On Monday 01 September 2003 09:52, Alan B. Pearce scribbled:
> Not sure that lowering the value of R2 will have the desired effect.
> However we are not told the ratio of R1:R2 which will have an effect, a=
nd I
> suspect is why the second example does not work as expected.

You're right.  R1 =3D 1k, R2 =3D 10k as drawn up.  But changing R1 to 10k=
 also,=20
gets the circuit to work properly.


> Having the BC junction forward biased when RB0 is driven low is an inhe=
rent
> part of the circuit design IMHO, and if this did not happen then the
> circuit would never work at all, as then there is no way to pull the
> R1/base junction down to a low voltage to allow detection of the other
> switch's true state. The base-emitter junction does the same job as the
> diode in series with the switch.

Ding!  This is what I was expecting, which is why in the original (workin=
g)=20
circuit, I removed the diode, and the circuit functioned just the same.


> I suspect the reason the second example does not work is that R1 is a m=
uch
> lower resistance than in the first example. Personally I would prefer t=
o
> see R2 a little higher in value (probably around 33k), and R1:R2 in the
> region of 1:10, so R1 is then around 3.3k to 4.7k. This will then allow=
 the
> voltage to pull up to above the RA input threshold when RB0 is high, bu=
t
> allow RB0 to pull right down to ground without going into current
> saturation. I suspect the faulty unit has R1 around 100 ohms or so and =
is
> then causing RB0 to go into current saturation which means it will be
> pulling down to about 3 to 4 V, leaving the RA input above the high
> threshold.

Okay, I get it.  I always felt that the base current was just to get the=20
transistor to switch on, and anything higher was *only* wasted as heat.  =
But=20
I guess I have much to learn still.    I'll go do the math to fully=20
understand this now.

Thanks,
-Neil.


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