I am currently reading a book titled “Make: Electronics”. This is an electronics book whose target audience is supposed to be complete beginners.

One of the book exercises (Experiment 18) includes a resistor capacitor connection that is giving me some headaches. A picture is worth a thousand words so if you would, please take a moment to check out the partial circuit diagram below paying special attention to the connection between IC7 pin 3 and IC6 pin 4.

enter image description here

EDIT: Per comments request, IC7 and IC6 are 555 timers. Here is the pin-out for the 555 timer (sorry, I can't believe I totally missed calling out the ICs types):

enter image description here

As you can see from the picture above, the IC7 chip is configured as a monostable timer and the IC6 chip is configures as a bistable timer. The connection between IC7 pin 3 and IC6 pin 4 has a capacitor C4 in between. There is also a connection comming out from IC6 pin 4 to a resistor R10 all combined to form what appears to be your typical RC connection.

The gist of how the circuit is supposed to work is like this: You press the S4 push button, this makes IC7 pin 3 to go high and after a couple of seconds it goes low. The transition between high and low on IC7 pin 3 is supposed to send a signal to IC6 pin 4 that should reset it (vial pin 3 going low).

In case it helps, below is a link to a video that someone made describing this experiment. On the video, the chip all the way to the left is IC7, and the next one to the right is IC6.


Unfortunately, no matter what I do this is not working for me because pin 4 on IC6 never triggers no matter what size capacitor or resistor I use. I would like to get this to work, but before I can attempt to fix this, I need to understand what exactly is supposed to be going on here and this is where I need your help.

So far, what I think is supposed to be happening is the following:

EDIT: If anyone is still reading this post, what I am looking for (as the answer to this question) is for someone to verify whether the four steps listed below accurately describe what is going on in the circuit. If yes then please let me know, if no, then if you don't mind, please let me know were did I go wrong.

  1. When the circuit is first powered on, pin 3 in IC7 is low so capacitors C4 begins to charge via resistor R10.
  2. After some time, C4 charges up and current stops flowing. At this point, pin 4 on IC6 become positive.
  3. When you press S4, pin 3 on IC7 becomes high. This causes the C4 capacitor to discharge via R10 until both sides are equally positive. Here again, pin 4 on IC6 remains positive.
  4. Eventually, IC7 (the monostalbe timer) changes state and its pin 3 goes low and this is where things get interesting (I think). When this happens, the impedance on C4 momentarily drops down to zero making C4 momentarily appear to the circuit as a direct cable connection. It is at this very precise moment that pin 4 on IC6 should be negative and pin 4 should trigger.

Assuming that the logic stated above is sound, the process does not work for me. I think the problem may be that in step 4, pin 3 on IC7 only reaches about 7.7 volts when the pin is high (the source voltage of the circuit is 9 volts by the way). This I think is a problem because step 4 expects pin 3 on IC7 to ideally be 9 volts. If the voltage I lower than 9 volts the transition will not generate a low enough voltage to trigger pin 4 on IC7.

Sorry for the long post. Any help is appriciated.


  • \$\begingroup\$ It would help to know what "IC7" and "IC6" are. A good schematic would say, as well as label the pins according to their function so the schematic can be read without constantly referencing the datasheet. \$\endgroup\$
    – Phil Frost
    Jun 23, 2015 at 17:50
  • \$\begingroup\$ I think these are standard 555's (IC6 and IC7), so pin 3 is output and pin 4 is reset. C4 essentially acts as a time delay for IC6 to start. If it never works, no matter what size R and C you use, have you checked to make sure there is output on IC7 pin 3 when you press the start button? Measure it without anything connected to the pin. \$\endgroup\$ Jun 23, 2015 at 17:59
  • 3
    \$\begingroup\$ With power and ground going every which way, it's really hard to see what's going on in that diagram. If that came from the book, it's doing you a major disservice. Yes, it looks sort of like how you might wire the circuit on a breadboard, but that comes after understanding what the circuit does, which depends on having a readable schematic. \$\endgroup\$ Jun 23, 2015 at 18:06
  • 1
    \$\begingroup\$ It's not a matter of being able to follow the picture, but of learning good habits. A good circuit diagram has power at the top, ground at the bottom, and a flow from left to right. That makes it easier to understand the circuit, and much easier to communicate. If the book isn't teaching you that, it's a major omission. \$\endgroup\$ Jun 23, 2015 at 18:31
  • 1
    \$\begingroup\$ @T555 LOL, well obviously the circuit has been proven. So you have this put together on your breadboard, and you've double checked your connections? You could disconnect C4 from IC7 pin 3, and then quickly touch C4 (the disconnected end) to V+ and then to ground to mimic the pulse received from IC7 pin 3 and see if IC6 starts up (LED on). This might prove/disprove your voltage level theory. \$\endgroup\$ Jun 23, 2015 at 18:34

2 Answers 2


What if you put R10 to 10KOhm and inserts an 82KOhm instead of C4? The time between starting and the light switch is controlled by IC7, so the connection from IC7 pin 3 to IC6 pin 4 is only to turn on the light, there is probably used a 555 because it is easy, as IC6 really only works as a flip flop.


This is from a simulation. Hope it gives you an understanding of the signals.

0 = LED off

1 = LED on

S4 = enabled

S3 = enabled

Yellow = IC7 pin 3

Blue = IC6 pin 4

Red = IC6 pin 3

Green = IC7 pin 6/7

enter image description here

  • \$\begingroup\$ The inclusion of the C4 capacitor is part of the original circuit design and it is supposed to be there for a reason. Although I am sure the circuit could be changed around to make it work, I think the idea of the exercise is to first try to understand how things work as originally design and then later on try to improve on it. Could you let me know if the the 4 step listed on the original post sound correct to you? \$\endgroup\$
    – T555
    Jun 24, 2015 at 15:26
  • \$\begingroup\$ Quite obviously, but now it is a book about "tasks" and although errors often can appear in the diagrams, then it might just be to see if you learned something and understand the error. 1 and 2 charging takes place very fast, is a ceramic capacitor, guess 10 or 100 nf? 3 what you mention here also tells me that I was wrong when a resistor, instead of C4, would only work if pin 3 goes low. The shift from low to high and from high to low occurs during 100ns, but as C5 times more, so I think you should try to put the R10 to - rather than +? \$\endgroup\$
    – JustMe
    Jun 24, 2015 at 16:52
  • \$\begingroup\$ What volt lies on IC6 pin 4, both when IC7 pin 3 is low and when it is high? \$\endgroup\$
    – JustMe
    Jun 24, 2015 at 19:20
  • \$\begingroup\$ Just tried and tested your layout around IC6 and it works as it should. Mounted then the whole layout and it works as well. First time voltage is connected, the light is on, otherwise the LED is reversed, resets the S3 and goes off, and pressing the S4 turn it on as soon as pin 2 gets minus. So either faulty components or connections. \$\endgroup\$
    – JustMe
    Jun 25, 2015 at 12:40
  • \$\begingroup\$ Thanks for following up! I really appreciate your help. Out of curiosity, what is the max voltage that you get out of IC7 pin 3? I get 7.7 volts but I bet you get more than that (my guess minimum 8.3 volts). Also, out of curiosity, how low does pin 4 in IC6 has to go in order for the pin to trigger? My 555 timers have to go as low as .7 volts, perhaps yours need less than that? By the way, I am using the NE555P (Precision Timers) IC, are you using the same component? \$\endgroup\$
    – T555
    Jun 26, 2015 at 15:26

If the voltage does not reach the desired level it is likely that the time constant is too long. Reducing the size of the capacitor or the resistor or both so the capacitor reaches the proper voltage before any changes occur will probably work.

  • \$\begingroup\$ Per op, "Unfortunately, no matter what I do this is not working for me because pin 4 on IC6 never triggers no matter what size capacitor or resistor I use." Not know what sizes OP used, you could expand your answer to describe what possible values should be tried. \$\endgroup\$
    – Passerby
    Aug 6, 2015 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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