ICL7611 op-amp (was: Motorola MC14081B)

Adrian Graham witchy at binarydinosaurs.co.uk
Sat Dec 24 06:07:25 CST 2016

On 24/12/2016 03:48, "Brent Hilpert" <hilpert at cs.ubc.ca> wrote:

>> On a whim I managed to solder new legs onto the old LM385Z and it works,
>> giving 1.2V at IN-, but the output is still only 0.2V.
>> I don't mind admitting I'm stumped :)
> The op amp is configured as a schmitt trigger or comparator with hysteresis:
> There is no negative feedback so it is operating at full gain and functions
> like a comparator.
> However there is positive feedback via R412 (*1), this adds hysteresis to the
> trip point(s).
> (Brief hackneyed, not rigorous, theory of op: As the input differential varies
> past the trip point, the output pulls the + input
> further above or below the point at which it just tripped, so the inputs now
> have to 'overcome' a greater differential
> to trip as the input differential varies in the opposite direction.)

OK, that explanation helps a bit. My brain still has trouble 'slowing down'
how these components work, like when I first dropped from 3GL programming to
assembly. Fortunately there are a lot of tutorials out there.

> One input to the (now) comparator is the 1.2V from V407 regulated down from
> some power rail.
> The other input is the ~ 1/4 voltage divider down from +5V formed by R398 &
> R406 (netting 1.34V @ 5V).
> It appears the idea is that as the +5 supply ramps up at power-on the
> comparator trip (and hence release of reset) is delayed till the +5 reaches
> something around +4.5V.

That's what I figured yesterday which is why I was surprised that
reinstating the LM385Z didn't make it all spring into life.
> The ICL7611 is, I expect, a very-low-power (Intersil's niche) op amp. Together
> with the CMOS 4081 the circuit appears tailored for low-power operation.
> Is it supplied by the battery?
> It may require the battery presence for stable supply at time of power-up to
> get reliable reset operation from this power-on-reset circuit.

That's a very good point so now I'm going to have to scrabble around to find
the battery to find out what voltage it was since I removed it ages ago and
may have recycled it. I don't recall any markings on it though, it looked
like a 'normal' 3 terminal NiCAD wrapped in blue plastic. I did take hi-res
pictures of it however.

The 5V input at R395 does head off towards the battery location before it
hits the resistor so I'll trace that out too.
> The diode-resistor pairs at the 4081 AND gate inputs turn the AND gates into
> 'asymmetric edge delay gates': the resistor together with some capacitance
> delays the switching of the gate for an edge of one direction, while the diode
> shorts the delay for an edge of the other direction.

Gotcha, I wondered why the diode was there.

> If there is no cap at the input, they must be relying on the gate capacitance
> of the CMOS inputs, making for a pretty short delay.
> *1: Are you sure that's a 220‡ resistor? It's awfully small compared to the
> impedance of the voltage divider feeding it.

My bad, it's a 1.4Mohm and tests ok.


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