1
\$\begingroup\$

This question refers to the book Make electronic's experiment 18.

I have included a picture below.

R7,R9,R10,R12: 1K
R8: 2K2
R11: 330K
C1: 100uF
C2: 68uF
C3,C4,C6: 0.1uF
C5: 10uF
S1,S2,S3: tactile switches
IC5,IC6,IC7: 555 timers
IC1: 4026 counter

quote from the book:

When the output from IC7 goes from positive to negative, it will trigger the reset of IC6, flipping its output negative, which allow the count to begin
The capacitor C4 communicates the sudden change from positive to negative, but the reset of the time it blocks the steady voltage from IC7 so that it won't interfere with IC6

My question: How does IC6's pin 4 get negative voltage or grounded? I thought C4 would only accumulates positive current from IC7's pin 3 and release it to IC6's pin 4.
I thought the capacitor should be hold fixed to connect one pin to positive, and another to negative. From what i can see the IC7's pin 3 is switching from positive current and then to ground on one of the pin of C4.

Thank you

enter image description here

\$\endgroup\$
5
\$\begingroup\$

How does IC6's pin 4 get negative voltage or grounded?

enter image description here

If pin 3 is high (+5V) and C4 is discharged, both sides of C4 are high (+5V), The right hand side of C4 is held high by R10. The left hand side is held high by pin 3. There is 0V across C4.

If pin 3 goes low (-5V), C4 cannot instantaneously charge from 0V to 10V so taking it's left hand side down to -5V takes the right hand side to -5V. This -5V is seen at pin4 on IC6.

R10 then slowly charges C4 bringing the voltage at it's right hand side towards +5V.

enter image description here

The circuit would function the same if I arbitrarily changed the labels +5V and -5V to, for example, +10V and 0V (or GND). In which case you might or might not want to replace the word "negative" with something else.

Do you mean if one pin of C4 is grounded, then the other pin of C4 would suck current from this pin as well?

No, See the "Differentiator" example in this tutorial

enter image description here

No current needs to flow through or across the capacitor because the voltage across it is initially unchanged. If the voltage with respect to a fixed reference point (ground) at the left hand is changed, the same change will be seen at the right hand side. No current need flow. Subsequently there are changes but these involve external flows of current.

\$\endgroup\$
  • \$\begingroup\$ Hi RGBrick i am sorry for the late reply. i was really busy for the past week. Thank you for your great effort for helping me. But i still have some question about it. "If pin 3 goes low (-5V), C4 cannot instantaneously charge from 0V to 10V so taking it's left hand side down to -5V takes the right hand side to -5V. This -5V is seen at pin4 on IC6." Do you mean if one pin of C4 is grounded, then the other pin of C4 would suck current from this pin as well? am i missing some basic understanding of capacitor? Again Thank you \$\endgroup\$ – mathwannabe Jan 30 '14 at 11:47
  • \$\begingroup\$ @matwannabe: See new material at end of answer. You may find it useful to follow the tutorial material referenced and to do simple experiments with a battery, capacitor, resistor and oscilloscope (or large value resistors to slow down changes and view using logger or multimeter(s)) \$\endgroup\$ – RedGrittyBrick Jan 30 '14 at 12:11
  • \$\begingroup\$ Excellent explanation, thanks! I saw such capacitors between IC1:pin3 and IC2:pin2 in a number of schematics (eg TI's NE555 datasheet) for daisy-changing 555s (end of one 555 triggers next 555), but I couldn't figure out how that was actually working. \$\endgroup\$ – Duoran Jun 15 '14 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.