Date   : Wed, 21 Jun 2006 20:34:18 +0100
From   : Sprow <info@...>
Subject: Re: Watford Electronics EPROM software

In article <000e01c69341$f20ea080$384b0954@...>,
   Mick Champion <RS432@...> 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.

[snip potential divider]

> R1=680 ohms and R2= 910 ohms (680/910) gives a 1 to 0.74 differential 
> ratio at 30ma 
>
> As the 27C128 takes 25ma at 12volts, I think R1 will need to be a half 
> watt resistor.

One minor problem with this approach is that the potential divider equation
assumes no (or negligable) current flows. If you have 

   21 -+
       |
       /   680R
       \
       /
       |
  out -+
       |
       /
       \   910R
       /
       |
      ---
       -

then out = (21/(910+680)) * 910
         = 11.97

but if 25mA are flowing through the 680R resistor then there must be 17V
across it, which means out must be 4V under load.

Also, I believe we should be aiming for 12.5V not 12V, for the 27C128 and
27128A EPROMs.

I personally would go and look how the 21V is generated (presumable some
sort of charge pump or an inductor arrangement) and modify that, or possibly
insert a 12.5V regulator in the way (a 7812 with a diode in its ground leg),
Sprow.