Overclocking the DC and where to get the Increase

[flaming_carrot]

Doh! how could i forget such a simple thing!?

To clarify, i read V. by using the ground coming from IC 104, which i assumed would be the ground fed to the SH-4 therefore the multimeter was in parallel. Perhaps im wrong?

Anyway, flaming_carrot, thats a good point but the problem is that if i was bothered to find a resistor that was in series with the output of pin 3 of IC 104 then i would have to replace it with something so close to the original value as the V. is only increasing so slightly (eg. V=IR 1.99/0.23= 8.65, so to increase that: V'= 2.01 so: 2.01/0.23= 8.74 - i know that these values for resistance arent true but for an example) i dont think i could find a resistor so close to a value as the ones used are gonna be nice round no.s such as 100 then im gonna need something like 101.7. The other way would be potential dividers but thats even messier on an already packed board.

Maybe i'm wrong and over complicating it, this is only GCSE physics, and me doing A2 i should know this stuff

No, you read the voltage properly, but like semicolo said, you need to connect the meter in series (ie desolder the pin and route the signal through your meter) to get a correct current reading.

And yeah, I think you are overcomplicating things. First off, a resistor after pin 3 (by the way, you have switched the numbering of the pins; pin1 is Vcc, pin 5 is ground) wouldn't work that way, it would limit the current, not raise the voltage. Here's a diagram I threw together quickly:

I may have switched R133 and R134, I'm at work right now and this is from memory. Pin 4 is the Vref pin and it should have a voltage of 1.25V over R134. R134 is prolly 1kohm. R133 will control Vout according to:

Vout = Vref*(1 + R133/R134)

So R133 is prolly 600ohm now. If you replace this with a 619ohm resistor (use 0.1% tolerance) you will raise Vout to a little over 2.02V.

Unfortunately resistors only come in 600, 604, and 619 values so if you want values in between, solder two on top of each other in parallel fashion, eg use 1k580 and 1k to get circa 612 ohm, giving Vout ~= 2.015V (umm..uncertainty is really greater than that but whatever...)


Important notice! Caveat emptor! I take no responsibility for any damages that may result from this!

[ptr.exe]

Yeah, i mixed up the no's in the text and so thought it easier to change the diagram. don't know why.

You obviously know so much more than me its ridiculous. That method makes alot more sense than before. yeah i remember how to use resistors in parallel to have a lower value, so i'll have a go at that.

Thanks for all the help

[Morph]

Sorry about being gone. Been busy as all get out, and haven't had the opportunity to post. if everything goes well, I should be able to try this by this coming Monday. I see someone else posted with new data. So intercepting the lead via lifting a pin is not necessary? Because the diagram you drew up is all like...

EDIT: nvm, now it works

But yeah, I would like to hear about ur second DC ptr. Wonder if the voltage has to do with the speed or whatnot.

[ptr.exe]

Can't you see the pic?

Yeah, i dont know why we assumed the easiest method would be a completely new circuit for making the 2.01V increase, i may give this a go soon as im on holdiays soon

[Morph]

Been too busy to give this much thought myself. been working on a different undiscussible project on the side, and really haven't had the chance to tinker with the hardware.

[flaming_carrot]

No one tried this yet? I'm bored, I wanna hear some stories about SH4s bursting into flame, muhahahaha!

Anyway, R133 and R134 are correct in the diagram, so any bold ones can just replace R133 appropriately. It's located on the same side of the PCB as the SH4 and is pretty close to it.

[ptr.exe]

I'll try this as soon as i can find a cheap DC. At the moment i have one stock one that i don't wanna screw up as its mint, my MF DC which i don't wanna screw up as it took too long to build. I bought a DC the other day for £4 and got it working (F1 fix) and then sold it for £15, i'll find another cheap one then try this out.

[flaming_carrot]

Understandable. I was thinking more of the other guy who seemed so eager before

[ptr.exe]

Yeah, morph this was pretty much all your idea, since you hijacked the topic from cross, who is probably still patiently waiting for his original questions to be answered 8 pages on

Any luck yet? you must have bought a soldering iron by now

[Morph]

Well, yes and no. I have a new soldering tip now, yes. Problem being I have to find the time to do it. Hopefully this coming week will allow me a chance to have a go over it.