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I am currently using a schmitt trigger to turn off power to non essential circuits when the battery voltage dips below a certain threshold.

The problem with this circuit is that it is dumb, i.e., it cannot tell the difference between a temporary voltage drop across the battery due to a high current load being switched on, and when the battery is actually low.

What kind of analog circuit can I add to differentiate between these two events?

Edit: I have a 12V LA battery that dips to around 8V for approx. 1 sec when I turn on a high current load.

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    \$\begingroup\$ Why do you insist on it being an analog circuit? How about throwing a tiny microcontroller and some simple software at it? \$\endgroup\$ – vicatcu Jun 14 '12 at 16:36
  • \$\begingroup\$ Micrcontrollers are expensive when you include components such as the crystal, and the cost to create and maintain the software. \$\endgroup\$ – Dan Jun 14 '12 at 16:43
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    \$\begingroup\$ Dan, this is a repeated discussion here on Electronics.SE. Olin will come and say you have to use a PIC10F220 :-). Most of the times he's right. In software you can construct filters which are hard to do in hardware, and the PIC10F220 doesn't need a crystal. All it needs is a decoupling capacitor. \$\endgroup\$ – stevenvh Jun 14 '12 at 17:20
  • \$\begingroup\$ I think what we need here is an 8 pin microcontoller programmed to act like a 555 timer, then configure it as monoflop ;o) \$\endgroup\$ – jippie Jun 14 '12 at 17:37
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    \$\begingroup\$ Why are you all still talking about microcontrollers? This is a basic low pass filter application. \$\endgroup\$ – Rocketmagnet Jun 14 '12 at 19:04
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What you want is a low pass filter.

A temporary dip is a high frequency event, while a real low battery is a low frequency event. If you pass this signal through a low pass filter, it will remove the high frequency events, and allow the low frequency one to pass.

Low pass filter

Increasing the value of the capacitor and resistor increases the time constant, thus decreasing the frequency threshold.

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Filter with fast reset:

Assuming

  • CMOS Schmitt

  • Schmitt sees high = OK input

Feed Schmitt input via a resistor.
Add capacitor from Schmitt input to ground. Optional - add diode across resistor, Cathode to Schmitt, Anode to voltage source (ie diode "arrow" points at Schmitt).

Diode adds "fast recovery" which simple RC lacks. If Vin is low 51% + of time for periods < RC time constant a simple RC will slowly discharge and trip. The diode "resets" the capacitor rapidly and means dropouts of ~= t = RC+ are needed to trigger output.

enter image description here

If Schmitt is already fed by a resistor divider just add cap + diode to input resistor.

When input goes high cap charges via diode to Vdiode < Vmax then cap tops up via Rin.

When input blips low cap starts to discharge via Rin with time constant R x C.

If Vin restores high then diode will allow charging faster than if just via Rin.

Time constant = Rin x Capacitor value.

eg if C=0.1 uF, R=10k. Tc = R.C = 10,000 x 10^-7 = 1mS = not long.
100k , 1 uF = 0.1s.
100k, 10 uF = 1 s
etc. Cap best to be NOT electrolytic due to leakage but depends on cap and brand.

Actual delay till triggering depends on how far Vin is above Vtrip but 1+ timje constant usual.

As always, component types and an actual circuit diagram would help.

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  • \$\begingroup\$ Wouldn't the diode just bypass R1? \$\endgroup\$ – Dan Jun 15 '12 at 19:12
  • \$\begingroup\$ @Dan - Yes - but only when Vin >= Vcap + Vdiode - that's the reason for having it. When the occasional dip lowers Vcap and then Vin restores to full value the diode gets Vcap back there faster. On a real low the diode is reverse biased and stays that way until trip point is reached. \$\endgroup\$ – Russell McMahon Jun 16 '12 at 12:00

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