Making a filter that remains the same when loaded

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SUMMARY

This discussion centers on designing a filter using a BJT (Bipolar Junction Transistor) that maintains stable poles under load conditions. A buffer stage, such as a unity gain MOSFET or an emitter follower, is essential to prevent pole shifting when loading the filter. The conversation highlights the importance of circuit stability, gain characteristics, and the potential pitfalls of using primitive biasing techniques. Participants emphasize the need for precise specifications and stable designs to ensure reliable performance in real-world applications.

PREREQUISITES
  • Understanding of BJT (Bipolar Junction Transistor) operation
  • Knowledge of filter design principles and specifications
  • Familiarity with buffer stages, including emitter followers and MOSFETs
  • Experience with circuit simulation tools like LTspice
NEXT STEPS
  • Research "BJT filter design" for specific methodologies and best practices
  • Learn about "emitter follower" and its applications in buffering
  • Explore "Sallen-Key topology" for advanced filter designs
  • Investigate "LTspice" for simulating circuit behavior and performance analysis
USEFUL FOR

Electronics engineers, circuit designers, and hobbyists focused on filter design and stability in amplifier circuits will benefit from this discussion.

  • #31
LvW said:
Comment: Sorry - I forgot that you did require a 40dB/dec roll-off. Is this really important?

I don't see how this changes things? You can just use two RC filters cascaded and each will provide 20dB/dec. In total there would be 4 RC filters.
 
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  • #32
Yes - of course, It is not a huge problem. I have just forgotten to mention that the passive sections are each of 2nd-order.
But it is imopratnt to decouple low and high pass section in order to avoid interactions between the stages.
 
  • #33
interactions like Q-point and harmonics?
edit: nevermind, that's only for LC circuits. I'm not sure what interactions you mean then.
 
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  • #34
Does your client/boss insist on BJTs? I know some audiophiles have strange ideas about what should and shouldn't be in an amplifier.

Anyway, here's a quick and dirty design process:

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).

Op amps usually contain several transconductance amplifiers usually in a push-pull configuration which give very linear response except very close to the rail voltages. These are followed by a low impedance output stage, usually some sort of current mirror. Because of all the stages, they tend to have a fair amount of parasitic capacitance which limits their frequency response to the Gain-Bandwidth product. This is specified by the datasheet. The gain times the top bandwidth needs to be less than the gain-bandwith product (GBP). Of course this could be a problem if you want your roll off to be exactly 40dB/decade rather than simply greater than 40dB/decade. But that seems unlikely.

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'm not sure where you stand on economics. If this is a mass produced unit, keeping to a lower number of cheap parts may be very important, in which case I would stretch the 10dB requirement and build two 13dB stages. I'll go with this assumption. (Higher quality products require more parts and more work.)

So we need an op-amp with; 20 (13dB) times 10,000 = 200,000 GBP. We want two circuits on the chip for low chip count. Digi-Key has a nice database display, so we go there and find a part. I'll choose GPB of 1,000,000 to be conservative. (This isn't an unreasonable number and shouldn't add much cost/complexity.) There are hundreds of possible op-amps. Pick one with your design in mind (voltage levels, price, etc.) Perhaps an AD8542ARUZ-REELCT-ND? (BTW, I would find another chip supplier for mass produced stuff. Digi-Key is great on customer service, but often weak on price.)

Google "op-amp filters" to find a basic circuit layout. I got: http://www.electronics-tutorials.ws/filter/filter_7.html

Next, get the data sheets on your parts. Read them. Read them again. When you realize they don't work, rinse and repeat this procedure until you find what you need. Select your caps and resistors (remembering your input and output impedances).

Next enter all your parts in the BOM (bill of materials) and your cad software. Draw your schematic and lay out your board. Generate your gerber files and send them to the board fab. Wait for the mail to bring your boards. Populate them. Test them. Rework until it works.

Congratulations, you now have a working pre-prototype. Your client/boss will no doubt send it to some third world hell hole where illiterate peasants will change all your parts selections to something cheaper that may or may not work, but that's business, not engineering.
 
  • #35
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.
 
  • #36
Ry122 said:
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).
 
  • #37
LvW said:
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.
 
  • #38
LvW said:
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.
 
  • #39
@ 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
 
  • #40
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
 
  • #41
meBigGuy said:
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.
 
  • #42
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?

TTCm0hQ.jpg
 
  • #43
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?
 
  • #44
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..
 
  • #45
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.
 
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  • #46
And what about the main purpose of your activities: Filtering with the required properties?
 
  • #47
I imposed those limits because it was something I didn't know how to do.
 
  • #48
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.
 
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  • #49
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.
 
  • #50
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.
 
  • #51
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.
 
  • #52
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.
 
  • #53
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?
 
  • #54
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)?
 
  • #55
LvW said:
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?
 
  • #56
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.
 
  • #57
Ry122 said:
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.
 
  • #58
Jeff Rosenbury said:
Your design is more advanced and better, at least for learning purposes.
May I ask (for clarification purposes only): ...better than what?
 
  • #59
LvW said:
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.
 
  • #60
No problem - we have two different definitions for power and voltage (factor 10*log resp. 20*log).
 

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