Making a filter that remains the same when loaded

[LvW]

interactions like Q-point and harmonics?
edit: nevermind, that's only for LC circuits. I'm not sure what interactions you mean then.

Interaction means: The 2nd passive stage must not load the 1st stage because this makes calculations much more complicated.
It is much better to design lowpass and highpass separately (indpendent on each other) and to combine them with a buffer in between.
Perhaps the finite input resistance of the buffer must be included in the design of the 1st stage (depends on the buffer realization).

 

[Jeff Rosenbury]

May I give some general remarks?

"Do you care about phase shift? It wasn't in the specs, so I'll assume not. If you do, you need to study the many different filter types. They mostly give similar frequency performance, but there are subtleties I don't fully understand having to do with phase shifts and the like. If these characteristics are important to you or you just want to explore the math, there are lots of filter types. For example if you want a flat frequency response in the pass band, look at the Sallen-Key topology (which isn't the one I found, BTW). "

Phase shift is closely related to the amplitude response and has nothing to do with the topology of the circuits. Flat response means "Butterworth" characteristic and a passband response with ripples belongs to a Chebyshev response.

"A 40dB filter is usually 2 poles. But I was taught good design practice is to limit gain to 10dB/stage and you want 26dB across midband. So typically you want three stages, two filters and a gain stage to round it out. "

I am afraid, here is a confusion between a filters slope of 40dB/dec and the required midband filter gain.

Thank you. I'm no expert.

I assumed the midband gain of 400 was a linear measure, i.e. not in dB. Four hundred = 26dB (I think). Two 13db stages = 26dB = a gain of 400.

Where the confusion might arise is between power and voltage. A first order filter (1 stage) has 6dB per octave or 20 dB per decade. But I think that's in terms of power. In terms of voltage I think it's half that (power is proportional to the square of the voltage, which in dB is a factor of 2.). If you need more roll-off, you might want more active stages (they make chips with 4 op-amps) or use two active and two passive filters. Or 3 and 1? Lots of choices.

I was under the impression that different filters placed the parts in different places (which is what I meant by topology). I was also under the impression that there was no ideal filter. A Butterworth gave a flat amplitude while a Bessel gave a flat phase response (i.e. the group and phase velocities matched). I have no idea what a Chebyshev is for. There are eliptical filters as well. Of course I could be wrong about all of this. Filters aren't really my thing.

 

[Jeff Rosenbury]

Interaction means: The 2nd passive stage must not load the 1st stage because this makes calculations much more complicated.
It is much better to design lowpass and highpass separately (indpendent on each other) and to combine them with a buffer in between.
Perhaps the finite input resistance of the buffer must be included in the design of the 1st stage (depends on the buffer realization).

One of the advantages to op-amps is their infinite(ish) input impedance. Of course the extra components you add will make set the stage's input impedance. Similarly they typically have a low (<10Ω) output impedance. The other components will determine that as well. This simplifies design considerably. If you choose high value resistors on the input side and low value on the output, loading will be minimal.

These basic designs have been used successfully for decades.

 

[Baluncore]

@ Ry122.
In LTspice you can select a window, drag it to size, then use “Tools”, “Copy bitmap to Clipboard”.
No cropping is required and the file can be smaller.

Use the standard multipliers a, f, p, n, u, m, k, MEG, g when specifying values in LTspice.
For example 4k7 = 4700 ohms. See; http://en.wikipedia.org/wiki/Metric_prefix

In LTspice, the C, R or the L symbols are place fillers that should be removed once a value has been assigned to a component. Avoid the units symbol, especially F for farad which is interpreted by LTspice as f for femto the multiplier.

Take a look at the BJT bias pictures and text here; http://en.wikipedia.org/wiki/Bipolar_transistor_biasing

 

[meBigGuy]

I would have removed the emitter bypass capacitors if it was my drawing, not just a lazy copy from google images.

Sorry. That capacitor is probably not what you want. I should have at least said that when I posted the image. It increases the high frequency gain (or lowers the low frequency gain). With a bypass capacitor the high frequency gain is controlled by the internal re of the transistor. If you are OK with its parametric variation and want the low frequency rolloff, then leave it in. Or, you can also split the emitter resistor and bypass one of the resistors.

Re provides high input impedance for the stage, and the ratio with Rc controls the gain. You can think of it as follows:
Whatever current flow in Re flows in Rc (minus base current) so the voltage drop across Rc is Rc/Re higher than the drop across Re.

As for input impedance, the effective input impedance of the transistor is approximately (beta)*Re. SO that in parallel with R1 || R2 is the amplifier input impedance.

Here is yet another writeup:
http://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

 

[Jeff Rosenbury]

I would have removed the emitter bypass capacitors if it was my drawing, not just a lazy copy from google images.

Sorry. That capacitor is probably not what you want. I should have at least said that when I posted the image. It increases the high frequency gain (or lowers the low frequency gain). With a bypass capacitor the high frequency gain is controlled by the internal re of the transistor. If you are OK with its parametric variation and want the low frequency rolloff, then leave it in. Or, you can also split the emitter resistor and bypass one of the resistors.

Re provides high input impedance for the stage, and the ratio with Rc controls the gain. You can think of it as follows:
Whatever current flow in Re flows in Rc (minus base current) so the voltage drop across Rc is Rc/Re higher than the drop across Re.

As for input impedance, the effective input impedance of the transistor is approximately (beta)*Re. SO that in parallel with R1 || R2 is the amplifier input impedance.

Here is yet another writeup:
http://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.html

The wacky parameter on a BJT is usually the ß, so having the input resistance determined by the ß is sort of bad. Lowering the bias resistor values can mitigate that, but at the cost of a higher bias current. BJTs sometimes work better with higher bias currents, but that also means more heat, less battery life, etc.

Betas will be specified on the data sheet with a typical value and a minimum value. For a limited production run there's nothing wrong with hand matching the ß, but it becomes expensive on large (or even medium) production runs. If the minimum ß is around 50, disregard this note. R2 will dominate. But if it's less than 10, R2 will still dominate, but ßR4 will have a distinct effect.

For anyone who cares (it's not relevant for this discussion) by work better I mean have better frequency response. This only really matters when using a transistor near its rated maximum frequency, which will be much higher than 10kHz.

 

[Ry122]

Having the resistor at Re not bypassed by a capacitor means I have much lower gain in my midband. The bode plot with the capacitor provides an output much closer to what I require.
However, if the output is going to vary a lot with temperature in that design, it's of no use to me.

To retain the benefits of an Re but also have a high gain, I made this circuit. Would this be better than the last in terms of temperature variation affecting the output?

 

[LvW]

Ry122, do you really consider the 5-transistor circuit as an acceptable solution?
Again my question: Are you forced to use transistors instead of opamps?

 

[LvW]

Quote: "Where the confusion might arise is between power and voltage. A first order filter (1 stage) has 6dB per octave or 20 dB per decade. But I think that's in terms of power. In terms of voltage I think it's half that (power is proportional to the square of the voltage, which in dB is a factor of 2.)"

Just to avoid confusion: No - 20db/dec for a first-order filter is in terms of voltage..

 

[Ry122]

Yes, I do consider it acceptable to use 5 transistors, and I'm doing this to self-teach myself op-amp design using discretes, so it would defeat the purpose if I just used an op-amp.
The question is, does this provide a better/more stable output that's less dependent on transistor temperature than the diagram that mebigguy posted.

 

[LvW]

And what about the main purpose of your activities: Filtering with the required properties?

 

[Ry122]

I imposed those limits because it was something I didn't know how to do.

 

[meBigGuy]

Once you go to multiple stages, there are much better architectures.

Cascode, complimentary, and differential techniques provide better performance. This paper has lots of good ideas.
http://wiki.analog.com/university/courses/electronics/text/chapter-10. 10.2 is what I consider to be a basic 2 stage amplifier.
Chapters 9,10,11,12 are all about what you are trying to learn.

You are missing the point about what makes opamps easy to control.
The opamp, by itself, has wildly variable gain and bandwidth, but it is then controlled with external feedback. Read about opamp open loop gain.

 

[Ry122]

yeah, I've already implemented a mosfet cascode mirror source and mosfet differential amplifier. The gain stage is the final part I'm needing to design.

 

[LvW]

Here is my design recommendation (in case you want to use BJT`s only). This design was verified by circuit simulation.
* Two RC sections R1C1=R2C2 form a second-order lowpass (R2=100R1 and C2=C1/100) with fc=10kHz
* Two CR sections C3R3=C4R4 form a second-order highpass (C3=100C4 and R3=R4/100) with fc=100 Hz.
* Lowpass a nd highpass are decoupled by a common-collector stage (emitter follower) which has an input resistance of app. 50k (at least). Of course, a large coupling capacitor is necessary (100uF)
*This gives a bandpass response as desired.
* The required gain must be provided separately with two stages: (a) Emitter follower plus common emitter gain stage or (b) two gain stages.
In both cases, the finite input resistance of the first stage (after the highpass section) must be taken into account in parallel to the highpass resistor R4.

 

[Ry122]

Yeah, but what about the arguments made in the first 2 pages of this thread about Re? Would you implement that in the same as I have above, or would you implement it in the way that its done in mebigguy's diagram.

 

[LvW]

There is only one way to "implement" Re - as a resistor between the emitter node and ground. I suppose, you know the circuit called "emitter follower"?

But I forgot to mention the relevant time constants: R1C1=R2C2=11.3µsec and C3R3=C4R4=2.3msec.

 

[Ry122]

yes, but in mebigguy's diagram it's bypassed by a capacitor. In mine only 50ohms is present at AC due to the other 450ohms being bypassed by a capacitor. The question is, what are the trade offs between these. Is 50ohms enough or too little? Should I increase it, and use more transistors to make up for the loss in gain?

 

[LvW]

To me, this is a secondary problem - and it depends on your accuracy requirements regarding the gain value.
Bypassing the emitter resistor in the gain stage increases the gain value but reduces its sensitivity to temperature and tolerance parameters.
Of course, a trade-off (bypassing a portion of Re only) makes sense and is used quite often.
But - why not first concentrate on the filter sections and the decoupling stage (emitter follower)?

 

[Ry122]

the finite input resistance of the first stage (after the highpass section) must be taken into account in parallel to the highpass resistor R4

Why not just throw another unity gain mosfet buffer between the 1st gain stage and the filter stages so it doesn't even see the impedance of R4?

 

[LvW]

Up to now - I was of the opinion that you want to use BJT`s only. Of course, with FET`s you have other options.

 

[Jeff Rosenbury]

yes, but in mebigguy's diagram it's bypassed by a capacitor. In mine only 50ohms is present at AC due to the other 450ohms being bypassed by a capacitor. The question is, what are the trade offs between these. Is 50ohms enough or too little? Should I increase it, and use more transistors to make up for the loss in gain?

Your design is more advanced and better, at least for learning purposes. Someone, (meBigGuy?) pointed to this emitter configuration earlier. (Sorry, I'm on a satellite connection and reloading pages is a pain, so I can't check the thread.)

The low value resistor gives gain stability, while the high value, bypassed resistor gives DC stability.

I haven't run the numbers, but copy pasta stages rarely match impedances for common emitter configurations. A low impedance feeding a high impedance is sometimes acceptable for signal work, but the opposite is not good. As a learning lesson, you need to match them. This becomes important in power work where loss of power gain is heat and smoke. Of course changing the resistor values means changing the caps, means recalculating the filters... Good practice! And also a good motivation for using op-amps.

Op-amps are frequency limited, so this sort of work goes on all the time at higher frequencies. It's not a wasted effort to learn various configurations of discrete transistors.

 

[LvW]

Your design is more advanced and better, at least for learning purposes.

May I ask (for clarification purposes only): ...better than what?

 

[Jeff Rosenbury]

Quote: "Where the confusion might arise is between power and voltage. A first order filter (1 stage) has 6dB per octave or 20 dB per decade. But I think that's in terms of power. In terms of voltage I think it's half that (power is proportional to the square of the voltage, which in dB is a factor of 2.)"

Just to avoid confusion: No - 20db/dec for a first-order filter is in terms of voltage..

Sorry. Oddly, my dyslexia keeps me from remembering which way the voltage/power 10dB vs. 20dB thing goes. It's great for recognizing symmetries; not so good for remembering which side I'm on though.

 

[LvW]

No problem - we have two different definitions for power and voltage (factor 10*log resp. 20*log).

 

[Jeff Rosenbury]

May I ask (for clarification purposes only): ...better than what?

Better than other configurations. No emitter resistor is somewhat unstable. A simple resistor costs gain. Bypassing it makes the gain dependent on the ß. But bypassing most of it gives the best of all worlds.

Remember there's a cost to each added component. I've seen professional designs with a simple base bias resistor (plus the collector resistor). These have lots of drawbacks, but they are cheap. Sometimes cheap matters more than other considerations.

But for learning purposes, extra parts are free and doing the calculations is good practice.

 

[donpacino]

(for some mysterious reason that I cannot always pin down)

story of my life

 

[LvW]

Better than other configurations. No emitter resistor is somewhat unstable. A simple resistor costs gain. Bypassing it makes the gain dependent on the ß.
....
.

Yes - that`s what I also have tried to explain in my posts#50,#52 and#54.
However, I doubt if Ry122`s five.transistors circuit (5 emitterstages in series) is really "better than other configurations".
Example: Where is the bandpass circuitry?

 

[Jeff Rosenbury]

Yes - that`s what I also have tried to explain in my posts#50,#52 and#54.
However, I doubt if Ry122`s five.transistors circuit (5 emitterstages in series) is really "better than other configurations".
Example: Where is the bandpass circuitry?

It's not a contest. He is a student needing encouragement.

His coupling capacitors will act as high pass filters. He had some low pass filters at the beginning. As I said, I didn't run the numbers and have no idea if they are the right values. I suspect I could find lots more wrong with it as well. (Why put the filter where the signal strength is lowest, for example?) But this is a teaching moment. Besides, most of the "wrong" things aren't wrong except in specific contexts. In many (most?) designs the filter is fine at the beginning of the circuit. But there are some cases where that would bite him.

If he wanted a working circuit, he would have used the op-amps. He wants to learn. I'm all for that.

I do have to agree that if I were his boss and he handed me this five transistor stage design for a two pole, 26 dB gain filter, I'd want an explanation.

 

[LvW]

It's not a contest. He is a student needing encouragement.
His coupling capacitors will act as high pass filters. He had some low pass filters at the beginning. As I said, I didn't run the numbers and have no idea if they are the right values. I suspect I could find lots more wrong with it as well. (Why put the filter where the signal strength is lowest, for example?) But this is a teaching moment. .

OK - I got it and I agree with you. It is - perhaps - a teaching event.
And from my own teaching experinece I know that, of course, there are always different teaching approaches.
However, in particular, because of this background I have recommended to the questioner a SYSTEMATIC approach: Starting with the desired filtering action (bandpass) - and shifting the design of the gain stages to the end of the design process. For my opinion, this is the best way to solve the problem.
To be honest - do you really think it would be possible to design the various coupling capacitors of the five stages so that the meet the highpass requirements?
More than that, also the by pass capacitors in the emitter legs exhibit a highpass function!
I really have severe doubts that this approach is a good one.

In short: I think, the best/most simple/most systematic approach would be a series combination of
* A passive 4-pole RC bandpass - decoupled with an emitter follower, followed by
* a two-transistor gain stage (emitter follower and common-emittere stage).

 

[Jeff Rosenbury]

I agree that his approach isn't the one I would use. But we have pointed that out to him repeatedly.

He doesn't seem to want to learn proper design methodology. Nor should he unless he's planning on working as an engineer. Only he knows what he wants to learn from this. He has repeatedly stated he wanted to use common emitter BJT transistors. I have to accept that.

Since this is a Physic Forums, I assume he's a physicist and knows what he wants. So I'll try to help him as I can.

Scientists don't like learning engineering. It's a point of pride with some. It seems to interfere with their ability to ignore reality. Had Einstein been an engineer, he probably would have learned Newtonian mechanics so well he never would have found relativity.

Of course I wouldn't want to drive over a bridge built by a scientist. To each cat his own rat.

 

[Ry122]

Because you have ac-shorted the emitter resistance in both stages the input resustance at the base node as well as the signal gain depends considerably on transistor parameters. Bad design.

Contrary to what you said, I found this at allaboutcircuits"The fact that we have to introduce negative feedback into a common-emitter amplifier to avoid thermal runaway is an unsatisfying solution. Is it possibe to avoid thermal runaway without having to suppress the amplifier's inherently high voltage gain? A best-of-both-worlds solution to this dilemma is available to us if we closely examine the problem: the voltage gain that we have to minimize in order to avoid thermal runaway is the DC voltage gain, not the AC voltage gain. After all, it isn't the AC input signal that fuels thermal runaway: its the DC bias voltage required for a certain class of operation: that quiescent DC signal that we use to “trick” the transistor (fundamentally a DC device) into amplifying an AC signal. We can suppress DC voltage gain in a common-emitter amplifier circuit without suppressing AC voltage gain if we figure out a way to make the negative feedback only function with DC. That is, if we only feed back an inverted DC signal from output to input, but not an inverted AC signal.

The Rfeedback emitter resistor provides negative feedback by dropping a voltage proportional to load current. In other words, negative feedback is accomplished by inserting an impedance into the emitter current path. If we want to feed back DC but not AC, we need an impedance that is high for DC but low for AC. What kind of circuit presents a high impedance to DC but a low impedance to AC? A high-pass filter, of course!"

That high pass filter mentioned above is the bypass capacitor posted in mebigguy's image

Or were you referring to the AC gain being dependent on the beta value and not that thermal runaway is still occurring?

 

[meBigGuy]

He is referring to the AC gain depending on the transistor characteristics. Either you can live with that or not, depending on your application's requirements.

One solution is to get as much gain as you can in a multistage amplifier and then stabilize with negative feedback. This both controls the gain AND reduces distortion. It really is required in high quality audio circuits using discrete devices.

Look at google for "discrete negative feedback audio amplifier design"

http://www.allaboutcircuits.com/vol_6/chpt_5/13.html
http://circuit-diagram.hqew.net/Feedback-Amplifier-Using-Transistors_2749.html

http://www.pearl-hifi.com/06_Lit_Archive/14_Books_Tech_Papers/Self_Douglas/Small_Signal_Audio_Design.pdf Do you understand open loop gain in an opamp? Why it is important, and how it decreases distortion? You can accomplish some of that in a well designed multi-stage negative feedback amplifier.

 

[LvW]

Contrary to what you said, I found this at allaboutcircuits

Ry122 - where is the contradiction? I spoke about the DEPENDENCE of the signal gain on transistor parameters - not about the VALUE of the gain.
Of course, the gain drops without the capacitor because you now have not only dc but also ac feedback. But this is a desired effect!
All good amplifier stages work with signal feedback - BJT, FET, Opamp.
(remember: Opamps do not work at all without signal feedback).

 

[Jeff Rosenbury]

To put it another way, maximum gain is dependent on the ß of the transistor. But since each transistor has a different ß it is hard to know just what that maximum gain will be.

Adding a small AC emitter resistance lowers the gain to a fixed and therefore predictable level.

If your goal is maximum gain, hand pack your transistors and ditch the resistor.

All of these variations have uses in differing applications. The key is to choose the one you want for your particular application. Knowing how to do that is why EEs get paid the big bucks. After all, anyone can look up a few schematics online.