Suppose I have a power supply the puts out 14Volts and I have a processor that needs 6V and I want to run say, some '373 data latchs off the same 6v regulator using the ground from the rectifier but I want to create a seperate ground starting at the first 6 volts or processor six volts and regulate the 6 to 14v as a seperate 6 volt source. How would I drive the upper 6v with the lower six volt logic (processor). I am looking for cheap and dirty method. I want an 8 line bus I can control on both power levels with one processor. Any suggestions?
Posted on 2004-09-15 20:30:44 by mrgone
I hope I understand correctly what you want.

Take a look at the picture. U1 is a 6V regulator of your choice.
U2 is a 5V regulator. Due to the diodes, its output voltage will be approx 5.6V. You can replace each diode with 2 diodes in series and the output will become 5+2*0.6=6.2V approximately.

To turn "off" this regulator you turn on the transistor (using your 373) and it will now produce 6V with respect to ground, not with respect to +6V. So the +12V now becomes +6V, that is 0V with respect to your second "ground".

If your 373's are CMOS you can eliminate the transistor and drive the diode directly with a 373 output.
Posted on 2004-09-16 12:08:34 by VVV
What was that with the diode? I've used that trick but they do have 7806's. Actually I want to control the upper six volt data bus with the lower 6 volt logic. I have three 74hc373's and I don't want to use 18 transistors. Probably have to but just thought I would throw it out there. Maybe if there is an 8 transistor linear IC that could be biased individually or someting like that.
Posted on 2004-09-16 13:31:50 by mrgone
Sorry, I completely misunderstood your question. I thought you wanted a 6V power supply "on top" of the existing one and which could be turned on or off by a ground-referenced signal.

Now I understand what you need, but I do not have a simple solution for that. The only simpler thing that comes to mind is to use open collector gates (such 7406) to drive all the input signals of the upper 373's. But that still requires some resistors and will not be particularly fast.

Sorry...
Posted on 2004-09-17 11:27:04 by VVV
Any input is appreciated. That's still a slick circuit you posted. Yeah I'm still thinking it out. I may abandon it depending on the amout of time I have. I was just thinking that this Xformer secondary puts out alittle more than I want (Radio Shack special). I hate wasting power and thought to myself that would be a highly efficient option but not if I need 9000 parts to do it. :lol:
Posted on 2004-09-17 13:07:20 by mrgone
A transformer will only put out as much power as the load is demanding. The current ratring is a maximum level, not the constant drain.

IE, a 1 A secondary can be used with loads UP TO 1A.

Try pulling more and something might catch fire.

Try pulling less and you'll get less. And safely too.



(PS: Don't stack the power supplies, just get one that will drive everything and une one ground.)
Posted on 2004-09-18 09:18:43 by Ernie
Finally I got around to post this, for whatever it's worth.

The two regulators are stacked and the IC's are powered as follows: the 7407's from the lower 6V, the 373's from the upper 6V.

If you use 8-resistor resistor networks you will need eight of them plus five 7407's to drive three 373's on the upper bus (30 lines 3x8 data + 3x2 controls). The resistor networks should have separate resistors, not bussed (the upper ones can be bussed).
Of course, you will need only two regulators to power the whole circuit.

The data bus can be shared by the 373's, maybe the /OE signal, too.
In that case you need only two 7407's (12 gates: 8 for data, 1 for /OE's and 3 for the strobe inputs) and 4 resistor networks.

This circuit is not going to be very fast because of the resistors, but it will work. The resistors are all equal and (if you use HC devices) they can be from 1 to 100K.

Hope this helps. :roll:
Posted on 2004-09-25 00:43:43 by VVV