555 PWM circuit questions

[FMICW]

I am designing a circuit to pulse width modulate speech through a circuit consisting of an Astable 555 and a Monostable 555.

The modulating signal input is fed through a capacitor into Pin 5 of the monostable 555.
Now, my question is regarding the capacitor on the input to PIN5.

What does this value control?

I have read that the control input must be AC coupled to ensure this cap is charged.
I have also been told it may have something to do with the 600 ohm impedance of the signal generator.

Any tips here much appreciated.

 

[colin55]

Pin 5 of a 555 actually needs a VOLTAGE to change the mark-space output of the chip. The capacitor on pin5 is just to hold the pin at a voltage that is 66% of rail voltage as the chip has three 5k resistors in series and the join of the top resistor and the middle resistor is connected to this pin.
It "holds" the voltage on pin 5 and does not have any effect on this voltage.
But it does prevent spikes entering the chip and causing timing problems.
You will not achieve anything by modulating a low value capacitance on this pin and it will need a high value cap to produce any change in voltage.

The BC 108 transistor is connected as an emitter-follower. This is the same as saying COMMON COLLECTOR STAGE. Basically the emitter follows the base exactly, but the voltage on the emitter is about 0.65v lower than the base. The purpose of this type of stage is this: You can move the base up and down "with your little finger" and the emitter will move up and down with about 100 times more strength.

Now the value of the electrolytic on the base:
If you move the left side of the 10u electrolytic up and down, very slowly, with your finger, the electrolytic will start to charge during the "up cycle" and not all of your movement will be transferred to the base. The same applies when moving down. To make sure the base moves up and down as much as possible, the value of the electrolytic is made as large as possible. The 100k is there to pull the base up a small amount so that you can move the base up and down via the electrolytic. The actual amount of "set" produced by the 100k depends on the gain of the transistor and the value of the emitter resistor. You simply change the value of the 100k until the transistor sits at about mid-rail.
Can't be described any simpler.

 

[FMICW]

thanks a lot for your reply!

You will not achieve anything by modulating a low value capacitance on this pin and it will need a high value cap to produce any change in voltage.

Do you think the capacitor in the diagram is too small?

I have been led to beleive the capacitor prevents the output impedance of the generator having a bias over the timer... now i do not fully understand much of this, so that may be within your answer but i didn't spot it. Is this correct? Is this circuit direct coupled? And in what way does it bias it?

thanks again for a comprehensive reply!

 

[colin55]

No. The emitter-follower stage (BC108) is AC coupled at both the input and output.

The statement that the capacitor on pin 5 "prevents the output impedance of the generator having a bias over the timer... " is correct. But the size of the capacitor has to be large enough to produce a voltage-change on pin5. The amount of change on pin 5 will depend on the frequency of the signal. As I said before; if the frequency is low, the cap has time to charge and not all the amplitude is passed to pin 5.
As the frequency is increased, pin 5 sees more of the amplitude of the signal. Try your circuit and see how it performs.

 

[FMICW]

thanks, that is a big help.

the circuit certainly works, although i have not posted the whole thing for you to see.