Date : Mon, 19 Jun 2006 02:44:41 +0100
From : "Mick Champion" <RS432@...>
Subject: Re: Watford Electronics EPROM software
Joel wrote;
>>>>>Can I ask the electronics gurus on here, is a programmer conversion
from 21V to 12.5V as simple as putting a resistor in there, or are
the actual electronics different please?<<<<<
Have you tested Vcc to see if it is only putting out 5volts?
In programming mode, normal 27128s use 21volt at Vpp. 27C128s(CMOS) use
12volts at Vpp You most likely now this already.
I am very rusty on this subject but with the lack of replies from gurus, I
have decided put my 10 pence worth in. A single resistor won't drop the
voltage on it's own. It has to have a flow through it (a load) in order for
it to have anything to resist. To use a single resistor, you would not only
need to know the load on VPP (25ma), but the current would have to be
constant for the voltage to be stable. I don't know if Vpp contantly
requires 25ma. I suggest you try using two resistors in series instead.
Please Note: The following is NOT guananteed to work on your Programmer, it
is just a simple way to reduce voltage when input voltage is constant.
Connect two resistors in series, (values discussed next). Attach R1 (the
lower value of the two) to the 21v line, and R2 to ground. Take your output
from where the two resistors join each other. In theory, if both resistors
were equal, voltage at the join would be halved. In this case, the one
joined to 21v must have a lesser resistence than the one attached to ground,
so output voltage will be over half than the input. The differential between
the two resistors sets the voltage in proportion to the input voltage.
>From the best of my fading memory, here is how I think you work out required
resistor values. Others please correct me if wrong. Stepping down from
21volts to12volts means you need a drop of 9v across R1, and 12v across R2.
Divide 9v by 12v and you get 0.75. That is means your required resistor
differential rate is 1 to 0.75. Now you need two values that have a similar
ratio and are readily available giving enough current.
Examples;
Divide R1 by R2 to find differential, to calculate current, use I=V/R where
V=21 and R=R1 value.
R1=820 ohms and R2 =1.2K (820/1200) = 1 to 0.68 differential. Max 25ma -
maybe not enough.
R1=680 ohms and R2= 1K (680/1000) = 1 to 0.68 differential. 30ma max
R1=680 ohms and R2= 910 ohms (680/910) gives a 1 to 0.74 differential ratio
at 30ma (910 ohms are available but NOT what I'd call readily available)
As the 27C128 takes 25ma at 12volts, I think R1 will need to be a half watt
resistor. If I am wrong, please excuse me but.12volts X 0.025Amperes= 0.3
of a watt.
A point to note with this resistor set up : Whatever voltage you put in, it
will reduce it by the same ratio. So when you are not in program mode, the
5volts that would normally be applied to Vpp would reduce to less than 3.
You could use a switch to bypass if experiment works. If you were to use a
LM317 variable voltage regulator instead, a 5volt input would put out around
3volts. A zenor diode circuit is possible. The best overall option would be
to find someone who knows your programmer, and how to modify it.. I hope
this helps.
Mick.