Discrete Pulse width Modulator...

[Rajagopal87]

Can anyone help me design a pulse width modulator using discrete components without using an IC. I have found one circuit in the web and I cannot understand how it works.

**broken link removed**

I guess a square wave is produced in the capacitor(am I right?) if its value is low enough. what I cant understand is the square wave is fed into one input of differential pair and the signal ( inc or dec width) is fed into another input, Will it not cause a change in frequency instead of the width in the output terminal?
Please explain in detail.

Thank You.

 

[dknguyen]

I'm still trying to figure out that circuit myself. Since you already tried looking at the circuit, I will try looking at it too and say what I think about it (but remember, I could be wrong)

Something about what you said first: A square wave is not fed into the input. A DC voltage is fed into the input and the PWM comes out of the output. The duty cycle depends on the DC voltage that was inputted (or at least that's what I think it going on.)

From reading what the author has said:

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PARTS OF THE CIRCUIT:
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The top 3 transistors are biased to be current mirrors, so they are current sources and make it so the same current is always running through their collector/emitter and everything in series with their collector/emitter. Because the differential pair is in series with a current mirror, the total current flowing through them is always the same. When the input voltage or current is higher, it makes more of the total current in differential pair flow through the left transistor, and less current flow through the right transistor. If you increase the input current to the left transistor, you need to increase base current on right transistor to make same fraction of total differential pair current flow through the right transistor as before.

The capacitor charges up, and when it's voltage get's high enough, it turns on the two-transistor circuit on the right side of circuit. And this causes the capacitor to connect to bottom-left 100ohm resistor and discharge through the resistor, causing capacitor voltage to drop. Capacitor voltage is connected to right transistor of differential pair. Remember all this,

The output and the output transistor are open collector. So you have to attach a pull-up resistor to the output to make it be able to go high. The output transistor pull the output to ground when enough current is flowing into the base. This resistor is not drawn in circuit schematic.

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ENTIRE CIRCUIT WORKING:
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When the two transistors right side of the circuit are off (because capacitor voltage is too low), the capacitor is disconnected from the resistor and charges up. WHen capacitor voltage is high enough, the current from capacitor goes to base of right transistor of the differential pair. This makes right transistor of differential pair conduct more current to the base of output transistor.When there is enough current flowing into base of output transistor, the output transistor will turn on and pull output to ground (the rest of the time it will be high because of the pull-up resistor you have to put on output). When capacitor voltage gets too high, it turns on the two transistors in right part of the circuit and this connects the capacitor to the 100ohm resistor on the bottom right so the capacitor will discharge. As the capacitor discharges, the capacitor voltage drops and this makes current at the base of right transistor of differential pair drop. When this happens, less current will flow through the right differential pair transistor to the base of the output transistor. When the current drops low enough, the transistor will turn off, no longer pull output to ground, and the pull-up resistor on output will make the output high again. Also, when capacitor voltage drops low enough, the two transistors on right of circuit will turn off and disconnect the capacitor from 100ohm resistor on bottom right. This will make the capacitor start to charge up again. This repeats over and over with output going high and low to make the PWM.

Remember that the total current flowing through transistor differential pair is always the same because of current mirror. When you apply higher voltage to input, it makes larger portion of this total current flow down left side of differential pair, and you need more base current on the right differential transistor to have same proportion of total differential pair current flowing on right side of differential pair. This means that when the capacitor voltage turns on the right transistor in differential pair, less current will flow to output transistor. This will make output transistor take longer to turn on and make it so that it takes longer to pull output to ground. This will make output spend less time low, and spend more time high.

From what I said, I think in this circuit, if you change the duty cycle the frequency also changes...but I could be wrong. I hope this helped you (you put some effort into your problem before asking the question).

 

[Rajagopal87]

Something about what you said first: A square wave is not fed into the input. A DC voltage is fed into the input and the PWM comes out of the output. The duty cycle depends on the DC voltage that was inputted (or at least that's what I think it going on.)

I think you have misunderstood what I said. Sorry for the confusing words. What I meant to say was that the square way at one i/p of the diff pair and dc voltage in the other as you said. However I am still not sure if its a square wave in the capacitors terminal.

Thank You for the reply.

 

[Rajagopal87]

Sorry. I posted reply before reading the full post. Yes now I can get a clear idea .
Again Thank you for your reply.

 

[dknguyen]

I edited post while you were reading to make things more clear. Please read it again. The capacitor voltage is not exactly a square wave. It is voltage of capacitor charging up and discharging over and over. It is more like a triangle wave I believe.

It is a smart circuit, hehe.

 

[Russlk]

Making a current mirror with discrete transistors does not work very well unless the transistors are matched.

 

[dknguyen]

Making a current mirror with discrete transistors does not work very well unless the transistors are matched.

Yeah, same thing for differential input transistors too. Make sure all the current mirror transistors are on the same IC. And make sure the differential transistors are both on the same IC.

 

[Overclocked]

It seems easier to just use a 555 Timer.

 

[dknguyen]

But after an apocalypse with mountains of BJTs everwhere and no 555s in sight, however will you build a PWM circuit?!