Design of a Pulse width modulation circuit.

Designing a PWM unit with manual / electronic speed control.


I have been doing some reading up on how 'pulse width modulation' works, and now I feel like I am ready to design my own version of it using all discrete components.

Step 1. Design a 'astable multivibrator circuit' for symetrical (50%) duty cycle.

1,) VCC = 6v.
2.) Chose RC = 200 ohms
3.) Fo ~= 1Khz.
4.) Chose RB = 470 ohms.
5.)Chose control (manual) pot. ~= 5K ohms.
6.)therefor 1/2 of pot setting = 2.5K ohms.
so RBtotal for each transistor to run symetrically,
with eachother (50% duty cycle), means RBtotal = (2.5K + 470) ~= 3K ohms.
7.)Using this formula, [ (Fo = 1 / (1.38 x RBtotal x C) ]
8.) rearanging to solve for C then becomes [ (C = 1 / (1.38 x RBtotal x Fo) ] ~= 0.24uF.
7.) so using a standard electrolytic 0.22uF capacitor will work.

Now building the circuit and testing it, shows around 10% to 90% duty cycle, with a slight change in frequency. Drops from 1.5Khz to 1.25 Khz, at 50% duty.


Here is a video of an 'astable multivibrator working in PWM'

So, good pulse width regulation. With a Vout of around 2v. peek.

Using a LED at the output also helps square up the output from the original leading edge round over.

Now unhooking the capacitors, and checking voltage drops across the 200 ohm resistor, and the 470 ohm resitor, gave a collector current of ~ 29mA. and a base current of ~ 11mA.

So now I remove the manual control (5K ohm pot). and will now design a electronic control to vary the duty cycle of the waveform.

First I will use a PNP transistor and bias circuitry to replace the RB resistor (470 ohm), by using the collected values of base current from the test measurements.

(11mA.) = IB which now becomes( IC ) for this PNP transistor.
1.) I will choose 3v. to be across the transistor (VCE= 3v.)
2.) That calculates an emitter resistor (RE) to be [(VCC - VCE - Vbe) / IB]
[(6v. - 3v. - 0.7v.) / 11mA.)] = 200 ohms for (RE).
3.) Choose base to positive supply bleeder resistor to be (10 x RE) = 2K ohms.= RBA
4.) Voltage across (VRE) = (IC x RE) = (11mA. x 200 ohms) = 2.2v.
5.) (VRBA) is the Voltage across RBA = ( VCC - VRE - Vbe) = (6v. - 2.2v. - 0.7v.) = 3.1v.
6.)Divider current (ID) = (VRBA / RBA) = (3.1v / 2k ohms) ~= 1.5mA.
7.) Bias resistor (RBB) = [(VCC - VRBA) / ID] = [(6v. - 3.1v) / 1.5mA] ~=1.8K ohms. Choose 2K ohm resistor.
This gave a good 50% duty cycle.


Now added a pot for the input, so I changed the 2K's to 1K's and put a 2K pot trimmer across the base resistors.

Built the circuit and found out it works good, except for the duty cycle is less than before it starts at around 30% and goes to about 70%,


so now that needs to be worked on to get a higher percentage of change in the duty cycle.

First I will put back the 2K resistors and keep the base bias resitors in place, and switch out the two 200 ohm emitter resistors (RE), and swap them for one single potentiometer of 500 ohms.

Now connect the emitter of each PNP transistor to each side of the pot, and connect the center tap of the pot to the positive supply.

This now gives me a much greater change in duty cycle starting from 12% to 80% duty cycle much better output, across the LED, the LED dims very much than gets brighter.
But also the frequency changes too.


Now it's time to start designing it towards electronic feedback control.

So keeping the same configuration, using PNP drivers to run the multivibrator, with manual control at the emitters, I'll now remove the manual control from the emitters, and place two more PNP transistors into the circuit.

First I know that the PNP drivers along with there emitter resistors of 200 ohms, are acting like a 470 ohm resistor at the base of the oscillator (astable multivibrator).

So now moving farther up the chain, I will replace the 200 ohm emitter resistors with a PNP transistor whose collector is connected to the emitter of the PNP driver, and the emitter of this transitor connected directly to the positive supply.

First to get some stability as well as bring the base voltage up above 0.7v.

I'll place a 47 ohm resistor in the emitter lead, it was already established that this transistor needs to supply ~ 11mA. of current to it's PNP driver load,
so (11mA. x 47ohms) = 0.517v. dropped across the emitter resistor, add the Vbe of 0.7 v. and the total voltage drop from positive supply to the base = 1.217v.

I'll choose to make the bleeder base resistor, to be 10 x emitter resistor. which is 470 ohms.

So 1.217v. is across the bleeder resistor 470 ohms. called (VRB470)
The current in this resistor will be (1.217v. / 470 ohms)= 2.58mA. called (ID470)

Therefor the bias resistor for this PNP stage will be [(VCC - VRB470) / (ID470)] = [(6v. - 1.217v.) / 2.58mA.] ~= 1.8K ohms.



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