30 Jul
2013
30 Jul
'13
10:33 p.m.
On 2013 Jul 28, at 11:09 AM, Tony Duell wrote:
The input transition is coming from a slow mechanical process (along
with typical LDRs not being exactly speedy), and with the 18K
resistor in feedback it's essentially a varying-gain amp (from some
gain to near unity) during the transition region, so some of that
slow transition will appear on the output. Plus the output is on the
low end of the TTL output level when it doesn't need to be, there's
2V of unnecessarily unused noise margin there.
It may work the way you drew it but it begs the question of why a
designer would do it that way when it has poorer performance than the
'typical' (and actual) way.
Speaking of response times, going from Rik's info that a CdSe cell
was originally used, I wonder if faster response time is the reason
HP selected the CdSe over CdS LDR. Some web info suggests the CdSe
are faster although I haven't found categorical specs.
Earlier, I
could have said: the circuit as you (Tony) drew it is so
obviously unusual for the task you should have double-checked it as
you drew it, and the error never should have made it into your
schematic. But I didn't. Out of politeness.
Yes, agreed I should have checked it. But, actually, is it totally
ridiculous as I've drawn it (meaning 'should my bogometer have hit the
end stop'>).
I know nothign about the characteristics of the LDR. So suppose
it's very
high resistance. We agree that the +ve input of the op-amp is helt at
2.5V. So if there was a -ve feedback resistor as I drew and the LER
was
high resistace, the output would eb essentially at 2.5V so as to
make the
-ve input at 3.5V also (it would be a voltage follower, followign the
2.5V on the +ve input). As the LDR resistance dcreases, the output
would
have to swing towards the -ve reail so as to keep the -ve input at
2.5V.
The (zener) diode on the outptu would prevent it going very far -ve.
So the ouput of the circuti as I drew it (albeit not the circuit HP
used)
could have given a TTL signal with a suitable LDR.
So I feel there is nothing that should instantly have said to me 'This
must be wrong'. But anyway. It is wrong. The resisotr goes form -ve
input to
ground, not
from -ve inptu t ooutput.