Appendix E of The Art of Electronics, 3rd Edition (LC Butterworth filters) starts by saying that "active filters are convenient at low-frequencies but impractical at higher frequencies". They go and say that "at frequencies of 100kHz and above, the best approach is passive LC filters" (paraphrased in both cases).

My first question: really? A mere 100kHz is already too high for active filters to be practical?

I understand that op-amps with high bandwidth and HIGH slew-rate can be pricey, making it "impractical" in the general case --- however, a low-pass LC filter with, say, 1MHz cutoff, T topology with a 1kΩ load ends up requiring inductors in the order of hundreds of μH --- if I need to avoid distortion (magnetic core saturation and hysteresis), an air-core inductor in that range makes the whole thing rather impractical.

Question 2 would be: is a cutoff frequency of, say, less than 10MHz too high for a Sallen-Key 2nd-order low-pass filter?


simulate this circuit – Schematic created using CircuitLab

Analyzing it from the perspective of the ideal case (assuming op-amp always within linear operation), all three pins of the op-amp will be subject to the low-passed output signal --- at < 10MHz cutoff frequency that's certainly not an issue (neither bandwidth nor slew rate). Input capacitance shouldn't be a big issue --- with R in the order of 1k, the capacitors are in the order of a few tens of pF to a few hundreds of pF --- high enough to make the op-amp's input capacitance negligible.

Are there any other practical issues that I'm overlooking? Am I being realistic if I want such an active filter with cutoff in the order of a few MHz? (pricing is not an issue --- if I need an op-amp in the $10 or $20 range, that's fine)

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    \$\begingroup\$ Can you define source R and load R and cable capacitance? and if possible phase shift at 10MHz @ -3dB and rejection -dB @ 20MHz. Linear phase, maximally flat or ?? Usually GBW must be much greater than signal BW to reduce 200 Ohm or so Ro by gain. There is a reason why it is limited and it depends on these parameters What is the purpose? \$\endgroup\$ Jun 5, 2019 at 19:27
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    \$\begingroup\$ The main reason is unity gain instability with >= 100GBW on capacitive cable loads, high output impedance unless impedance matched as 1pF stray capacitance can cause peaking. \$\endgroup\$ Jun 5, 2019 at 19:34
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    \$\begingroup\$ Walt Jung of ADI warns us "To achieve -40dB stopband performance in an active lowpass filter, the opamp needs to have 40dB gain surplus everywhere in the stopband." Additionally, opamps often have inductive Zout (rising resistance and a 90 degree phaseshift provided by the opamp rolloff), and the capacitors in the filter provide a high-frequency path AROUND the opamp; with rising Zout, the opamp cannot attenuate that high frequency energy. Thus if you REALLY need excellent stopband performance, have a passive RC LPF as the first pole, and be generous with opamp specs. \$\endgroup\$ Jun 6, 2019 at 2:08
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    \$\begingroup\$ The book is probably right if you equate "op amp" with "741". But not if you actually use an op amp instead :) \$\endgroup\$
    – alephzero
    Jun 6, 2019 at 9:39
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    \$\begingroup\$ @analogsystemsrf -- good point; I was precisely thinking that it wouldn't hurt doing a 3rd-order Butterworth (1/H(s) = (s+1)(s²+s+1) if I remember correctly). In any case, the filter gets an initial input stage that is just an RC. \$\endgroup\$
    – Cal-linux
    Jun 6, 2019 at 11:59

4 Answers 4


I believe your analysis to be good. I've made sallen-key 4th order filters that cut-off around 3 MHz with absolutely no worry about performance. I don't see that 10 MHz is unachievable.

It's all about op-amp choice. For a unity gain stage it's easy to ascertain where the gain starts dropping below (say) 0.99 and regard that as the limiting frequency. On the other hand, the output impedance of an op-amp usually gets worse as it enters the MHz regions so you have to be sure it can deliver the peak current without clipping or going too sloppy.

You also have to consider slew rate limitations but, as far as I'm aware, that's about it.

It's quite possible that The Art of Electronics, 3rd Edition didn't make any updates on that section since it first came out in 1980.

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    \$\begingroup\$ That's the 7th down vote today - any ideas anyone? \$\endgroup\$
    – Andy aka
    Jun 5, 2019 at 18:28
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    \$\begingroup\$ I get the same too. Must be newbies that do not appreciate free expertise and don't know how to write a comment \$\endgroup\$ Jun 5, 2019 at 19:15
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    \$\begingroup\$ @Andyaka. The down votes have got to be frustrating. But your answers always get positive votes in the end . \$\endgroup\$
    – Marla
    Jun 5, 2019 at 19:38
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    \$\begingroup\$ indeed 10MHz is perfectly possible. I recently made an active low pass for 10 MHz and even let the OpAmp drive a 50 Ohm load. It works fine, however these OpAmps are pricey and also produce a bit of heat. \$\endgroup\$
    – T. Pluess
    Jun 5, 2019 at 21:40
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    \$\begingroup\$ It is often very frustrating to deal with posts that go hot on the stackexchange. All kinds of undesired behaviour creeps in. I wish it didint. \$\endgroup\$
    – joojaa
    Jun 6, 2019 at 17:47

My first question: really? A mere 100kHz is already too high for active filters to be practical?

No, 100kHz is nothing, but it all depends on the opamp. At some point the Gain Bandwidth Product is going to cause problems. If you had an op amp with a 1MHz or 10MHz GBWP (which may have been typical at the time of the first edition of AofE, maybe they didn't update it is my thinking so I'd compare editions) then 100kHz doesn't sound too unreasonable, because you'd only get a magnitude or two of filtering and then the bandwidth goes below unity gain. Then your lowpass filter looks more like a bandpass.

Are there any other practical issues that I'm overlooking? Am I being realistic if I want such an active filter with cutoff in the order of a few MHz? (pricing is not an issue --- if I need an op-amp in the $10 or $20 range, that's fine)

If you really do need filtering past 50MHz then parasitics need to be modeled as ESR and ESL in capacitors will start to affect the filter poles and create their own filter poles at high frequencies. Use a spice package if possible. Make sure the GBWP is high enough, these days it's not hard to get op amps that work in the +100MHz range.

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    \$\begingroup\$ This is right to the point. High GBWP OP Amps were not as performant, cost efficient, or even available in 1980 when AoE was first published. In 1980 the 8086 was cutting edge and 10MHz on an IC was blazing fast. Now we can buy an LMH6881 for $3 with a 2.4GHz bandwidth, or the LMH5401 for $7 with an 8GHz GBWP - that would have been unthinkable in 1980. The book just hasn't been updated. \$\endgroup\$
    – J...
    Jun 7, 2019 at 13:11

The main problem with that Sallen Key topology at high frequency is that the output impedance of op-amps rises, so fails to control feedforward of the input signal through the 2C capacitor, trashing the stopband.


TI has a 10MHz design App Note. It is based on their THS4001 low-cost 270 MHz -3dB Op Amp.

Op Amps have an open loop output impedance much higher than your 50 Ω signal generator. This makes them stable with their short circuit protection. The higher GBW is used to lower the Zout = Zoc/GBW. The breadboard ESL (0.5nH/mm) and stray capacitance will need to be minimized.

With 150 MHz GBW you can use 1k R's with 5 pf, 10pF.

I did not read their design.


To design any filter, you should consider these specs 1st;

Source impedance \$Z_S(f)\$   
Load Impedance \$Z_L(f)\$   
Gain   -3 dB passband \$f_p\$    
Loss   @ \$f_s\$stop band edge   e.g. \$  ~-dB~ @ ~2*f_p, 10*f_p\$    
 ..  or order of filter    
% load regulation error = % Output/Load impedance ratio ( for low % )    
Phase shift in passband, group delay  
Noise, supply power  
Output swing and slew rate limit  

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