Square wave generator won't work

Question

I'm trying to generate a square wave at 21kHz (+13V,0V) using this schematic:

I used formulas and this website to find the correct values for my components but it doesn't seem to work on Simetrix.

Do you see any problem?

Answer 1

Your positive feedback attempts to move the non-inverting input above and below ground. This would work with a dual supply system - ie negative rail below ground.

In a single supply system you need to provide hysteresis above and below some intermediate voltage - this may be Vcc/2 or some other value to allow for eg the fact that input common mode range is typically ground to Vcc-1.5V.

As Jasen notes - the resistor values are on the small side.
The LM2904 will source and sink 10+ mA but it is best to limit I/O current to usefully less than I_out_max.

In this case try eg at the non-inverting input

• Remove R2

• Add 15k to V+

• Add 10k to ground.

• Change R1 to 10K.

• Increase R3 and decrease C1 - or reduce the hysteresis range (smaller R1) to increase frequency.

That should provide an OK starting point for playing.

LM2904 data sheet

Answer 2

Let's first look at the output divider pair, \$R_1\$ and \$R_2\$, you provided. This takes the opamp output and divides it in half. (We are assuming you aren't over-loading the output's current compliance for the moment.) If you assume the opamp is a perfect rail-to-rail output then the output may either be \$V_\text{CC}\$ or else \$0\:\text{V}\$. So that means that \$0\:\text{V} \le V_{\text{IN}_+}\le \frac{V_\text{CC}}{2}\$. But clearly, given \$R_3\$ and \$C_1\$, it follows that under all circumstances \$0\:\text{V} \le V_{\text{IN}_-}\le V_\text{CC}\$.

Note the fact that both ranges have \$0\:\text{V}\$, inclusive? Not so good.

What you want is that the range for \$V_{\text{IN}_-}\$ to completely envelope, clearly and unambiguously, the range for \$V_{\text{IN}_+}\$. Let's leave the basic concept presented by \$R_3\$ and \$C_1\$, so that we continue to allow \$0\:\text{V} \le V_{\text{IN}_-}\le V_\text{CC}\$. But we must now restrict the range of \$V_{\text{IN}_+}\$.

It's really convenient to set a range such that \$\frac{1}{3}V_\text{CC} \le V_{\text{IN}_+}\le \frac{2}{3}V_\text{CC}\$. Then you may keep your resistor values all the same; for both \$R_1\$ and \$R_2\$, as well as a new one I'm adding to your circuit. (But I'd recommend increasing their values, somewhat -- perhaps to \$10\:\text{k}\Omega\$ to lighten the load on the opamp's output.) And then simply add one resistor from \$V_{\text{IN}_+}\$ to \$V_\text{CC}\$, using the exact same value for it that you use for \$R_1\$ and \$R_2\$.

If you sit down with the simple RC exponential decay equation and work out the timing required to go from \$\frac{1}{3}V_\text{CC}\$ to \$\frac{2}{3}V_\text{CC}\$ you will have half of the total cycle time. And you can use that equation to solve for a value for \$R_3\$ given some value for \$C_1\$, pretty easily. In that way, you can achieve your \$21\:\text{kHz}\$ value.

You should be able to write that equation very quickly in this fashion:

$$\frac13 V_\text{CC}+\left(V_\text{CC}-\frac13 V_\text{CC}\right)\cdot e^{^-\frac{t}{R_3\:C_1}}=\frac23 V_\text{CC}$$

That reads as, "The capacitor voltage starting at \$t=0\$ is one-third of \$V_\text{CC}\$. To that, we add the charging rate which is driven by the difference between \$V_\text{CC}\$ and the starting point voltage, that difference then times the exponential decay rate over time. We want this result to reach two-thirds of \$V_\text{CC}\$." Or something like that, anyway.

You can work out that half the time is \$t=\frac1{2\:f}\$ where \$f=21\:\text{kHz}\$. Just drop in your value for \$C_1\$ and you should be able to compute a value for \$R_3\$.

It's really no more complex than that.