A PROTECTION DEVICE FOR THE COMMODORE 64 04-05-06.
A PROTECTION DEVICE FOR THE COMMODORE 64 Latest updates and/or corrections 4-15-06 This over-voltage protection unit is designed to prevent damage to the Commodore 64 computer due to a failing power supply. One of the ways the Commodore "black brick" supply can fail is from a shorted internal 5 volt regulator. That fault puts excessive voltage into the computer and damages chips such as RAM very quickly... and silently. The protector is essentially a fast acting electronic circuit breaker. It functions to quickly cut off the 5 volt supply line to the computer if the voltage exceeds about 5.5 volts. Built from readily available parts, the protection circuit is installed in series with the 5 volt line between the power supply and the computer. The use of a power hungry add-on like a RAM expansion unit is not recommended with the original black brick power pack. That RAM unit originally shipped with its own power supply, one that provided 2.5 Amps for the 5 volt line. That beefier supply makes the computer saver unnecessary since it's a switcher and not vulnerable to failure like the brick. If you have an expander but no supply for it, consider getting a C128 power supply and installing a power connector that fits your C64. With that hookup, again, you don't need a saver. There are several ways to add the protection device to your system. One way is to install the components inside the computer itself. That way, it is protected from any supply that is plugged into it. If you have more than one computer, a protector can be installed in each one. Cost of parts should be less than US$10 for each unit. Since the parts count is low, I didn't bother making a circuit board for my internal protectors. I used some silicone rubber sealer as glue to hold the relay in place on the motherboard, then wired the other components on top of it. Wires to and from the added components complete the installation. Look on my schematics site for photos of three installations in several versions of the C64. They are located in the C64 and C64C directories at http://staff.washington.edu/rrcc/uwweb To install the protector circuit inside a C64, it is necessary to open the +5 volt line on the circuit board. Input and output lines of the protector are wired to each end of the now open circuit. If your C64 board differs from the pictures on my site, you will have to locate the correct area of your board to make the connections. In the brown case C64, look for a small coil called L5. Disconnect one end of that part and install the input and output connections of the saver across the open circuit as shown in the picture. On a C64C short board, a PC board trace must be cut to install the saver. As shown in the picture, the copper trace near capacitor C58 is the place to cut and install the new connections. It is important that the input and output lines of the protector go to the appropriate places on the board. The easiest way to identify which is which: disconnect L5 (or cut the trace if a C64C short board) and measure continuity (zero ohms) with an ohmmeter from pin 7 of any of the RAM chips and one of the open connections at L5 (with the power switch on and PS disconnected, of course). The -output- of the protector (the normally open relay contact) goes to that connection. The -input- side of the protector gets +5V directly from the power supply through the added fuse. As an alternative, the protector can be built in a small box as a stand-alone device. That way, you can use it with any supply or computer, but that has the disadvantage of needing two DIN connectors and an enclosure. If the protector box is wired directly onto the power supply cable, that eliminates the need for the added connectors, but it will work with only that supply unless physically transferred. You can choose the implementation that works best for you. The circuit is the same regardless of how it's installed. The stand-alone protector requires a 7 pin male DIN plug on one end that fits the power socket on the computer, and a 7 pin female DIN socket on the other to receive the plug from the power supply. Note that although a 7 pin connector is used, only four pins are active. The socket can be an in-line or chassis mount connector, but the plug should be wired to a short length of cable which allows the protector box to be moved out of the way of the user. My prototype model used a cut off 2' length of cable from an old dead supply. I found a chassis mount female DIN socket from a junked piece of electronics gear and installed it in the protector box. A male DIN plug was pigtailed out of the other side of the box. If the protector is going to be added directly to the supply cable: route the cable through the box by notching the ends of the box to allow a loop of cable to go through it when the cover is reattached. Cut open the outer jacket of the loop of cable in the box. The red wire (+5V) inside the cable needs to be cut and the ends wired to the protector input and output lines. The end from the supply goes to the input of the protector and the one to the computer goes to the output end. The uncut black wire (ground or negative lead) should have a small bit of insulation removed so it can be attached to the protector circuit ground. The other two wires of the supply cable (usually blue and brown) are the 9VAC lines and should not be opened. How it works: The circuit for the protector is simple and strightforward. A 4.7 volt zener diode on the 5 volt line continuously monitors that DC level. The resistor in series with the zener diode limits the current to the base of the first transistor Q1, and the resistor from base to ground keeps the transistor cut off until the threshold voltage is exceeded. The zeners 4.7 volts plus Q1 transistors 0.6V B to E drop totals slightly over 5 volts, the trip point of the device. If the supply voltage rises above that level, the zener diode conducts, turns on the transistor and pulls its collector lead to ground. That lead is wired to a normally conducting second transistor Q2 which drives a relay. The diode across the relay coil is there to protect the transistor from back EMF "spikes" from the coil which could destroy the transistor. The relay contacts are used to pass the +5 supply voltage to the computer when the relay is energized in normal operation. The relay is de-energized only during an over-voltage condition. If the supply fails in that manner, the relay drops out and opens the previously closed contacts cutting off the +5VDC supply voltage to the computer. The circuit is designed to keep the 5 volt line open as long as the fault persists. It is self-resetting. There is one fuse inside the C64 that protects the 9VAC line but neither the C64 or standard Commodore PS "brick" 5VDC lines are fused. One has been added to the protector circuit. That fuse will open if an external overload (short) is accidently applied to the computer ports or if an internal component such as an IC shorts in the computer. The new fuse protects the saver relay (too much current can weld the contacts together) as well as the power supply from excessive current draw. Some notes about the parts used: The relay is a SPDT (single pole, double throw) but an SPST (single pole single throw) can substitute if the indicator LEDs are not used. It has a 5 volt coil and contacts rated at 2 Amps. Since the computer only draws about 800mA, a 1 Amp relay would also work but a margin of safety is a good idea. This design is different than a previous one that turned on the relay during an overvoltage fault condition. One problem with that design is that a 5 volt relay will overheat and eventually burn out if left connected to a failed (internal regulator shorted) power supply. That is because the voltage goes upwards to 12 volts which will be more than twice normal operating voltage. In such a design, if the coil burned out, the relay would deactivate and the computer would not be protected. That's why the design was changed. It's best to use a relay with a 5 or 6 volt coil, but one that has a high DC resistance. Relays used in my early protectors had coil resistances of about 60 ohms. One drawback to that is it puts an additional load on an already marginal power supply. The higher the DC resistance of the relay coil, the less load it represents, so a coil resistance of 200 ohms or more is preferable in this application. The transistors can be any low power silicon NPN type rated at 500mA or more. The voltage rating is not critical either. Generic types such as the 2N2222 or 2N3565 are OK as is their equivalent such as the old Sylvania ECG123A or NTE123A. A higher power NPN transistor will also work, such as tab mount or TO-220 "flatpack" types although they are physically larger. Note also that the metal tab or case of such a transistor is normally connected to one internal element... usually the collector, so its metal case must be isolated from surrounding metal or wiring. The zener diode can be a 1/2 or 1 Watt but its voltage rating is important since that sets the "trip" point of the device. The silicon diode across the relay coil is a standard power diode (1N2004) but can be a small signal type such as a 1N914. Its current and voltage ratings are not critical as it's only there to absorb "spikes" from the coil when it de-energizes. Polarity is critical for any diode. They usually have a band around the cathode (K) end. The resistors are all 1/4 to 1/2 Watt carbon or carbon film types. 1 Watt resistors of the same resistance value can also be used although they will be physically larger. Space is limited inside the computer, especially in the C64C which has a metal shield over the PC board. Make sure nothing shorts out when you put that cover back on!!! If desired, LEDs or a buzzer can be used to monitor the relay status and indicate the fault condition, but their use is optional. A green LED may be used to indicate normal operation and a red LED (or alarm buzzer) indicates PS overvoltage. You can substitute a buzzer or other 5 to 9 volt alarm device for the RED LED if desired. In such a case, eliminate the series resistor as its only function is to limit the current to the LED if one is used. 1.5A fuse ___~~~____________________________________ +5VDC | | | C | com from \ | K_|_ O 0 PS / 470 | A /\ I \ relay \ ohms | | L 0 \0 4.7V K | / |___| NO| |NC zener \__|__ \ 2.2K | |--------- +5V out to diode A /\ \ / | | | computer | \ | \ / | | | / \ two | | C__| \ / 470 ohm | | |/ / \ resistors | |____| Q2 \ / | __| B | | | | |/ C |\__ |A |A |_______| Q1 E | green _\/_ _\/_ red | B | | |K |K / |\__ | | | indicator 470 \ E| | | | LEDs ohms / | | | | | | | | | GND (-) __|___________|________|_________|_____|_____ GND Ray Carlsen CET Carlsen Electronics