1
\$\begingroup\$

I'm designing a circuit consisting mainly of 74LS series logic. I am clamping and buffering several input signals using a zener diode and a 74HC7014 Schmitt trigger buffer. A relevant excerpt of the schematic is below, for the simplest case of the problem I'm seeing. (The rails on the 74HC7014 and the 4066 are T-GND and T-VCC. T1-7 and B1-7 are connected. The 2N3904 part is just a discrete inverter.)

When I construct precisely this circuit on the breadboard using through-hole components, everything works as expected. When DIVSW is closed, points A, B, and C all measure pretty close to T-VCC.

However, on the PCB I had manufactured with SMT components, when DIVSW is closed, point A measures at T-VCC as expected. However, point B measures approximately 2.8V, and point C measures 0V. I understand neither why the buffer input would be at 2.8V in the first place, nor why the buffer output would be at 0V given that fact (74HC7014 high trigger level is 0.65V). Edit: that should read 0.65 x VCC, which does make sense.

I can't see any significant differences between the through-hole components and their SMT counterparts, and all the traces appear to be correctly connected. I am using the same power supply in both cases. I also tested this on two separate SMT PCB prototypes with similar results (one has since been hacked up in an attempt to diagnose).

I am stumped as to what could possibly be going on. Any suggestions on where I might start looking? Happy to provide any additional details needed.

schematic

\$\endgroup\$
6
  • \$\begingroup\$ Use a multimeter to ensure that both gnd and Vdd are at the appropriate voltages. \$\endgroup\$ Commented Jul 17, 2015 at 0:39
  • \$\begingroup\$ T-GND and T-VCC, along with the corresponding power rails on the ICs, show 0V and 4.98V respectively. \$\endgroup\$
    – ezod
    Commented Jul 17, 2015 at 0:55
  • 1
    \$\begingroup\$ For Vcc = 5 V, the datasheet shows the Vt+ (positive-going threshold voltage) as 3.1 volts, and the negative-going threshold as 2.9 volts, so your 2.8 V should be taken as a low. \$\endgroup\$ Commented Jul 17, 2015 at 1:08
  • 1
    \$\begingroup\$ Please note that 74LSxx is bipolar TTL logic, and 74HC is CMOS logic. The two logic families have different logic thresholds, and care must be taken when combining the two logic families in a common circuit. 74HCT or 74ACT are CMOS logic with input thresholds adjusted to work correctly with the bipolar 74LS (and other non-C) parts. \$\endgroup\$ Commented Jul 17, 2015 at 1:18
  • 1
    \$\begingroup\$ Another thing to consider is that using a zener with a 100k input resistor can lead to all sorts of unexpected results. Your input current is on the order of the zener's leakage currents. \$\endgroup\$ Commented Jul 17, 2015 at 1:39

2 Answers 2

3
\$\begingroup\$

This behavior is expected based on your component selection. All zener diodes leak, but typically that leakage current is negligible. Since you measured 2.8V at point B, then the resistor is experiencing a 2.2V drop. The leakage through the diode is getting its current from node B1-7, so it's passing through the 100k resistor first, creating the 2.2V drop. Let's use Ohm's Law to determine the apparent leakage current: $$I=\frac{V}{R}=\frac{2.2V}{100k\Omega}=22{\mu}A$$ So the zener diode appears to be leaking 22uA. Totally believable. Essentially, the resistor and diode are together acting like a voltage divider.

The reason this happened was because your resistor value is very large. When the current in the circuit is in the same order of magnitude as leakage currents, strange behaviors like this will appear. Suddenly those negligible currents aren't so negligible compared to the signal current.

To avoid this problem, either choose a zener diode with a much lower leakage current (the hard way) or choose a smaller valued resistor (the easy way). This problem would go away if you used, say, a 1k resistor instead (i.e., 22uA through a 1k resistor would drop 22mV instead of 2.2V).

\$\endgroup\$
2
\$\begingroup\$

The positive going threshold for the 74HC014 is typically 3.1V at 5V VCC, not 0.65v. That is above the voltage you measured so the output could be zero (what you measured).

It sounds as if the zener is leaking. What voltage is it? Low voltage zener are notorious for having high leakage. I would suggest changing the 100K resistor to something like 1k.

\$\endgroup\$
3
  • \$\begingroup\$ It's a 1N4733A, 5.1V zener. The 100K resistor is there to present high impedance to the inputs. Since the 74HC7014 inputs already include clamping diodes, maybe I don't need the zeners at all? (Raw inputs can theoretically swing between -12V and +12V.) \$\endgroup\$
    – ezod
    Commented Jul 17, 2015 at 1:17
  • 1
    \$\begingroup\$ Better yet, leave the 100k resistor alone and just yank the Zener diode. \$\endgroup\$ Commented Jul 17, 2015 at 1:26
  • 1
    \$\begingroup\$ I agree - get rid of the zener. \$\endgroup\$ Commented Jul 17, 2015 at 2:16

Your Answer

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

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