I’m working on what I hope will be the final hardware changes for the Yellowstone 2.1 PCB. Among these are changes to my capacitor usage and placement strategy. In my previous posts I discussed intermittent problems that looked to be caused by ground bounce, VCC sag, or both. Apple II expansion cards only have a single ground pin on the motherboard connector, so they’re prone to ground noise when signals with fast rise and fall times cause large transient currents. Among several methods that I’ve used for mitigating this problem, one not-so-obvious strategy is to add bypass capacitors on the +12V, -12V, and -5V power supplies, even if the card doesn’t use those supplies.

Apple IIgs Technote #68 says:

Right now Yellowstone doesn’t use or bypass the +/-12V supplies; it just routes them through to the disk drive connectors. -5V is completely unused, and there’s nothing at all connected to that supply pin. I had actually forgotten there even was a -5V supply. (Maybe this could be leveraged somehow for current-limiting of Macintosh drives?)

Why does bypassing unused supplies help anything? Even after doing lots of reading about current loops and ground bounce and impedance, I have to admit I don’t really understand it. I read a T.I. paper that discussed how bypass capacitors help reduce the area of current loops, but I don’t see how that’s true for bypass capacitors on unused supplies, where there’s no path back to the chip that’s sourcing or sinking current. The Apple technote mentions “improved ground by providing additional AC ground paths”, which makes a little more intuitive sense to me. I imagine the bypass capacitors as acting like tiny batteries that work to maintain the voltage between their two terminals. As long as +12V, -12V, and -5V are stable, then bypass capacitors on those supplies will work to keep local ground at 0V by sourcing or sinking small amounts of current as necessary.

I probably wouldn’t really care why it helps, and would just trust that it does, except that I need to know what types and values of bypass capacitors to use, in what locations. Without a better understanding of the theory of why it helps, I’m uncertain how to proceed.

The Apple technote specifically says electrolytic or tantalum capacitors, but doesn’t mention ceramic, which would normally be my go-to for this. I’m not sure why they omitted ceramic.

For the capacitor value, Apple doesn’t say anything. Is a standard 0.1 uF appropriate? I believe the 0.1 uF rule-of-thumb is related to the rise and fall times of typical ICs, and an assumption that the capacitor is placed within a few millimeters of the IC supply pin. But when there’s no specific chip being bypassed, how do you determine the appropriate value? Would a larger value like 1 uF make more sense?

Then there’s the question of placement. I can’t place the bypass capacitors near the IC supply pins, because there are no ICs using +12V, -12V, or -5V. Should I put the capacitors near the board’s supply pins on the motherboard connector? Or maybe in the middle of the board?

My best guesses are that a ceramic capacitor would be fine as long as its voltage rating is high enough, the capacitor should be placed close to the motherboard connector, and the value should be something like 1 uF or 10 uF. Or maybe multiple values with multiple capacitors per supply. I don’t want to rain down tons of extra capacitors though, because the number of discrete components on the PCB is already higher than I’d like.

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I’ve been searching for a simple and reliable way to prevent a short-circuit on Yellowstone’s -12V supply when a particular type of Macintosh disk drive is connected. Connecting this drive results in a +5V to -12V short circuit, but if the short is removed, the drive otherwise works just fine. One obvious solution is to put a series resistor on the -12V supply connection. For example a 1.7K ohm resistor would limit the short-circuit current to a safe 17V / 1.7K = 10 mA. But a series resistor on the -12V supply for Macintosh drive protection would obviously screw up other types of drives, by making the supply appear to be something other than -12V volts. Obviously. Or maybe not so obviously.

When I actually tested it, I was very surprised by the results. As far as I know the -12V supply is only used by Disk II drives. A 1.8K series resistor on the -12V supply for a Disk II drive resulted in an apparent supply voltage of -8.7V, but the drive still worked fine. I then tried larger value resistors resulting in “-12V” supply voltages of -6.5V, -5.0V, and -0.8V, and everything continued to work. Even at +3.7V the drive continued working, but I/O was slightly slower than normal, suggesting that a high number of retries were happening. The drive finally failed after I reached 100K ohms and +6.9V for the so-called -12V supply.

How can this be? How can the -12V supply voltage accuracy matter so little? What is it used for?

Looking at the schematic for the analog board of a Disk II drive, -12V is only used in one spot, at one terminal of a 10K potentiometer at the bottom-right. It’s part of an analog reference adjustment for the MC3470, the drive’s floppy disk read amplifier circuit. Pages 7-119 to 7-120 of the MC3470 datasheet describe how it adjusts the differential amplifier that’s used to detect peaks in the AC signal from the drive’s read head.

As best as I can tell, if the -12V supply voltage is incorrect, it will result in “peak shift”. The boundaries between bits will appear to shift slightly forward or backward from where they’re actually located on the disk. A small degree of shift won’t cause any problems, but eventually it will reach a threshold where bits begin to be read incorrectly, resulting in errors.

For the past week I’ve been researching and discussing current limiting circuits, current mirrors, and supply switching circuits. Can I forget about all that stuff and simply put a 1.8K resistor in series with the -12V supply? These test results say “yes”, but I’m still somewhat hesitant.

Hesitancy #1 is because I don’t know what the worst case might be. Maybe my Disk II drive isn’t representative of others. Maybe there are other types of drives (DuoDisk? Two daisy-chained Unidisk 5.25 drives?) that use the -12V supply in ways I don’t know about, or that will result in bigger voltage errors for the same series resistance.

Hesitancy #2 is because even if it works in tests, any error in the -12V supply voltage will cause some amount of peak shift, which might be the difference between a badly-degraded floppy disk working and not working. It’s a little like throwing away part of the noise margin for digital signals. It may still work fine for most situations, but when things are already working poorly for other reasons, it could be enough to tip the result into failure.

On the other hand, maybe the simplest solution is best. If it works in tests, even with series resistor values far larger than the ones I’d actually need for current limiting, then what’s not to like about this solution?

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I’ve returned to the question of how best to limit the current from Yellowstone’s -12V supply. I discussed this earlier here and here, and had more-or-less settled on the idea of using a current mirror circuit. But now I’m taking a second look at possibly using JFETs or a purpose-designed CCR (constant current regulator) intended for powering LEDs instead.

Here’s a review of why this is needed. Yellowstone has two connectors for disk drives, and each drive is supplied with +5V, +12V, -12V, and GND. Some types of drives don’t actually use the -12V supply, and other types have the -12V supply pin internally connected to the +5V supply pin. Ouch! A picture may help clarify:

The idea is to insert some new component or circuit inline with pin 2 (the -12V supply pin) of each Yellowstone drive connector to prevent a short circuit when a Macintosh drive is connected. Keeping in mind there might be two Disk IIs connected, or two Mac drives, or one of each, or any combination with other drive types. The anti-short-circuit protection should happen automatically, without requiring any user switch settings or other manual configuration.

Ideally the solution would just disconnect pin 2 when a Macintosh drive is connected, but I don’t know any good way to do that. The next best solution is to limit the current through pin 2, to a level that’s high enough for a Disk II to operate normally, but low enough to prevent problems with the Macintosh drive.

Here are the requirements for the -12V supply connection, as I see them:

1. When a Disk II drive is connected, it should see a supply voltage as close to -12.0V as possible, since it uses the supply in an analog reference circuit. This means any resistors between the drive and Yellowstone’s -12V supply should be small, to minimize the voltage drop.
2. When a Macintosh drive is connected, the current should be limited to about 10 mA. The exact value isn’t critical, and anything from 5 mA to 20 mA is probably OK.
3. The two drive connectors should be isolated from one another, so a Mac drive at connector 1 doesn’t affect the -12V supply for a Disk II drive at connector 2.
4. The current limiting function should work automatically, without any control or reset logic.
5. Power dissipation in the current limiter should be as low as possible.
6. The current limiter should be built from commonly-available parts.
7. The current limiter should use the fewest possible number of parts, that are small and inexpensive.

 
Current Mirror

In my last post, I considered using a current mirror. In this circuit, a reference current is passed through a transistor, with two other transistors controlling the -12V supply current, one transistor per drive connector. It can be built with three matched BJTs or FETs, plus a single resistor for setting the current limit.

Here’s an example using FETs, with a Disk II drive plugged into the first connector and a Mac drive plugged into the second. The -12V supply current is shown for each drive connector. The current for the Disk II is nearly identical to the 2.4 mA that would be expected if no current-limiting circuit were present. But the current for the Mac drive is limited to 9.5 mA, instead of the many amps that would otherwise flow due to the short circuit.

A current mirror requires each transistor to have the same voltage threshold, gain, temperature, and other properties, else the current may be incorrect. This is where my difficulties begin. I went looking for a single IC package with 3 FETs, or maybe 4 FETs, in order to help ensure they’d all have matching properties. I learned that these aren’t common: you can find discrete FETs or dual FET packages, but beyond that it’s slim pickings. With the global shortage of semiconductors, I’d prefer not to rely on a part whose availability might be an issue in the future.

I could easily use two dual FET packages, or three discrete FETs, but then their temperatures and other properties might not be well matched. Given that I can tolerate an error of least 2x in the current, maybe this isn’t a big worry. I need to take a closer look at the FET datasheets to get an idea of how much the current might change if one transistor’s threshold voltage is 0.1V off from another’s, or one’s temperature is at 25C and the other is at 85C. Maybe it’s not a big enough issue to worry about for this application.

One small drawback of the current mirror is that it wastes 10 mA from the -12V supply to create the reference current, regardless of whether any of the connected drives actually use the -12V supply.

 
JFET

Another approach to current limiting is to use a self-biased JFET. A commenter on one of my previous posts mentioned this option, but I never followed up on it. The basic idea is that a JFET is a “default on” device, but its drain-source channel can be partially closed by making the JFET’s gate-to-source voltage negative. The more negative you make Vgs, the further the channel is closed, until reaching the Vgs cutoff threshold where the JFET turns off completely. By adding a resistor at the JFET’s source terminal, and connecting it back to the gate, this behavior can be leveraged to make a current limiter. There’s a nice tutorial about this here.

The JFET’s I_DSS (its maximum drain-to-source saturation current) must be larger than the desired current limit. If the circuit can’t reach the current limit, because there’s a large external resistance like in the case of the Disk II drive, the behavior is less obvious to me. From my scrutiny of graphs in the datasheets, it looks like a typical JFET in this situation will behave like a resistor of about 80 ohms, in series with the external resistor used to set the current limit.

Yellowstone would use two independent JFETs for the two disk connectors, with no reference current needed or wasted. That’s nice.

The higher “on” resistance compared to a normal FET is less nice. Combined with the external resistor, it would create a larger voltage drop, so the Disk II drive would see a voltage like -11.7V instead of -12.0V. Maybe that’s still fine. I did a test earlier with a Disk II and a 100 ohm series resistor on the -12V supply, and it seemed to work OK.

It’s slightly complex to calculate what value of external resistor is needed to achieve a desired current limit. The formula depends on the current, the JFET’s I_DSS, and its cut-off voltage. The tutorial linked above has the derivation. Or you can eyeball it from the graphs in the datasheet, if a precise value isn’t needed.

I couldn’t find many widely-available dual-JFET packages, which is too bad. So a Yellowstone solution using JFETs would probably be built with two discrete JFETs and two resistors, to make a separate current limiter for each drive connector.

 
Constant Current Regulators

The JFET sounds intriguing, but then I stumbled across the concept of constant current regulators. From what I can tell, these CCRs are just the JFET plus resistor circuit described above, all wrapped up into a two terminal package that looks like a diode. CCRs are available in different types with different fixed current limits, and are typically used in LED driving applications. As far as I can tell, their behavior is identical to the JFET circuit with the discrete resistor.

Since a CCR is a two-terminal device, it would be very easy to use: just put one in series between the drive and the -12V supply for each drive connector. Boom, done. It’s too easy, I think I must be missing something.

This specific NSI50010YT1G CCR from ON Semiconductor looks nearly perfect, with a built-in current limit of 10 mA. The trouble is that right now it’s unavailable just about anywhere. There’s a 15 mA version in stock at Newark, but that’s still only a single supplier, and I’d rather have 10 mA than 15. I really like the idea of using a CCR, but it’s not going to work unless I can find one with an appropriate current limit, with good availability from multiple suppliers.

 
Power Dissipation

For any of the above solutions, when a Macintosh drive is connected and the current is limited to about 10 mA, the power dissipation must be considered. With +5V at the Mac drive end and -12V at the Yellowstone end, that’s 17V times 10 mA, or 170 mW of power. Most of the packages that I looked at have power dissipation limits close to that, or slightly higher. I’m uncertain how to optimize this, but probably I need to choose packages that are physically bigger (violating my requirement 7), and/or make the PCB pads very large so they can double as heat sinks. These kinds of power considerations are mostly foreign to me. I don’t want to make the PCB pads any larger than absolutely necessary, since board space is at a premium in the area near the -12V connections. If anyone has recommendations for package types or connection methods that would be better for dissipating a few hundred milliwatts, I’d love to hear them.

 
And the Winner Is…

I’ll do some more searching for CCR availability. If I can find one with the right properties and that’s widely available, I think that’s my preferred solution.

Otherwise I’ll probably stick with the original plan of using a current mirror and three FETs. I can build it with two dual FET packages plus a single resistor, leaving me with an extra FET available for some future purpose. I’ll do some more analysis and testing to try to estimate how much current error I might see if the reference current FET is in a different package than the FET that limits the current. As long as the answer is closer to 20 percent than 400 percent, it should be OK.

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I’ve written several times about Yellowstone’s need for in-system FPGA reprogramming. Once the Yellowstone card is in the user’s hands, the user needs a way to update the FPGA firmware when new versions are released, without requiring a JTAG programmer or some kind of USB interface built into the card. The current Yellowstone prototype was designed to support in-system FPGA programming, but it had never been tested until today. The photo here shows a BASIC program on an Apple IIe successfully communicating with the FPGA’s SysConfig port and reading its device ID. I’ve verified that 01 2B A0 43 is the expected ID for the Lattice MachXO2-1200HC FPGA on the Yellowstone board, so everything looks correct. It will still be a long road from here to reach the point where the FPGA is actually reprogrammed, but this proves that the basic communication mechanism works, which was the part I was most concerned about.

Yellowstone’s in-system programming support doesn’t require any extra hardware on the board, which is great! It’s all implemented through PCB routing, connecting the right signals to the right pins. The MachXO2 FPGA supports communication with its built-in SysConfig hardware through many different interfaces including JTAG, I2C, and a Wishbone interface implemented in the FPGA logic, but I’ve chosen to use the SPI interface. This is a “hard” SPI port, so reconfiguration and reprogramming via SPI should work even if the FPGA is in a fubared state or is completely blank.

How do you connect the Apple II peripheral bus to the SPI port, in a way that avoids accidental transitions on the SPI I/O signals, but allows for full bidirectional SPI I/O when needed? Here’s how I did it.

CS: The SPI SysConfig port has a chip select input pin, which is normally high, leaving the SPI port disabled. As long as CS remains high, it doesn’t matter what’s happening on the other SPI I/O signals, and the FPGA will ignore it. The CS input is connected to another FPGA pin that’s normally in a Hi-Z state, but that can be driven low by FPGA logic. This makes it possible to programmatically enable the SPI port by having the Apple II CPU write a magic value to a special address, which the FPGA logic watches for, in order to drive CS low. For situations where the FPGA is blank or in a bad state, the programmatic enable may not work. In this case there’s also a hardware jumper on the PCB that can be used to force CS low.

DI: The data input is address line A0.

CLK: The SPI clock is the Apple II /DEVSEL signal. This means an SPI data bit will be clocked into the FPGA from A0 whenever there’s a low-to-high transition on /DEVSEL. This signal is normally high, but it goes low for about 500 ns whenever the Apple II CPU makes a memory reference to the “device” region of the peripheral card’s memory. For Yellowstone, this region is where the virtual IWM lives, in address range $C0E0 to $C0EF (assuming the card is in slot 6).

DO: The data output from the SPI port is connected to another FPGA input pin. Yellowstone’s FPGA logic includes a special behavior that makes use of this. If the SPI port is enabled and the CPU reads from address $C0EA or $C0EB, the FPGA will return the SPI DO bit that it sampled on its other pin, rather than the normal return value from the IWM.

If this last piece sounds somewhat complicated, it is. The DO bit needs to get on the data bus somehow, so that the CPU can read it. But DO can’t be directly connected to the 5V data bus – it must connect somewhere on the 3.3V side in the Yellowstone logic, and then the ‘245 bus driver must be enabled at the proper time to drive the DO value onto the 5V bus. The ‘245 enable timing is under control of the FPGA logic. That means when the FPGA is blank or in a bad state, there’s no ‘245 enable, so there’s no way for the CPU to read DO. In short, a correctly-configured FPGA is a requirement for reading DO. Without this, the program running on the Apple II will need to proceed with blind SPI communication in which it can send data but not receive it. Fortunately this seems to be sufficient to perform FPGA reprogramming. I don’t see any way around this inability to read DO when the FPGA is blank without adding extra hardware to Yellowstone, which I’m loathe to do for such a rare situation. So be it.

When all of this is put together, the SPI communication looks like this: The CPU writes a magic value to a special address to force CS low. Then it begins a series of reads from $C0EA or $C0EB. Every read from $C0EA transmits a 0 bit because A0 is 0, and every read from $C0EB transmits a 1 bit. In either case, D7 of the byte that’s read is the reply bit from the SPI port. The rest is all software to transmit the correct bit sequences and make sense of the replies.

Now onward to the FPGA datasheet, where I can learn what sorts of magic SPI incantations are required to actually reprogram this thing.

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The Yellowstone disk controller’s whole reason for existence is its ability to control any type of Apple II disk drive, but for a few Apple II programs, this universality causes problems. Yellowstone handles standard 5.25 inch drives just fine, but some software gets confused when it doesn’t find the ROM for a standard Disk II Interface Card in slot 6. For these rare cases, Yellowstone has an option to enable a “compatibility mode” where it emulates a plain vanilla Disk II Interface Card without any extra bells and whistles. The question is how does the user enable compatibility mode?

In the current version of the Yellowstone prototype, the choice is determined by a DIP switch. But as I’ve been refining the design, I’ve developed a strong dislike of DIP switches. They’re big and awkward, and not at all user-friendly. I’ve already managed to find ways to eliminate two of the four DIP switches on the board, and if I can find another solution for enabling compatibility mode then I can eliminate a third DIP switch.

My first thought was to check the keyboard immediately after the computer is powered-on, before loading anything from the disk. If the ‘C’ key is held down (c for compatibility), I could enable compatibility mode, otherwise I could proceed normally. On the Apple II it’s easy to check for keypresses by examining memory address $C000. If there’s a keypress waiting, the MSB will be 1 and the lower seven bits will hold the ASCII value of the key. So I could check for the value $C3 ($80 plus $43) and everything would be peachy.

Unfortunately, I quickly discovered that one of the first things the Apple II ROM does after a reset is to clear the keyboard buffer. This happens before control is transferred to Yellowstone’s ROM, so by the time my code checks the keyboard buffer, it’s too late. What to do?

 
Rely on the Buffer’s Lower Bits

From testing on my Apple IIe, the keyboard buffer isn’t actually zeroed when it’s “cleared”. The MSB is set to 0, but the other seven bits still retain the ASCII value of the key. So perhaps my code could just check $C000 for $43 instead of $C3 to detect the ‘C’ key, and it would work as originally intended.

This seems slightly dubious, since I’m not sure if this is documented behavior, or if other models of Apple II will clear the keyboard buffer in the same way. There’s also a small bug this introduces: if ‘C’ was the last key typed, and you then turn off the computer and turn it on again within 1-2 seconds, $C000 will still hold the value $43 during power-up. This may lead to accidentally detecting non-existent keypresses and enabling compatibility mode when the user didn’t intend to.

 
Delay Booting and Poll the Buffer

Another option is for Yellowstone to sit in a busy loop after the computer is first powered-on, and continuously poll the keyboard buffer looking for the value $C3. If the keyboard has a key repeat behavior, then $C3 will appear in the buffer after roughly 0.75 seconds if the ‘C’ key is continuously held down. Or the user could be instructed to quickly tap-tap-tap the ‘C’ key when the computer is powered-on, instead of holding it continuously down. This would eliminate any dependency on the keyboard’s key repeat behavior. At least one modern Apple II peripheral card uses this method.

The main drawback of this approach is that it would introduce a busy-waiting delay every time the computer is turned on. To reliably detect a key repeat or a tap-tap-tap, Yellowstone would need to insert a startup delay of about one second, every time the computer powers-on or resets. It sounds like a small thing, but I think I’d find this delay very annoying. Furthermore, the key repeat behavior might not work on the original Apple II or Apple II+ keyboard, so tap-tap-tap is probably the only reliable solution.

 
Check the Open Apple Key

A third alternative is to check whether the Open Apple key is held down, instead of the ‘C’ key or any other standard key. The behavior of the Open Apple key is very different from the other keys. Open Apple is just a duplicate of the first button on the first game controller, and its current status can be read at memory address $C061. As long as Open Apple is held down, the MSB of $C061 will be 1.

The Closed Apple key works identically, and its state can be read at address $C062. But holding the Closed Apple key during power-up will trigger the Apple II’s built-in diagnostics, so that key isn’t a good candidate for triggering Yellowstone behavior changes.

For now I’ve implemented Yellowstone’s compatibility mode enable using the Open Apple key, and it works well on my Apple IIe. Unfortunately, earlier models of Apple II computers don’t have an Open Apple or Closed Apple key. Users with those computers would need to attach a game controller and press its button during power-on, which is far from ideal. There’s also a risk that some other third-party peripheral cards might use Open Apple in the same way, which would make it impossible to trigger Yellowstone’s compatibility mode without also affecting the other card. But I don’t know of any specific cards that use Open Apple, so maybe I shouldn’t be concerned.

 
Something Else

I could stick with the DIP switch, despite its awkwardness. Or I could ask users to perform some voodoo in the Apple II monitor, and write a magic value to a special address in the card’s memory to enable compatibility mode. Or some other solution I haven’t thought of yet. Opinions?

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I’ve fallen into a deep hole involving current-limiting circuits and current sources, in an attempt to solve a minor problem with the Yellowstone disk controller. In the big scheme of things, this isn’t a very important problem, but it has me intrigued. I’ve created a simulation of a circuit that may solve the problem, but it exhibits some transistor behavior that I don’t understand. Specifically, the base currents of the transistors don’t seem to follow the rules that I thought I understood.

You can see a working simulation of the circuit here.

The are two SPDT switches at the top of the diagram, representing Yellowstone’s two disk connectors. On pin 9 of this connector, some types of drives will have a 10K resistor connected to the +12V supply. This type of drive needs a -12V supply input on pin 9. It will use about 3 mA at most from -12V. Other types of drives will have a direct connection to +5V on pin 9 (+5V is also provided on pin 11 to all drive types). Yellowstone can either provide an additional +5V supply on pin 9 for this type of drive, or more simply, just leave the pin unconnected.

The basic idea here is to establish a current limiter of about 10-20 mA for pin 9. That’s more than enough for normal operation with the -12V supply, and it prevents dangerously high amounts of current flowing if -12V is directly connected to +5V as in the second type of drive.

Here’s the puzzle: the three NPN transistors have all their bases connected, and all their emitters connected. As shown, they all have the same base-emitter voltage Vbe of 0.668 volts. My mental model is that the base-emitter connection is essentially like a diode. With identical diodes, identical Vbe, and a single shared 470 ohm current-limiting resistor, I would expect the base current Ib to be identical for all three transistors. Yet they’re not. The middle transistor’s Ib is 7.2 mA while the others are only 0.165 mA.

The fact that one transistor is in saturation must be relevant. But the “base-emitter connection is a diode” assumption is failing here, and I can’t explain why. I need to read more transistor theory.

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