I have assembled a square wave generator following this schematic from an old book.


R1 and R2 are both 1 kohm, C1 is 68 pF, and C2 is 2.2 nF. The author of the book claims that this combination is suitable for working frequencies of 2 to 3 MHz. Capacitor C3 is to be selected during adjustments.

Instead of the К155ЛА3, I used a КР1533ЛА3, a later and improved version, differing in faster switching time and lower power consumption. Online shops and references give SN7400 and Texas Instruments SN74ALS00AN as analogues for those ICs. As for the structure, both of them contain four TTL two input NAND elements.

The assembled generator works, and the frequency is more or less stable 2 MHz, in accordance with the crystal resonator BQ1.

What I don't like at all is the output voltage going negative, as low as -1.2 V. Below are two plots from an oscilloscope, the second one showing the lower part of the signal zoomed in.

zoomed out plot zoomed in plot

I tried fiddling with resistor and capacitor values, like decreasing R1, R2 to 470 ohm or increasing them to 1.2 kohm, setting C1 to 100 pF and C2 to 10 nF (that was another combination given in the book, stated to suit 2 MHz exactly), or selecting C3 between tens of pF and 68 nF. All of that only slightly affected the rise and fall time, but the low and high levels remained nearly constant, about -1 V and 4.7 V respectively.

Can I get the lower output level to correspond to TTL levels, say, something between 0 and 0.4 V? If yes, what should be changed about the circuit?

UPD: I have soldered a 0.1 uF ceramic capacitor right to the IC's power pins. Below are two photos of my circuit from two points of view. Sorry if the setup is hard to understand, it is actually pretty tight.


The circuit is powered by a 7805C regulator (in the corner of the breadboard,) with two 20 uF capacitors connected between its input / output and its ground pin (middle one.) The input power (yellow and green wires) is regulated 9 V from a lab source. I also tried setting the source to 5 V and connecting it directly to the IC, and that didn't change the circuit's behavior.

I connected the oscilloscope's input probe to the point labeled 1, and the ground probe to the point 2, next to the short blue wire going to the ground bus. Before that, I used to connect the ground probe to the bus, at point 3 for example. The probes are short hard pins held by crocodile clamps, did not photo them because they cover most of the circuit when in place.


The output is now about -0.5 V at its worst, which is not perfect but much better than before. I also tried connecting a 3.3 kohm resistor (the first one available at hand) from the output (pin 8) to the ground bus (the row marked by 3) and connecting the probes across that resistor. The numbers I got were nearly the same, so the load must not affect the circuit much, I suppose.

UPD2: The overshoot stays nearly the same (-480 to -560 mV), no matter if I change the capacitor on the power pins from 0.1 to 0.47 uF or disconnect it altogether. So the positive effect was mainly from repositioning the probes.

UPD3: I remembered about having a probe like that available and tried it. probe It has obviously the shortest connectors there can be. Also, it acts as a 10X voltage divider. After the calibration I got the following signal shape for a 0.47 uF capacitor at the power pins.


I tried increasing the capacitor to 1.5 uF, but got nearly the same result. It seems also that the signal's shape changed depending on how exactly I placed the probe, and the least overshoot I have seen was about -120 mV.

UPD4: Before I experimented without that RC couple between DD1.2 and DD1.3. Now I have added it, and the circuit behaved a bit better. I also tried adding a 220 ohms or 470 ohms resistor parallel to C2, and a similar one between IC's pin 1 and the quartz, as suggested by other similar schematics. Anyway, the best value I have seen was some -40 mV.

What I am thinking about is couldn't I influence the operating point of one or more logic elements, so that the signal "shifts" up?

The final update: The idea to separate DD1.3 from the first two elements worked well, thanks once more. I also added some negative feedback to DD1.3, since there was none. So the circuit is looking like this now. C5 is situated near the voltage regulator, and C6 at the IC's power pins.


simulate this circuit – Schematic created using CircuitLab

And this is the output signal. The minimum value (200 mV) was measured by those two downward "teeth", and the value at the horizontal cursor is about 320 mV.

final plot

Of course, the "high" level also got decreased because of the feedback, but I hope 3.2 V is pretty enough for TTL. Maybe after I reassemble the circuit using soldering it will need some more tuning, but now I know at least what to change.

  • 3
    \$\begingroup\$ What you're seeing is usually called "overshoot", and correcting it first requires that we ensure it exists at all--how are you probing the circuit? It's possible that poor probing technique or an improperly compensated scope probe can cause this problem with fast enough fall time. But it's also quite possible this is a real effect happening. It's also likely, however, that it's not a problem given that you're using TTL, which is relatively robust and probably doesn't care if the input spikes slightly negative for a few nanoseconds. \$\endgroup\$
    – Hearth
    Commented Jul 8, 2022 at 16:21
  • 3
    \$\begingroup\$ Try adding 0.1uF poly or ceramic caps across each gate's power pins. \$\endgroup\$
    – rdtsc
    Commented Jul 8, 2022 at 16:25
  • 2
    \$\begingroup\$ Try setting up your oscilloscope probe correctly. \$\endgroup\$
    – Andy aka
    Commented Jul 8, 2022 at 17:02
  • 1
    \$\begingroup\$ 7905 sounds bad.Your oscillator circuit ground ground is floating. There can be a capacitive path through your oscilloscope and power supplies which spoil your regulation.. Get a 7805. \$\endgroup\$
    – user136077
    Commented Jul 14, 2022 at 8:15
  • 1
    \$\begingroup\$ @user287001 sorry to have misinformed you. It is actually a 7805C. I knew I was wrong somewhere, since the out pin of the regulator is connected to IC's pin 14, and that's +5 V, not the other way round. \$\endgroup\$ Commented Jul 15, 2022 at 11:18

1 Answer 1


Add a ceramic capacitor C4, between 0.1uF to 1uF, placed directly between pins 7 and 14 under the chip (on the bottom of the circuit board). Or solder it on top of the chip. Trim the pins of discrete parts to make them as short as possible.

Connect R4 directly between pin 8 and pin 7. Again, make the pins short. Connect the scope probe directly across R4.

I've assembled the circuit using a 74F00 I had around, and a 12MHz crystal since I couldn't find anything less. I've used RC coupling from the oscillator to the buffer gate NAND3. This is probably unnecessary, but I wanted to reduce DC loading.

The board also includes the 78L05 regulator, running from 9V.


simulate this circuit – Schematic created using CircuitLab

This should improve signal integrity.

The oscilloscope trace of output from across R4

The top view of the assembled circuit

The solder side view of the assembled circuit

  • \$\begingroup\$ Thank you for your answer. I will provide a photo in a day or two, as soon as I get back to this circuit. For now, it is assembled in a breadboard, and if that matters, then I will reassemble it using soldering. Also, I did not use that load resistor (R4), so maybe it is really my measurement that was wrong. \$\endgroup\$ Commented Jul 9, 2022 at 18:17
  • \$\begingroup\$ I can put it together on a solderless breadboard and I almost guarantee the output will not look much worse – if the layout is very tight, with shortest wire jumpers possible. The soldered version is roughly the same size a breadboard version would be. The resistor helps with signal integrity but mostly it helps you connect directly to two pins (7, 8) without going through “ground buses” and such. \$\endgroup\$ Commented Jul 11, 2022 at 14:01

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