(Ic)/d(Vbe)=Ic/Vt.. The ... (more) |
— | over 2 years ago |
| Comment | Post #284538 |
When discussing feedback we always have two basic parts: (1) An active device and (2) a feedback path. Without such a feedback path (which closes the feedback loop) there is no feedback at all - we have a so-called "open-loop gain". In the case under discussion, the resistor Re constitutes the feedba... (more) |
— | about 3 years ago |
| Comment | Post #284538 |
Quote (last sentence): "You can think of an emitter follower as working on internal feedback in the transistor."
Negative feedback ist realized by the emitter resistor - not "internal...in the transistor". (more) |
— | about 3 years ago |
| Edit | Post #284558 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Find feedback of circuit I must admit that your question (the described problem) is an interesting one. Speaking about the circuit (without consideration of the parts values), there are two feedback effects: At first, the resistor R1 provides negative feedback resulting in a low closed-loop gain slightly below unity. S... (more) |
— | about 3 years ago |
| Comment | Post #283942 |
Quote Elliot Andersen: "The essence of the ideal current source is quite simple, and trying to convert it to a voltage source and some kind of magic dynamic resistor only makes it harder to understand."
I strongly disagree. Who can UNDERSTAND (from the technical/physical point of view) an element ... (more) |
— | about 3 years ago |
| Comment | Post #282363 |
Quote: "You can keep asking "why" questions indefinitely. "Why is grass green?" "Then why is chlorophil green?" "What makes it look green to us?" ..."
Sorry for my silly question. However, when you make a statement without sufficient evidence I think it is legal to ask for explanation - to the benfi... (more) |
— | over 3 years ago |
| Comment | Post #282363 |
Quote:"When VIN- is a little below VIN+, that is more current than Q2 can sink, so VOUT goes high". Of course, I agree - however, what is the clear explanation? WHY goes Vout high? We have two (non-ideal) current sources with a conflicting behaviour. How and why does this fact influence the voltage a... (more) |
— | over 3 years ago |
| Comment | Post #282434 |
You ave nothing to do than to solve for iout and set ic2=-ic1 and ic3=ic4 - that`s all. (more) |
— | over 3 years ago |
| Edit | Post #282434 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282434 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282434 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Help with differential to single ended voltage converter Question: What is the purpose of the current mirror? Answer: It allows to combine the current changes in BOTH transistors (T1, T2). Explanation In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or... (more) |
— | over 3 years ago |
| Comment | Post #281322 |
I must admit - a rather strange discussion...... (more) |
— | over 3 years ago |
| Comment | Post #279596 |
Quote:..."in the differential resistance, the difference is between two voltage values at adjacent values of the current." Did the questioner (Pacifist) spoke about resistances? I think, he has mentioned instead a differentiating circuit. In this case, we could speak about two adjacent voltages at t... (more) |
— | almost 4 years ago |
| Comment | Post #279588 |
OK - I was of the opinion that in a short comment it would be appropriate to use the well-known abbreviations for the open-loop gain Aol and the closed-loop gain Acl. The quantity beta was defined using the symbols shown in the drawing. Again, I like to point out that for sinusoidal signals it is, of... (more) |
— | almost 4 years ago |