Learning the Art of Electronics, Week 2

 2L1.1

Ok, I  form a simple low pass filter using unknown RC values.

Then I measure time constant using scope and 500Hz square wave. Time constant is a time it takes for a voltage to reach ~63% of its total value.

Measured time constant using RIGOL DS1054z is 0.268ms

( measured using manual cursors: I set Percent for Vert. Units, then Set range of the wave with vertical cursors and then measure time to rise to ~63% (bx-by) )

Measured time constant using Tektronix 2221A is 0.270ms

( I position the wave from 0 to 100% and then using cursors get ▲T for ▲V to reach 63.2% )

t = RC

Now I need to find R. And task is to measure R without removing C.

We can do that by imagining that RC is just R, but we have to eliminate XC. We can do that by increasing frequency unti we get no phase shift. The higher the frequency the less XC there is. We can see that XC is not effective anymore when we see no more phase shift.

I add a known 1K resistor in front of RC filter.

Apply sine wave and increase frequency to such that there is not phase shift – it means we are now observing R and not Xc ( Xc ~ 0 ).

In scope I see input 5.12v and output 4.12v.

Using Ohms law: we loose 1V on R1 so current is 0.001A.

Knowing the current I calculate R2 and it is R2=Vout/I=4/0.001=4K

So R2 is ~4k

0.270ms = 4k x C

C = t / R = 0.270ms / 4k ~ 68nF

Actual values were 3.7K resistor and 683 capacitor (68nF).

2L1.2 Differentiator

Yellow input, blue output. Differentiator shows rate of change. We can examine it by applying sine, square and triangular waves to the input.

Especially it is seen by applying square wave. On rising edge or falling edge rate of change is very high.

On triangular wave rate of change is constant, just different directions so on the output we get square wave.

Its a high pass filter, so at f=0 impedance should be infinite, and at infinite f there should be no impedance, or impedance should be minimal.

2L1.3 Integrator

2L2 Frequency domain view

2L2.1 Low pass filter

I made such low pass filter:

By changing frequancy I found when the output voltage is attenuated by 3db i.e. is 0.7 of its original value. That frequency was 160Hz

Calculation confirms that f3db = 1 / (2×π×10k×0.1uF) = 159Hz

Result at 10 x f3db

Result at 20 x f3db:

Result at 2 x f3db:

Result at 4 x f3db:

Voltage vs f, the higher frequency the bigger attenuation.

2L2.2 High pass filter

f3db should be the same: f3db = 1 / (2×π×10k×0.1uF) = 159Hz

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