# Art Of Electronics - RC Differenciator Circuits and High Pass Filters In the above image, take note of the line

dV/dt being much less than dVin/dt as a condition. I once asked a question about Integrators. I understand both the above image and my earlier question on integrators well enough (I think). A later section of the book in the frequency domain (above) references image 1.14 when it says "we can restate the earlier time domain condition for its proper operation (Vout being much less than Vin). I understand what follows when it mentions the 3dB point. My question specifically relates to finding the condition ....

Vout being much less than Vin

I do not see this in the section on differentiators. If anything, I see the condition of dV/dt being much less than dVin/dt in section 1.14. Vout as a condition, being much less than Vin was in an earlier section on integerators, not differentiators, which states their rates of change.

Can someone point out where the book finds the required conditon Vout being much less than Vin in section 1.14? I suppose that if the condition on their derivatives...dV/dt being much less than dVin/dt is met then necessarily, Vout being much less than Vin is also met, though in a differentiator, Vout can go negative while Vin remains positive. (high to low transition). Is this statement referencing magnitudes? Basically, I am trying to find how the above line on the condition for a high pass filter, which basically is an RC differentiator, is referenced in section 1.14. I am not able to find the exact wording or convert the meanings/statements properly. Can one necessarily state that a relation on the derivatives of V and Vin transfers to their functions?Thanks.

## 3 Answers

" I understand what follows when it mentions the 3dB point. My question specifically relates to finding the condition ....Vout being much less than Vin"

The simple C-R highpass in Fig. 1.36 is NOT a differentiator. However, it can approximate a differentiating block (with a nearly constant phase shift of 90 deg) for frequencies much lower than the 3dB frequency of the highpass. For this frequency range, the magnitude of the capacitive impedance is much larger than the value of the resistor. For this reason we can say: The shown circuit works (nearly) as a differentiator for frequenciues with Vout much smaller than Vin.

• Thanks, When you mention magnitudes and "Vout much smaller than Vin," you are only talking about the second image right? Also, when you mention "Vout much smaller than Vin," you are referring back to a section of High Pass filters that I did not print? If that is so, I see that and think the book is not clear enough. But, if the book were clear, ... I was stuck trying to relate the only thing in parenthesis in the second image to what was explicity contained in section 1.14. – Jeffrey Edward Messikian Jan 4 '17 at 15:12
• The second image seemed to restate the explanation for differentiators as "Vout much smaller than Vin" I could not find/derive that from section 1.14 (the top image). I tried to simulate this in LTSpice and found that you can make a differentiator (as in 1.14) where the output spikes are not that much smaller than an input square wave. In fact, when the square wave transitions from high to low, in a section 1.14 type differentiator, the output spike is negative. Still, if you are talking magnitudes, one can be made where this spike is not that much smaller than Vin. – Jeffrey Edward Messikian Jan 4 '17 at 15:12
• I alco considered that most of the time, these spikes were 0, or high, low, when the input remained at said level for a longer time and that might make their derivates smaller. But, I did not see that as "Vout much smaller than Vin," especially when Vout could go negative. Please advise. – Jeffrey Edward Messikian Jan 4 '17 at 15:12
• Jeffrey, when speaking about the 3dB frequency and the voltages Vout resp. Vin we are always in the FREQUENCY domain. This term (frequency domain) is mentioned in both images! Hence, you should not speak about spikes etc. (time domain). I hope you know the difference between both domains. – LvW Jan 4 '17 at 16:47
• Your first sentance is key. I am quite new to this. I thought I knew the difference between the two domains. I had understood talk of frequency and response put me in the frequency domain. In the second image, the author states "We can restate the earlier time-domain condition for its proper proper operation." In parenthesis, he immediately writes Vout being much less than Vin. – Jeffrey Edward Messikian Jan 4 '17 at 17:54

Can someone point out where the book finds the required conditon Vout being much less than Vin in section 1.14?

I=C(d/dt(Vin-V))=V/R

S0

I=RC(d/dt(Vin-V))=V

I=RC(d/dt)(Vin)-(d/dt)(RCV)=V

But R and C are defined to be small enough so it means the quantity (d/dt)(RCV) can be considered as approximately as zero. Or the quantity RCV is a approximately as zero.

So

I=RC(d/dt)(Vin)-(d/dt)(RCV)

is approximately equal to

I=RC(d/dt)(Vin)=V

thus

RC(d/dt)(Vin)=V=

C(d/dt)(Vin)=V/R

hence yielding the equation given in section 1.14

of the text book page shown in photo This implies the relation between Vout & Vin you consider

• Thanks Pete, I followed how you arrived at the above in section 1.14. My question had to do with what was said in a later section of the book. The later section of the book, which is in the second photograph, brought the readers attention back to section 1.14 and basically stated that what was described in section 1.14 AND what you illustrated above could be restated as .... "Vout needs to be much less than Vin" This is the only thing in parenthesis in the second photograph. How does what is shown in parenthesis in the second photograph relate to the above proof/section 1.14? Thanks – Jeffrey Edward Messikian Jan 4 '17 at 1:28 simulate this circuit – Schematic created using CircuitLab

A true voltage integrator assumes current is constant. but for small voltage changes like the 1st <30% of full scale this simple passive integrator is fairly linear.

for dV/dt<< Vin/RC we assume <30% is linear or <10% if really fussy. Did you want proof and error tolerance?

Can you guess what the differentiator signal looks like?

Here are 3 circuits that have the same total |Impedance| with a sine wave, but not with a square wave at same frequency due to impedance changes with harmonics.

For Integrate and differentiation, the goal with a simple passive circruit is to choose RC such that your output swing is 10 to 30% of full scale input to ensure reasonable linearity.

True integrator or differentiator circuits use constant current converter Op Amps to perform this task with linearity over the whole range.

Notice these RC values do not satisfy the linearity requierment of <30% full scale ( or <10% for better linearity) http://tinyurl.com/gwsg6ra