Thread Starter


Joined Feb 26, 2010
Hi all,
hope all ok, I'm Just started on this new VCO project.. and stuck as thick as it seems on how this Control Pot is wired to the Chip in the up-loaded picture..

Thanks for any help at all with it..




Joined Feb 24, 2006
At one end of its travel the pot behaves like a fixed 100K resistor. At the other end of its travel the 100K resistance is in parallel with the short circuit effectively removing the pot from the circuit. At points in between it looks like a fixed resistor in series with a short, which is the same as the fixed resistor.

Does that help?

Edit: It's hard to tell if the pot, R2 is 100K or 200K


Joined Mar 12, 2010
as for how it's wired into the circuit, the middle terminal on the pot is connected to pin 5 of the timer chip. One of the outside terminals is connected to ground and the other outside pin is connected to the 5 volt supply. If the circuit acts like the way you turn the shaft of the pot seems backwards, reverse where the 2 outside terminals are connected.

Thread Starter


Joined Feb 26, 2010
Hi bychon,
cheers for that, on Looking at the circuit, does it look any good for a VCO, reason for asking is that Ive never built this one....

Thread Starter


Joined Feb 26, 2010
Hi Bill,
well I never Just shows you dont it.. I would of burnt out my chip.. by the way bill Ive found a circuit diagram in my 130 in one kit thing, its a sawtooth generator as you mentioned to do..

/\/\/\/\/\/\< a rough sawtooth.. here's what the right up says,
The shape of this wave comes from the slow charging of the 0.1uf capacitor, through the control pot which is 50k) and the 100k resistor, and the capacitor's quick discharge through the PNP and NPN transistors.
I know you cant see the diagram.
the voltage divider- the 470ohm and the 100 ohm resisitors provides about 1.6volts to the transistors. The current flowing into the 0.1uf capacitor from the 9volt supply (through the control and the 100k res, causes the charge of the capaciotr to slowly increase.

When the charge of the capacitor exceeds the voltage of the voltage divider (1.6v) the transistors turn on and provide a path for the 0.1uf capacitor to discharge quickly. now the transistors turn off again, and the capacitor begins to slowly charge to repeat the Cycle..
hope this makes sense??

thanks bill.
I fell really stupid asking about a pot.. When I've done really hard equations on scentific calculator!!! dont make sense.. does it.. + Ive devised my own circuits.. but then some things just get in my way..
sorry that is a bit long for a short reply bill.


Joined Mar 24, 2008
I've started a project of tested circuits in a cookbook. It is still under development, but here is where I'm at...

Bill's Index

My Cookbook

Volume 6 of the AAC book has a bunch of experiments. I've released a bunch of 555 experiments there recently.


Joined Jul 17, 2007
You meant to say pin 7 instead of pin 5, I'm certain.

Pin 5 is the CTRL pin. It will not hurt the IC if CTRL is shorted to either Vcc or GND. However, the trigger/threshold inputs may not function as desired until the short is removed.

Pin 7 must never be connected directly to +V. Doing so will result in very high power dissipation in the IC, which will likely destroy it. I suggest that it is a good idea, wherever practical, to limit the current sink via pin 7 to 10mA or less.

To keep things really simple, always ensure that there is a resistor valued at least 100 Ohms per volt of supply between +V and pin 7.

For example, let's say that you were powering a 555 timer circuit from a 9v "transistor" battery, and you want to make certain that you are limiting the current between +9v and the 555 pin 7 sufficiently so that you don't ruin it.

Ohm's Law says: I=E/R, or Current (Amperes) = Voltage / Resistance (Ohms).
Expressed another way, R=E/I, or Resistance = Voltage / Current.

So, if you want a maximum of 10mA (0.01 Amperes) into pin 7, what resistance must you place between them?

Since the battery is 9v, and 10mA is the maximum, you calculate:
9v/10mA = 9/0.01 = 900 Ohms. That is the MINIMUM resistance you should use.

So, as long as you have 900 Ohms or greater of fixed resistance between the +9v supply and pin 7 of the 555 timer, you will not risk destroying that input, and perhaps the timer itself.

This is not a "cure-all" for 555 timer problems, but it is a good start.
Last edited:


Joined Mar 12, 2010
Oh yeah, Marsden is right. Pin 7 will smoke if R2 becomes zero. As for whether it will work as a VCO...What book is that, that you don't trust the labels?

ps, yes, it's a VCO.


Joined Jul 17, 2007
The schematic is the work of Forrest M. Mims III, a citizen-scientist who I think a great deal of.

It was published in an Engineer's Mini-Notebook titled "Timer, Op Amp & Optoelectronic Circuits & Projects" for Radio Shack in the year 2000. Sadly, Radio Shack no longer carries Forrest's books, but you can order them online.

Forrest's home page:
Publications list page:

Where you can order his books:

The Engineers' Mini-Notebook series:

Volume I and Volume IV are must-have books; lots of very handy reference material.

Forrests' books are very popular, and a number of them have been in print for many years. I have an "Engineer's Notebook" written by him that I purchased in 1981; it was a 4th printing edition.

Forrest's approach was to keep things as simple as possible, so that a newcomer to electronics could actually get some things working without having a lot of complexity involved. If you look for flaws, they are not hard to spot - however, the vast majority of the circuits are at least functional and won't blow up in your face.

This particular circuit has a couple of problems; the R2 being reduced to zero Ohms is one of them.

The other problem is the pot used for the control voltage. That pot is shown as 100k. Unfortunately, it will be very non-linear when connected to a standard 555 timer. The pot should be reduced to somewhere between 5k and 1k, depending upon the supply voltage.

Internally, a standard bjt 555 timer has three 5k (nominal) resistors connected in series to form a voltage divider; the upper side of the divider is 2/3 Vcc, the lower is 1/3 Vcc. These two levels are connected internally to the trigger and threshold comparators.

Pin 5, the CTRL pin, is connected to the 2/3 divider point. If you increase the voltage on that pin, you widen the internal points for trigger and threshold. In a typical astable configuration, the output frequency will decrease. If the voltage on pin 5 is decreased, the output frequency (when configured as an astable multivibrator) will increase.

Clear as mud?

Thread Starter


Joined Feb 26, 2010
Hi sgtwookie,
after close examination of that VCO, well its.. not that good a circuit, I know its a basic one.. so Im dropping that, and working now on a Ultrabright Led torch.. of very bright White Led's and make them rotate like a Star-trek phasor ...

thanks on all your help & other fellow members on here..
The Led's spinning around in a circle Matrix..!!