Commodore 64 Power Supply 902503-02, part 1.

In my Commodore 64 Power Supply magic post, I talked about the internal power supply and how not everything that you see when you quickly glance is as obvious as it seems. Yet again I’m faced with a similar situation. I needed an external power supply for one of the machines that I purchased recently, and I didn’t want to pay almost the equivalent of half of the cost of the computer for a new power supply. I searched eBay and come across a “faulty, for parts” unit. Perfect for me, as I’ve got a small “lab” at home and I’m able to troubleshoot and hopefully fix it. It was the 902503-02 part number, that as opposed to some of the “newer” version is not “potted” and can be opened and inspected. Before the auction was over I was “researching” the power supply. My “research” started with googling “902503-02 schematic diagram”. The result brings a few similarly looking diagrams, but here is one of the first hits that I found:

902503-02_PSU_baseline

Well, to me it doesn’t look like the original schematic diagram, it’s more of a reverse engineered diagram that someone has created looking at the real circuit of the power supply. We know roughly what to expect on the basis of what I covered in my previous post – two separate outputs: one is unregulated 9V AC, and regulated 5V DC. And at first glance, it sort of looks ok but in reality, this “diagram” is seriously misleading! Here is what I’ve noticed:

902503-02_PSU_errors

Let’s do the “simple” ones first: the diodes have some strange part numbers, in fact, those should be marked 1N5400. The “ground” symbol from the transformer goes nowhere, as there is only one of them. One of the outputs is marked 7.5A … 7.5A!!?? This should be corrected to 7.5VA – alternatively, the entire thing could be just marked 5V/1.5A. But the biggest and a major mistake is the “transistor” marked SK3052P. Yes, it’s a 3 legged device in a “TO-3P package” that looks like a transistor, but it would be impossible to create a 5V regulator based on a single transistor and a resistor. In fact, the “SK” in the part number is a logo for a Japanese company called “Sanken”, the full part number is “SI-3052P” – a 5V at 2A linear voltage regulator integrated circuit. So the author has done the following “mind shortcut”:

3052p

This is very unfortunate “shortcut” as I’ve seen a few posts on C64 forums where people are referring to that regulator as “transistor”! Well now you know it’s not, and the SI-3052P provides a few important functions: it regulates the output voltage at 5V DC independent of the load and it has a built-in overcurrent protection that will “kick in” when the load exceeds around 2.4A. Based on this new information and after “inspecting” one of those power supplies, I’ve re-created the diagram adding a few more details:

902503-02_PSU_my_v1.0

here is full resolution picture

Let me quickly describe what happens: mains voltage is supplied to the primary winding of our power transformer, notice the “protective earth” pin is somehow connected to the transformer (looks like it goes in the layer between the primary and secondary winding) but it doesn’t extend to the output and is not connected to the computer at all. There are two secondary windings:

  • Pin 9 and 11: provides 9V AC that goes directly to the output. Its protected by a 3A fuse F1, I’ve marked the voltage rating of the fuse (250V) but this is not as important on the secondary side, what is important is to know if this fuse is a “slow” or “fast” acting fuse. Unfortunately, the fuse itself doesn’t have any markings saying what type of fuse it is. Inspecting it physically and considering its function I think it’s a “fast” acting fuse. You may ask why do we need this fuse, isn’t there another fuse inside the C64 that is on the 9V AC power rail. It’s true but the fuse here inside the power supply is supposed to protect the transformer form accidental shortcuts on the output before the power supply is even plugged into the computer.
  • Pin 7 and 8: provides 18V AC with a “center” tab on pin 6. This center pin allows the use of two diodes in a full wave rectifier configuration, saving cost by not having to use a 4 diode full bridge rectifier configuration. Next follows a large smoothing filter capacitor C1. This part of the circuit is protected by 5A fuse F2, yet again there is no marking on the fuse, but considering it has to withstand a larger than standard operating conditions initial charging current of the smoothing capacitor and a clearly different physical appearance I deduct it’s the “slow” acting fuse type.

The unregulated voltage from C1 is regulated using U1, the SI-3052P integrated circuit. On the output of U1 we see an additional small electrolytic capacitor C2, that is recommended by the manufacturer to prevent oscillation. There is one more component, the 300 Ohm resistor R1, it’s used to “raise” the output voltage a bit over the nominal 5V to compensate for the losses on the long wires between the power supply and the computer. This is actually very surprising, as looking at the U1 datasheet, we see Note 3: “The output voltage may not be adjusted by raising the ground voltage (using a diode or resistor)”. This is related to the built-in “foldback overcurrent protection circuit” that will be affected if pin 3 of U1 is not at “ground level”. Looks like the engineers designing the power supply ignored the note. When I get a chance I will try to test and compare the behavior of the power supply in a fault condition, with and without that resistor.

Quick comment on the construction of the power supply. The diodes and fuses are soldered directly to the transformer “pins” (with the help of some additional internally unconnected pins 10 and 12). The rest of the components are mounted on a small printed circuit board (PCB). The transformer and the PCB are marked “Billion”, and I assume this is “contract” manufacturer Billion Electric Co., Ltd. of Taiwan, that Commodore tasked with the creation of this power supply.

Lastly, one should note this is a linear power supply, cheap but very inefficient. At full load (C64 plus accessories plugged into the expansion port drawing in total around 1.4), the input voltage on U1 is over 10V, so to “drop” the voltage to 5V on the output the linear regulator has to dissipate close to 8W of power (voltage on input minus voltage on output multiplied by current). U1 and its heatsink get extremely hot, and as we know, hot semiconductors don’t last long…

In part 2, I will cover the restoration/repair of my eBay purchased power supply, I will share some test results and implement few improvements… stay tuned.

Commodore 64 Power Supply magic.

Note:

At the beginning of this post, I cover some basic power supply stuff, so if you already know a lot skip to the later part where I cover the more interesting 12V DC rail.

Not as obvious as it seems.

While browsing “SAMS Commodore 64 Troubleshooting & Repair Guide” I came across the block diagram of the internal power supply system. It was interesting to me for a number of reasons. First, I’m “into” Commodore restorations those days so this generic knowledge is useful, also I hoping to develop a DC only power supply (so that you could run the C64 off of a 12V car power socket). As a reminder this is the block diagram that I’m talking about, from the book:

sams_guide_c64-powersupply_1

Something wasn’t right here… notice the diode on the +9V supply line. My background is electronics, but even though let’s just say I haven’t “practiced” for a while, the direction of that diode makes no sense, it wouldn’t work. Let’s look at the detail diagram from the C64 service manual:

PDU_ServiceManual_Path

Looks familiar? I’ve highlighted the section that I’m wondering about. Seem like the authors of the troubleshooting guide have made a mistake when creating the block diagram, in fact, the diode CR6 (1N4001) has nothing to do with the unregulated +9V power supply line. Since we are already discussing details of the power supply diagram let’s analyze all sections and how each of the power rails is generated one by one:

9V AC

PDU_ServiceManual_9VAC

The 9V AC power is supplied from the external power supply, passes via a common mode choke filter and a fuse. It is then used to generate all other rails plus is made available on the user port pins 10 and 11. Important to note is that the 9V AC rail is also used to generate the 60Hz/50Hz signal for the hardware real time clock built into the CIA chip. This is one of the reasons we can’t replace the external power supply with a modern DC switch mode alternative.

5V DC (+digital)

PDU_ServiceManual_5VDC_digi

Just like the 9V AC, the 5V DC is supplied from the external power supply, it’s filtered and used directly to power majority of the “chips” on-board. Important: since there are no additional components onboard of the c64 handling the 5V DC digital rail, you’re at the mercy of the quality of the external power supply – and we know those old original power supplies are failing a lot. Typically they fail with an over-voltage condition, resulting in “fried” memory and other components, so always check the condition of the external power supply before you use it for the first time.

5V DC CAN (Clock and ANalog ??)

PDU_ServiceManual_5VDC

You may ask yourself why do we need another 5V DC rail, we already have one from the external power supply. Separate, locally generated power is needed for the sensitive video and clock circuits of the computer. The configuration is “textbook” implementation of the 7805 regulator IC. The 9V AC is rectified using bridge rectifier CR4, smoothed using large electrolytic capacitor C19, decoupling capacitor C95 before getting into the regulator. Regulator output is decoupled with C102 and C103.

12V DC

Now we come to the interesting part of my article.

PDU_ServiceManual_12VDC_0

At first glance, you see the 7812 regulator, some bypass components around it (c59, C57, C89, C88) and fed from the 9VAC. But look closely and you see 3 extra components that are not instantly recognized by most: diodes CR6, CR5 and capacitor C90. What are those for? When I first noticed those I came to the conclusion that it must be used to boost the unregulated voltage coming out of the bridge rectifier CR4 and smoothing capacitor C19 to a level required for correct operation of the 7812 regulator – that is typically around 14.6V DC. What I could see from the diagram was:

  • clamper that consists of capacitor C90 and diode CR6, one should note that in our case this is a “biased clamper” type, as the anode of diode CR6 is not connected to ground but to the 9V unregulated dc supply.
  • peak detector (aka half-wave rectifier) that consists of diode CR5, followed by a smoothing capacitor C88.

Ignoring the fact that we are dealing with “biased clamper”, this combination of clamper and rectifier is a typical example of the voltage doubler.

I started “poking” around my Commodore with an oscilloscope (all captures are with 5V / division setting).

9_all

Higher resolution picture here. (btw one day I will learn how to implement a proper gallery in my posts).

Looking at the oscilloscope we see the AC supply entering C90 (bottom right). The amplitude is around 12V, but we can’t fail to notice we only see the positive half of the “AC” signal. This is due to the fact that we are measuring this in reference to the power supply ground, so one of the diodes of CR4 bridge rectifier is forming a half-wave rectifier.

The output from C90 can be seen in the top right and is “shifted” from the ground by the level of “biasing” voltage, in our case, it’s around 10V DC.

Finally, we can see what happens after our supply goes through the half-wave rectifier CR5 and smoothing capacitor C88. Note the output DC voltage that is present on C88 is over 20V. That means that C88 must be rated at 25V minimum. I mention that as I’ve seen people commenting that when “re-capping” capacitors on board of the Commodore you can stick to 16V rated electrolytic capacitors – as you can see this is not true.

All this “complications” mean that it’s not easy to replace the external power supply with a simpler DC switching power supply. Without modifying the internals of your Commodore you will need both DC 5V digital rail and separate 9V AC rail.

Final thoughts.

I hope this was informative. Note that what I described above is what I think is happening, and I’m around 95% sure, but feel free to comment if you think I’m wrong and I will fix it.

I’m working on more Commodore related “material”, so more will follow.