555 CMOS timer with potentiometer

Question

I'm trying to create an output frequency using a potentiometer and a 555 CMOS timer that can be adjusted from 200 Hz to 2 kHz. I only have a 25 kΩ and a 50 kΩ potentiometer.

Using the 25 kΩ potentiometer I've calculated R1 as follows:

f = 1.44/((R1 + 2·R2)·C)

R2 = 25 kΩ potentiometer

C = 0.01 μF

2 kHz = 1.44/((R1 + 2·25 kΩ)·0.01 μF

R1 = 22 kΩ.

However, I'm not sure this is correct. I've created the circuit in Multisim:

But changing the percentage on the potentiometer has little to no effect on the output frequency, which for some reason is around 2.5 kHz.

Answer 1

I'm glad GodJihyo worked with you to produce something that makes a lot of sense. Your own comments to that answer added some important details:

• You are using a toggle-flop at the 555 output. That relieves some ambiguity with respect to the output you require. By definition we now know that you want (or can accept) a 50% duty cycle.
• You are using CMOS.

This last detail made me want to revisit the CMOS variety. So I looked here at the LMC555. In skimming over it, I immediately saw their very simple circuit that doesn't even use the discharge pin and the approximate equation they supply for it. (It's something I've read about before, and seemed obvious at the time, but seeing it again made me want to write about it.)

The circuit would be something like this:

^{simulate this circuit – Schematic created using CircuitLab}

Here, they simply let the output do all the work. So still fewer parts.

You already know the two values you have for \$P\$. (You can use either.) But you need to find \$R\$ and \$C\$ given your frequency span. And this is pretty easy to do.

var('r c p fl fh')
eq1 = Eq( 1/1.4/c/(r+p), fl )
eq2 = Eq( 1/1.4/c/r, fh )
ans = solve( [ eq1, eq2 ], [ r, c ] )[0]
for i in ans: i.subs( { fl:200, fh:2000, p:25e3 } )
  2777.77777777778
  1.28571428571429e-7
for i in ans: i.subs( { fl:200, fh:2000, p:50e3 } )
  5555.55555555556
  6.42857142857143e-8

Suppose you wanted to gain a little bit of extra range, just to be sure:

for i in ans: i.subs( { fl:180, fh:2200, p:25e3 } )
  2227.72277227723
  1.45743145743146e-7
for i in ans: i.subs( { fl:180, fh:2200, p:50e3 } )
  4455.44554455446
  7.28715728715729e-8

By skimming the results above, you can see that you increase the range by increasing the capacitor value (\$C\$) and at the same time also decreasing the series resistor value (\$R\$.)

Another way of saying one part of this is to recognize that \$\tau = R\cdot C=\frac1{1.4\,\cdot\, f_{_\text{H}}}\$. Or, put another way, the potentiometer value has nothing to do with setting the high-end. That's entirely due to the product of \$R\$ and \$C\$.

However, the potentiometer value does impact the values of both \$R\$ and \$C\$ in just such a way that the dual affect cancels out when you multiply them both together. If you first set \$\alpha=\frac{f_{_\text{H}}}{f_{_\text{L}}}-1\$, the individual equations for each are: \$R=\frac{P}{\alpha}\$ and \$C=\frac{1}{1.4\,\cdot\, f_{_\text{H}}}\cdot\frac{\alpha}{P}\$.

The \$P=25\:\text{k}\Omega\$ looks nicer to me as it places the capacitor in a range of values I find 'comfortable.' In fact, the value of \$150\:\text{nF}\$ is readily available, too. So I might try making \$R=2.2\:\text{k}\Omega\$ to go along with that:

^{simulate this circuit}

An LTspice run looks like this:

The results are that the blue trace is slightly less than \$180\:\text{Hz}\$ and the green trace is just slightly less than \$2.2\:\text{kHz}\$. About as hoped.

I don't think the extra drag on the 555 timer output is a problem here and there are fewer parts and the mathematics is easy, too.

Answer 2

That circuit is for variable duty cycle. The frequency will stay roughly the same but the ratio of high time to low time will change.

Look for a variable frequency circuit, here's one you could try that supposedly lets you switch between constant frequency and constant duty cycle. I haven't tried it myself so try it in your simulator and see what it does.

You could use a simple astable multivibrator circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

This should go from a little below 200 Hz to over 2000 Hz so it will divide down to the range you want. Here's a 555 timer calculator you can use to find component values if you want to tweak it a bit.