[SOLVED] Any way to learn the values of elements?

[CalcoN]

Hello,

I was wondering if you could give me a little hint what values should I use to solder this circuit?

 

[Audioguru]

1) Select a frequency.
2) Look at Phase Shift Oscillator in Google to see the formula for the R and C values on the left side of the schematic.
3) Select a transistor current, maybe 1mA.
4) Allow +2V at the transistor emitter and half of the 7V remaining voltage (3.5V) to be across RL, then the collector is (9V - 3.5V=) +5.5V. Calculate the values for Re and RL.
5) The datasheet of the BC547 shows the base-emitter voltage at 1mA is about 0.62V so the base voltage must be about 2V + 0.62V= 2.62V.
6) The 3rd R for the phase shift is in parallel with R1 and the input resistance of the transistor. The input resistance of the transistor can be seen on a graph on the datasheet of the 2N3904 and is about 3.6k ohms. The minimum current gain of a BC547 the "C" is not available anymore) is about 100 at 1mA so make the current in R1 and R be 1/10th which is 100uA.
7) Now you know the voltage across R (2.62V) and the 100uA current so calculate its value.
8) You know the voltage across R1 and you know its current (100uA) so calculate its value.
9) Calculate the value for the other two R phase shift resistors so that their value is R parallel with R1 and parallel with the input resistance of the transistor.
10) Calculate the value for the C phase shift capacitors.
11) Calculate the values of the coupling capacitors so that they pass the frequency you selected. Calculate the value for the capacitor parallel with Re so that it has a low reactance at the frequency you selected.
12) Look at the datasheet of the LM386 to see its input resistance and select the value of the volume control to be the same. Look at the values for the RC at its output.

Simple, isn't it? Why didn't your teacher teach you these things? Since you were not taught the basics of electronics then maybe you should buy a kit with all the parts for you to solder together.

The output will probably be clipped because the simple circuit has nothing to limit the output level from the transistor.

 

[CalcoN]

Hey there Audioguru, nice to see you here too.

I really appreciate your post, thank you.

I tried to do the calculations, posting some pics.



your instructions.

I'm not sure about your 11) step, which capacitors you have in mind?

My circuit looks like that.

 

[Pjdd]

What Audioguru means is that the reactance of the coupling capacitor at the frequency of interest should be small compared to the impedance following it.

The following impedance Z is a combination of the pot and the input impedance (Zin) of the LM386. The combined Z will be lowest when the pot is turned fully up. Then Z will be two 50k's in parallel = 25k.

The value of C is not critical. It should simply be large enough so that its reactance 1/(2πfC) << 25k.

 

[Audioguru]

Your transistor circuit design is messed up:

 

[CalcoN]

Hey, I soldered the circuit with the values I counted, and you were right, my circuit doesn't work.

You wrote all the instructions for the 9V source, but now you write that 9V is too high, so I don't really understand is 9V good or not.

I appreciate all your help.

Attaching some photos just for visualisation of my first ever soldering work. (too bad not working )
**broken link removed**
**broken link removed**

 

[Audioguru]

The resistor values can be calculated so that the transistor works properly with any supply voltage. Your circuit uses wrong resistor values that makes the base voltage much too high that makes the emitter voltage too high.
I said to allow 2V at the emitter (yours is about 3.85V) and I said to allow half the remaining 7V to be across the collector resistor (your collector voltage is much too low because your collector and emitter resistor values are too high).

If you properly calculate the resistor values then it will work but will produce a lot of interference because your parts are much too far apart. All your long connecting wires are antennas that pickup the interference.

 

[CalcoN]

Would I be right if I changed the R1 value to 90k? Seems I counted it incorrectly before, because R1=U/I=9/100uA=90000 Ohms. As a result my base voltage should be normal now, and RL and Re should be okay with the previous values (3.5k and 2k). I wonder, is my assumption correct?

 

[Pjdd]

Your mentor on this thread Audioguru seems to have taken a leave of absence for the time being. So let's see what I can do as a sub

Your main point of confusion seems to be in calculating the value of R1, specifically about the voltage across it. (You already know ohm's law). R1 and the 3rd R are in series and the same current flows from +9V to 0V. That is, the total voltage across the series combination of the two resistors is 9V.

Now, you've already chosen to have 2.62V across the 3rd R. That leaves (9 - 2.62)V = 6.38V (not 9V) across R1. Since the same 100uA flows through R1, you should now be able to calculate the value for R1.

If you arrive at the correct value for R1, that will place the base at 2.62V and the emitter at 2V. Since Re is 2k, that fixes Ie at 2V/2k = 1mA. That in turn fixes the collector current Ic at 1mA.

Ic flows through RL so that there's a voltage drop of 1mA*3.5k = 3.5V. Dropping 3.5V from 9V places the collector at 5.5V.

The collector-to-emitter voltage Vce = Vc - Ve. It's vital that Vce be at least a few volts so that the transistor can operate properly. I've left some of the numerical calculations to you as it will be good practice.

Note: The currents through R1 and the 3rd R are not exactly the same, and the same goes for Ie and Ic. The (small) difference in each case is the base current Ib. But taking them as the same is a close enough approximation for most applications like this. So let's not go into that for now. Maybe later, if necessary.