Internal ram upgrade to 640k - continued

[1024MAK]

You'll only need a 250mA or 500mA 7805 - a 1 or 1.5A 7805 is overkill and will just make a lot of heat.

On this 500mA 7805 in SuperRAM I have used 2x .1uF ceramic and 2x 20uF 16V electrolytic, one each input to ground and output to ground. I also placed a 0.1uF ceramic at each IC, as close to the +ve pin as possible.

What is it with people getting confused with "linear" voltage regulators?
The current rating of the voltage regulator IC (78xxx) does not determine the amount of heat if used within the device specifications.
If a 7805 (1A rated) IC is used in place of a 78M05 (0.5A rated) IC, they will both dissipate almost the same amount of heat (there may be a very slight difference).

7805 regulators often get used in circuits where a lower rated device would work simply because 7805 ICs are cheaper.
Or 7805 or 78M05 types are used because using a 100mA 78L05 would take it close to or beyond its heat/temperature rating.

As to the capacitors, in most applications the values are not critical. Put 100nF to 220nF (0.1uF to 0.22uF) ceramic capacitors as close as possible to the input and output pins (each having the other connection to the ground/0V pin). Then use a 10uF to 100uF electrolytic capacitor with a voltage rating of 16V or 25V on the input. I use a "rule of thumb" of 0.5uF per 1mA, then round the capacitor value to the nearest available value. For the electrolytic capacitor on the output, again it depends on the current, but also the type of load. For small boards that have modern CMOS logic, I would use either a 47uF or a 100uF capacitor, with a voltage rating of 10V or 16V.
Why, well at small order quantities, there is not much difference in price and the larger value does help with the faster switching logic chips. Oh and use low ESR 105C types.

Also keep in mind that typical 7805 circuits often use electrolytic capacitors with values of 10uF or 22uF. But in most cases there is not a single "right" answer.

Oh, I do appear to have waffled on a bit

Mark

[Dave]

In theory, theory and practice are the same, but in practice, they're not.

Generally, you're right. I agree with you in broad terms. Heat production is mostly a factor of the voltage drop across the device. Hoooowever....

I look at the efficiency curve for each device, which varies quite a lot with load. A 1.5A 7805 that has 10% load runs with about 2/3rds the efficiency of a 500mA 78M05 at 30% load. The heat produced at the device is broadly the same. However, the upstream current draw is typically 220mA for the 78M05 and 290mA for the 7805. That is 70mA of extra draw on the QL's transformer, which does make extra heat.

Another factor is that at very low loads, regulators can get noisy.

I know, it's only 0.07A @ 5v, but, yeah. For the entire life of that product, it will consume that extra. It adds up.

Also, the 7805 in a TO220 package has a much lower c/W temperature rise compared to the S-PAK package, which has a much smaller interface area. It may not be as hot but can still be producing more heat.

When looking at heat, I am approaching it from the entire system point of view; I am not just looking at the 7805 in isolation.

I'm not disagreeing with you, or looking for a holy war or anything For this application, it doesn't take much thought. However, if you're making hundreds or thousands or millions of something, all those little 0.07A add up to a power station or two, eventually. So, I put a lot of thought and consideration into it that goes beyond "rule of thumb."

[tcat]

Dave, Mark,

Thank you so far.
I will use L78M05 with 350mA, and capacitors in the range you advised.

TCAT

[1024MAK]

Glad we could help

Just duck when you see two engineers throwing their tools out the tool box

Mark

[tcat]

This is part of the QubIDE schematics. Just connect a 512KByte SRAM chip to the bus and the 16V8 GAL.
Anything else?
Greetz from Brazil

Greetings to Brazil,

You have very simplistic design I looked at your GAL equations as well.
I'm sure it must work like a charm.

CHIP AMP512 GAL16V8A

A19=1 DSL=2 A18=3 ASL=5 WE=4
DTACK=16 DSMC=19 RAMCE=15 RAMOE=17

EQUATIONS

RAMCE = ASL + /A18 * /A19 + A18 * A19
RAMOE = RAMCE + DSL + /WE
DTACK = GND
DTACK.OE = /RAMCE * /DSL
DSMC = VCC
DSMC.OE = /RAMCE

How could you manage to feed GND and VCC into DSMC and DTACK outputs?
My GAL assembler does not allow this, to get HIGH LEVEL I had to use /OE inverted input wired to GND as the feed to DSMCL and DTACK.

TCAT

[1024MAK]

I look at the efficiency curve for each device, which varies quite a lot with load. A 1.5A 7805 that has 10% load runs with about 2/3rds the efficiency of a 500mA 78M05 at 30% load. The heat produced at the device is broadly the same. However, the upstream current draw is typically 220mA for the 78M05 and 290mA for the 7805. That is 70mA of extra draw on the QL's transformer, which does make extra heat.

Hmm. So 10% of 1.5A is 150mA.
30% of 500mA is also 150mA.
The typical current through the common terminal for TO220 package 78xxx regulators is 8mA or less. So the total current via the input terminal is the sum of the current flowing out of the output terminal plus the current flowing via the common terminal, that is a total of 158mA.
So where is the rest of this "upstream current" going?

Mark

[Dave]

*grins* I don't know. This was a specific comparison of an 80s 7805 (and datasheet) for a 1.5A part and a 2009 datasheet for a much more recent smaller part that may behave quite differently. Newer 7805s are possibly quite a lot more efficient than older ones of previous generations.

My focus is on improving the QL, and it's a quite narrow focus. I am starting to see other projects that can help Z88 owners, ZX80 owners, and etc too.

Like I said, I'm not looking to participate in a holy war. The part I chose is the right part to use for the profile of the expansion I am making - the profile includes power and heat elements, but also includes delivery target dates, prices and lack of development costs. Everything's a compromise.

The compromise comes out quite differently for UltimIDE, UltraQ and QL2.

[Dave]

I have a few hundred sets of components to do just that. The cost was not at all comparable, even buying in bulk. As I said before, the cost was >20x.

[Dave]

Ok, here's your opportunity.

Give me a parts list for a power supply with 8 or fewer parts, which if I buy kits for 100 pcs will cost 2€ or less per kit. The power supply must be able to supply a range of 75mA to 500mA with no appreciable RF noise, and must be >=85% efficient across that load range. A link to the regulator datasheet would be nice, too.

Show me the parts and price breakdown, and I will seriously consider it.

[Dave]

In an earlier post I specified "85% vs 70%" so I'm just asking you to demonstrate there is that level of efficiency gain. Most buck regulators these days are in the 90-96% range across 80% of their load range, so shouldn't be a problem. Remember, 85% vs 70% is a halving of waste.

I specified low RF noise because it's always an issue - the QL is a poor signal quality environment and I don't want to add to the woes. I just stated it because the cheapest buck regulators, etc, don't do well for noise below about 15% load - it just makes you put a little more effort into choosing the device load range carefully.

When SRAMs are not being accessed and the DIP switches disable this memory (we added the requested unexpanded QL kill switch) it only draws about 75mA plus what the 7805 burns for itself. Cutting the power to the board isn't an option because the GAL still needs to be powered to allow the thru-connector to manage downline expansion cards properly.

I'm just stating that your alternative regulator system needs to be able to handle a specific range of current draws, cleanly and quietly.

The point is: it is supposed to be a noticeable improvement to the product, not advanced technology for its own sake, right?