# Using Differential Amplifier to reduce required RC values

I need to implement a low pass filter to convert a 20kHz square wave into a 20kHz sine wave. It would be nice to be able to use lower values, because in CMOS large resistors and capacitors take up huge amounts of space.

Is it possible to reduce the value of RC required to achieve a given cutoff frequency?

This paper suggests that it's possible to hook it up so the cut off frequency is multiplied by the gain, allowing me to use smaller RC values. It states that the low pass filter makes the output of the entire circuit high pass. If I swap RINT and CINT, should this make the entire circuit low pass?

Below is my implementation, the output, and the expected output. I've tried it both ways round.

Is my idea possible, and where did I go wrong? Thanks.

• Step back and explain what you really want to accomplish and leave any imagined solution out of it. As it stands now, it would take too much to deal with existing preconceptions to answer this question, as apposed to just giving you a solution to the real problem, which is likely rather simple. – Olin Lathrop Dec 2 '14 at 16:36
• Edited it in. My main question is: "Is it possible to reduce the value of RC required to achieve a given cutoff frequency?" – Keri Dec 2 '14 at 16:47
• Lowpass of which order? Any number of opamps? Are OTA-C designs alowed? Capacitance multiplier for lower R values? ...so many options. – LvW Dec 2 '14 at 16:54
• What are the maximum allowed values for R and C (20 kHz pole frequency)? My first guess is a GIC based 2nd-order lowpass. – LvW Dec 2 '14 at 17:16
• You start by saying you want a sine from a square wave, but then dive into your imagined solution and ask about details of it. Again, give us the real specs. And what does it mean "because CMOS". Many opamps are CMOS, but many aren't either. – Olin Lathrop Dec 2 '14 at 18:42

Is my idea possible, and where did I go wrong?

Your idea won't work as per the circuit diagram you attached. The circuit diagram is a circuit that controls the offset voltage of a differential amplifier always forcing it to remain with 0 volts on the output. It's commonly used in instrumentation signal processing where the desired signal is AC but suffers with DC drift. It's basically a high pass filter and not a low pass filter system.

If you want to filter a sq wave into a sinewave then there is no alternative to choosing the correct value RC to do the job. If you have several sallen-key (2nd order LP filters) op-amp circuits cascaded then you have a little bit of leeway because the first harmonic doesn't occur until three times the fundamental frequency but, the closer you pitch the RC to the fundamental the better the sinewave purity will be.

Here is a GIC-based lowpass (2nd order):

### Design equations:

$$A_o=1 + \frac{R_5}{R_6}$$

$$wp=\sqrt{\frac{R_6}{R_1 R_3 R_5 C_2 C_4}}$$ with $$\wp=wc\$$ for Butterworth response.

$$Q_p=\frac{R2}{\sqrt{\dfrac{R_1 R_3 C_4}{C_2}}}$$ with $$\Q_p=0.7071\$$ for Butterworth response.

For desired R values you have enough freedom to "play" with.

With all due respect, I'm afraid that you'll find that capacitance amplification will probably not solve your problem. You'll find two things: First, along with capacitance amplification comes voltage amplification. So if a straightforward filter produces 2 volts across one of the big capacitors, a x10 capacitance amplifier will produce 20 volts across the internal equivalent, and you'll run into clipping problems. The second problem is Q. Since you are trying to turn a 20 kHz square wave into a 20 kHz sine wave, if you want a good sine wave you'll need a very selective filter, which translates into a high-Q filter. To get this performance you'll need a very high-frequency op amp, and you'll probably find you have stability problems. Just saying.

Additionally, your post stated "in CMOS large resistors and capacitors take up huge amounts of space." Well, that's sort of true, I guess, if you're rolling your own IC. Are you really rolling your own? If so, you need a lot more experience and expertise than you're demonstrating here. If you're using an off-the-shelf chip with external components, 20 kHz uses rather small values, so I really don't know what you mean.