Transistor rise fall time & frequency response - Page 1
Forum section: electronics / beginners

author: fonograph
author_rank: Frequent Contributor
timestamp: July 20, 2018, 06:04:33 pm
content: If we have transistor with 1 micro second 0% - 100% rise time and 1 micro second 100% - 0% fall time,at what frequency it starts rolling off? At what frequency there would be the "knee",I mean the point where it begins struggling to keep up and as result start acting like low pass filter,the point where harmonics start dropping off more rapidly than otherwise perfect waveform would becose transistor isnt fast enough.Lets pick square wave for example,perfect square wave harmonics decrease by amplitude by 1f,that is harmonic that is 90 times higher than fundamental is 90 times weaker than fundamental.Do they start rolling off by 1f^2 or 1f^3 or maybe its not linear but gaussian curve?1. At what frequency does 1 μs rise & fall time transistor start rolling off?2. Does transistor FALL time matter when it comes to high frequency roll off point?3. How rapidly does the frequency response start decreasing after the roll off point?

author: tggzzz
author_rank: Super Contributor
timestamp: July 20, 2018, 07:54:50 pm
content: Have a look at the references inhttps://entertaininghacks.wordpress.com/2015/08/11/measuring-digital-signal-edge-rates-without-an-oscilloscope/#more-652

author: fonograph
author_rank: Frequent Contributor
timestamp: July 20, 2018, 10:39:45 pm
content: Quote from: tggzzz on July 20, 2018, 07:54:50 pmThank you,I like that link but it doesnt answer any of my questions.I guess looking at the graph the roll off have gaussian shape?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 20, 2018, 11:17:16 pm
content: It's never 0 and 100%. Or actually it is, but the exact time depends on the amount of overshoot and noise.Why not 10-90%, or 63% (one time constant), or...?The least arbitrary case is a time constant, but strictly speaking, that's only applicable to a single order case (e.g. RC time constant).Also, what rises and falls are you measuring? If input current to output current, for small changes around a mean (bias) value, then it will be close to the fT figure. If not -- if there is any voltage swing, say -- then expect the times to be much slower.Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: July 21, 2018, 05:37:49 am
content: 0 - 100% becose its simple,I know 10 - 90% is more commonly used but for the purpose of this theoretical educational question,we will go with the former.Overshoot... Correct me if I am wrong,but when you switch to 100% it cant overshoot becose there is no more headroom,there is no additional voltage that can be provided for the overshoot by the source,any overshoot will be stopped instantly and completly by hard clipping.Lets ignore noise for sake of simplicity,in practice we must take it into account but this is just theoretical question,no need to complicate it.Imagine its gods personal magic noiseless mosfet or whatever.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: July 21, 2018, 06:30:40 am
content: Quote from: fonograph on July 21, 2018, 05:37:49 amGenerally speaking, it depends on the transistor configuration (common base/emitter etc). But there are a few physical mechanisms that introduce extra capacitance (and recombination time), and thus shift the pole of the device. These capacitance sources can be attributed to junction capacitances and potential excess minority carriers (charge) in the base of the transistor.In terms of a hybrid-pi model, the cutoff frequency is dependant upon Rpi and Cpi (BE junction capacitance), which in turn are influenced by quiescent bias conditions. The hybrid-pi model is a first order low pass response, and as such will roll off by 20dB/decade.The Gain Bandwidth Product also indicates at what point the transistor begins to "attenuate" at unity gain.

author: tggzzz
author_rank: Super Contributor
timestamp: July 21, 2018, 08:21:16 am
content: Quote from: fonograph on July 21, 2018, 05:37:49 amConsider a simple RC exponential risetime; itneverreaches 100%.Consider parasitic components, e.g. load capacitance and wiring inductance and how that rings when hit with a step or impulse. You will see that 100% is most definitely not the limit.

author: fonograph
author_rank: Frequent Contributor
timestamp: July 21, 2018, 02:27:15 pm
content: Quote from: tggzzz on July 21, 2018, 08:21:16 amIt can overshoot beyond source voltage? I didnt know that.Lets assume my theoretical transistor is criticaly damped,no overshoot.Or we can define the 100% as point in time where the rising voltage first crosses the source output voltage,so even through it might rise even more,beyond the source voltage due to overshoot,lets ignore that.Or we can imagine its magic transistor with perfect straight rising edge ,kind of like steep triangle.As for RC never reaching 100%,consider my 100% to be 99% then.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 21, 2018, 02:39:23 pm
content: The point is, your choice of criteria shows ignorance of the underlying principles. Indeed, 10-90% is both more common, and simpler, as you might've noticed from all the edge cases presented in reply.Likewise, you aren't learning anything by getting answers to this overly specific case.What you actually need to know are the underlying principles, and then you can answer any case yourself, whatever the percentage and whatever the frequency response.Do you understand?Tim

author: Benta
author_rank: Super Contributor
timestamp: July 21, 2018, 03:27:41 pm
content: What needs to be underlined here is, that this discussion only makes sense if we're talking about a transistor biased for linear operation. In that case, there is a direct correlation between maximum frequency and rise/fall time.If the transistor is operated as a saturated switch, it's a completely different story, and fTwill not tell you anything at all.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 21, 2018, 06:16:48 pm
content: Quote from: Benta on July 21, 2018, 03:27:41 pmExactly. Rise times are not measured in a linear circuit. A switching circuit is only partially representative of the true performance of the part, anyway, because of voltage swings. The fT parameter is more fundamental, but only true for small signals at the given operating point, so it's not exactly relevant to a large signal circuit.Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: July 21, 2018, 09:59:57 pm
content: Quote from: Benta on July 21, 2018, 03:27:41 pmWhy?

author: fonograph
author_rank: Frequent Contributor
timestamp: July 21, 2018, 11:14:22 pm
content: Quote from: T3sl4co1l on July 21, 2018, 02:39:23 pmHow does my choice of criteria show ignorance?Quote from: T3sl4co1l on July 21, 2018, 02:39:23 pmOfcourse it is more common,I never claimed anything else.I forgot to explicitly mention this before so I do it here,its simpler to ME,its question by me for me,I dont care what is simpler to you or anybody else.Quote from: T3sl4co1l on July 21, 2018, 02:39:23 pmThats ridiculous statement,completly wrong.Its perfectly valid question that I can learn alot from,you dont know jack sh!it what I am or what I am not learning from my question.Answers? What answers? I got 0 of my questions answered,all I got was "Oh,but what about overshoot though" and "Oh,but what about 100% though".Overly specific? What is that even to supposed to mean? Real problem are vague,unclear questions that arent specific enough ,the more I specify various variables,the clearer the question become.Asking very specific questions is normal and effective way to learn.Being highly specific is good and desirable,not faulty.Quote from: T3sl4co1l on July 21, 2018, 02:39:23 pmI agree but I dont understand how dismissing my question as being useless while not answering single thing I asked for helps me to understand the principles?

author: rhb
author_rank: Super Contributor
timestamp: July 22, 2018, 12:03:34 am
content: I have recently observed 3%, 7% and 10% overshoots on a 40 pS rise time square wave on some $20K 1 GHz scopes.The rigorous answer to your question can be obtained with Octave.Create one cycle of a square wave:a=[ones(1,1000000),zeros(1,100000)];A=fft(a);plot(abs(A));Then zoom in on the spikes. That's the amplitude of an infinitely short rise time. If you take the period as 1 second, then the spikes are at 1, 3, 5, 7, 9... Hz. And the resolution bandwidth of the plot is 0.000001 Hz.Now substitute a ramp for the leading and trailing edge. That will get fiddly to avoid introducing even harmonics. The FFT is defined on the semi-closed interval from -pi to pi or -1 to 1. The maximum frequency in the plot is the center. LHS is the positive frequencies, RHS is the negative frequencies.QEDEdit: The 0 to 100% ramp is a convolution with a short rectangular pulse. So the spectrum of the square wave gets multiplied in frequency by a sinc(f) function. The wider the pulse, the narrower the sinc(f) function. I'll let you look up the relationship between the period T and the zeros of sinc(f). I *should* be able to remember it, but after 30 years I still feel obliged to look it up if I'm actually doing something. But I do the same with the quadratic equation.

author: tggzzz
author_rank: Super Contributor
timestamp: July 22, 2018, 08:07:02 am
content: Quote from: fonograph on July 21, 2018, 11:14:22 pmSee the points in my previous message.I suggest you run a spice simulation of an abstraction of your question. I suggest a voltage step with a 1ns risetime driving the series combination of:a 10nH inductor (inductance of a 1cm wire)a 10pF capacitor in parallel with a 1Mohm resistor (a typical load)and observe the voltages and currents. Extra points for replacing the inductor with a 10ns transmission line.Understanding how and why those waveforms occur will be enlightening.

author: Benta
author_rank: Super Contributor
timestamp: July 22, 2018, 10:53:05 am
content: Quote from: fonograph on July 21, 2018, 09:59:57 pmBecause in saturation there is a surplus of charge in the base region of the transistor. When switching off the device, this charge needs to be evacuated before the transistor starts turning off. So the possible switching frequency in saturation is much lower than fT.