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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Learning the Art of Electronics, Week 1

Week 1. Ohms law, power, Kirchhoff`s law V,I.

Ohm`s law V=IR. So what is a Volt ? Potential energy per unit charge or work done to move a charge against electric field, form one potential to a higher potential.

Dynamic resistance Rdynamic= ΔV/ΔI it is the local resistance – the tangent to the slope of the device V-I curve.

Power P=I×V or derived: P=I²R or P=V²/R

Kirchhoff`s law: sum of voltages around circuit is zero, sum of currents in and out of node is zero.

Parallel resistance: Rtotal = (R1×R2)/(R1+R2)

Rule of thumb: if two parallel resistors differ by a factor of ten or more, then we can ignore the larger of two.

In other words: in a parallel circuit, resistor much smaller than other dominates. Ina series circuits, the larger resistor dominates.

Voltage divider. Rely on fact the I is the same on top and bottom.
Vout = Vin × R2 / (R1+R2)
Tip: since the current at the top and bottom are equal, the voltage drops are proportional to the resistances.
So if the lower leg is ten times the upper leg, it will show 90% of the input voltage ( or 10/11 if exactly).
Or if lower leg is 3 times than upper leg, output voltage should be 75% of input voltage ( or 3/4 if exactly ), lets check that:
Thevenin`s model. How to calculate circuit when it gets loaded? Model the actual circuit with the simpler circuit – Thevenin model – which is idealized voltage source with resistor in series.
Then we can rapidly see how that circuit behaves under various loads.
Vthevenin is just open circuit, the voltage when nothing is attached ( no load ).
Rthevenin is Vthevenin/Ishort-circuit 
The fastest way to calculate Rthevenin is to see it as the parallel resistance of several resistances viewed from the output.
Any non ideal voltage source droops when loaded. How much it droop depends on it output impedance. The Thevenin equivalent model, with its Rthevenin describes this property neatly in a single number.
Design rule of thumb: when circuit A drives circuit B. Let Rout for A be <= 1/10 Rin for B.
It means we will pass strong signal to B without big droop.
If RoutA is much smaller then RinB then the divider delivers nearly the all of the original signal.
Labs.
IL.1 nothing to do here, just some information how to use lab stuff.
IL.2 Voltage divider. I have used 10k resistors for voltage divider and 3.9k for load. Results:
So we have voltage divider with 10k resistors. We calculate Rthevenin 5k and Vthevenin 7.5v.
We measure that Vopen is 7.49V that confirms our calculations.
also we measure our Ishort-circuit to be 1.5mA
After measuring we calculate Rthevenin and it is 4.9K and that confirms what we have calculated previously.
Now we load circuit with 3.9V and measure that voltage droop or Vload is 3.2V
And our calculation later confirms our measurements.
Now, if we would like to deliver almost all of the signal to load, by our rule of thumb we should choose Rload to be at least 10 times Rthevenin. Lets choose Rload 56k and see what happens.
And that confirms our rule of thumb: that when loaded with at least 10 times higher resistance we got 90 percent of our signal delivered.
1L.3 I have used analog VOM, in 10v scale I have measured 200k internal resistance.
So it means it moves full scale on 10v with its internal resistance of 200k.
How could one use it to measure for example 50v or 100k ?
We must divide this voltage so that on 50v for example scale would behave same like on 10v, and same for 100v.
We have to add two voltage dividers and calculate resistor values.
We can calculate resistance in two ways: algebraic or just talk our way out.
The bigger the resistance, the more voltage we lose in a dividers leg. In case of 50v we now that we must have 1/5 (10v) on lower leg. Se we must loose 4 from 5 in upper leg. We must choose proportionally bigger resistor in upper leg. 200K times 4 is 800k. That is the value for upper leg resistor in case of 50v.
1L3.3 To get 10mA ammeter from voltmeter we fist calculate current when itsi reading maximum 10v using its internal 200k resistance.
It is I0=10v/200k=50uA
10mA that we need to measure is 200 time bigger that 50uA
So we would add parallel resistor to take 199 parts of current fro 200k and we get 1k resistor. As 1k is way higher that 200k, our internal resistance becomes 1k.

Learning the Art Of Electronics

I have decided to get my self together and systematize my knowledge of electronics. It is very similar situation as for this guy. Electronics is my hobby. And actually one of many hobbies. I would say I have a lot of knowledge it but there are plenty of gaps and places just for intuition. Especially in circuit design.

Once I have purchased a book “Learning The art Of Electronics” and used it as reference for this and that but never when through it.

But this is it.

My goal is to complete one lab at least once a week and post a report here.

Hopefully I will make it through 🙂