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So I want to trigger a monostable 555 to fire when a capacitor raises above 0.5 volts. My plan was to use a Zener diode to bias an NPN transistor that would trigger the 555. But I discovered the smallest zener voltage you can buy is 1.8 Volts.

Is there any way I can somehow bias the zener to react to a lower reverse voltage than its actual threshold?

-or-

Does someone have a better idea on how to get the 555 to trigger when intended?

ASIDE: I would also be satisfied with having the 555 trigger when the capacitor drops below 0.5 volts. I don't think it matters to me if it's going up or down, just so long as I can make the trigger at the stated voltage (very close to zero). The purpose is I'm trying to trip the 555 at some point where the cap is very close to zero volts.

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    \$\begingroup\$ Voltage drop of a normal (eg. 1N4148) forward biased diode is about 0.6 or 0.7 volts, maybe that is worth an experiment? A Schottkey diode (eg. BAT43) would be around 0.3V \$\endgroup\$ – jippie May 11 '13 at 9:04
  • \$\begingroup\$ Oh wow. Good call. LOL Just use a regular forward biased diode since it won't conduct strongly until the forward voltage is exceeded. So simple. I'm just used to thinking of regular diodes as polarity limiters. LOL Thanks! \$\endgroup\$ – JamesHoux May 11 '13 at 9:23
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The threshold triggering could be done using a rail-to-rail comparator, such as the Linear Technologies LT1711, using a 0.5 Volt reference voltage.

Depending on the precision required, the voltage reference can be obtained by using a sub-bandgap voltage reference IC such as the Analog Devices AD130 which has a 0.5 Volt output, or through a resistor voltage divider between the supply voltage and ground.

For a hybrid compromise, a Zener diode shunt reference of say 1.8 Volts can be used with a resistor voltage divider connected across it:

schematic

simulate this circuit – Schematic created using CircuitLab

The triggering slope (rising or falling) can be switched by interchanging the inputs to the comparator.

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Jippie pointed out in his comment this: Voltage drop of a normal (eg. 1N4148) forward biased diode is about 0.6 or 0.7 volts, maybe that is worth an experiment? A Schottkey diode (eg. BAT43) would be around 0.3V

My note: Brilliant! A regular diode won't begin to saturate with current until the forward voltage is reached. So just using a regular diode in the forward direction and connecting it to the base of a suitable NPN transistor would allow the transistor to be biased on strongly only when the diode voltage threshold was reached. Excellent!

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  • \$\begingroup\$ But what will the total voltage across the capacitor be at that point? Remember, the NPN transistor itself requires about 0.6V across its B-E junction before it will turn on anyway. \$\endgroup\$ – Dave Tweed May 11 '13 at 11:22
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This is hard and fast solution. You can use diode forward drop to bring down Trigger and Threshold requirements. Here is an example.

schematic

simulate this circuit – Schematic created using CircuitLab

There could be more elaborate schemes to manipulate Pin 5 (Control) to more accurate value for Trigger.

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protected by W5VO May 12 '13 at 17:58

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