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PICList Thread
'[EE:] V/I hysteresis circuit'
2005\08\08@131625 by Brendan Moran

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Hi,
I need a circuit that gives me a current hysteresis with respect to an
input voltage.  The catch is that I can use no powered parts (i.e. no
op-amps, no digital ICs) it has to be a two-pin circuit.  I'm using a
current regulator diode to limit the input current.  I've attached the
current circuit I've come up with.  The problem with it is that while
the turn-on voltage is nice and sharp, the current regulator diode is
causing the the turn-off voltage to be quite gradual.  I need something
sharper.

Here's the theory of operation:

The circuit should not turn on until the input voltage reaches VzD2 (8V)
+ VbeQ2 (0.6V, due to small current) + VzD1 (2V), so 10.6V

The circuit should turn off when Vin reaches VzD2 + VbeQ2 + VceQ1sat
(0.2V) + VCD1.  This is where the problem occurs; VCD1 is essentially 0
at turn-on.  But it rises to block more current from flowing.

CD1 is essentially a jfet with a resistor in a negative feedback
orientation, to form a two-pin device with a negative dynamic
resistance.  CD1, below a minimum operating voltage, is close to a pure
resistor, which does not practically affect turn-on.

This is where the soft turn-off is occuring.  As the voltage across the
circuit decreases, CD1 will reach its minimum operating voltage, and
reduce in current until IbeQ1 < VbeQ1/100k, when the whole circuit turns
off.

If R4 is changed to a lower resistance, turn-off becomes sharper, but
turn-on becomes softer.

Does anyone have an idea to sharpen up the turn-off point with out
softening the turn-on?

Thanks,
Brendan



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2005\08\08@155912 by Dave Tweed

face
flavicon
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Brendan Moran <spam_OUTannirackTakeThisOuTspamshaw.ca> wrote:
> I need a circuit that gives me a current hysteresis with respect to an
> input voltage.  The catch is that I can use no powered parts (i.e. no
> op-amps, no digital ICs) it has to be a two-pin circuit.

[snip]

> If R4 is changed to a lower resistance, turn-off becomes sharper, but
> turn-on becomes softer.
>
> Does anyone have an idea to sharpen up the turn-off point with out
> softening the turn-on?

How about adding a forward-biased silicon diode at the collector of Q2?
I'm thinking that this would force most of the current at low voltages
to flow Q1 E-C to Q2 B-E, forcing a sharp turnoff once this diode stops
conducting.

-- Dave Tweed

2005\08\08@180922 by Brendan Moran

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subject=Re:[EE:] V/I hysteresis circuit
source= http://www.piclist.com/piclist/2005/08/08/155912a.txt?

I tested adding a forward biased diode in series with Q2 collector. it
didn't seem to make much of a difference. I still managed to get down
below 10uA before it de-triggered. I'm trying to get the circuit to
de-trigger at 0.7mA or higher.
---

http://www.piclist.com/member/annirack-shaw-
PIC/PICList FAQ: http://www.piclist.com


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