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The Restoration of the PDP-12 Minicomputer (PAGE 2)


Capacitor Reformation -- 2003 11 14


Just as with the restoration of the PDP-8/I, I was assured by the seller that the machine had not been powered on in a long time.. This immediately means one thing -- reforming the power supply capacitors.

Having learned from the restoration of the PDP-8/I, I decided to reform the capacitors by using a trickle charge rather than the variac approach.

The PDP-12 power supply is very similar to the power supply of the PDP-8/I -- it uses the same big honkin' capacitors, and the same transformer. The capacitors are mounted in almost the same way -- on the PDP-8/I the connections were done directly onto the capacitors, but on the PDP-12 they are done to a PCB and the capacitors' screw terminals attach to the PCB. This makes it much easier to remove the PDP-12 capacitors. The capacitors were removed and labelled; reformation will take place over the next few days.


PDP-12 Power Supply Closeup S M L XL

In this shot, you can see the three main capacitors in the PDP-12 power supply (bottom-left to bottom-middle). Left to right, they are rated at 160mF/20V, 160mF/20V, and 57mF/50V (mF == milliFarad, or 1000 uF).

For comparison, here's the PDP-8/I power supply:

PDP-8 Power Supply Removed from Cabinet S M L XL

Capacitor Reformation Jig Schematic S M L XL

This is the schematic diagram for the "capacitor reformation jig" that I built on a small prototyping board. It's connected to 40V (tacky; I just connected it to the three 12V batteries of my UPS), and uses 4.7V zener diodes (1N750A or 1N4732 for example) to create a variety of voltages that can be used with various voltages of capacitors. There are series resistors for limiting the current to 5mA.


When reforming capacitors, it's a good idea to measure the voltage periodically. This way, you can draw a graph and determine the capacitance of the capacitor. For example, here are the values from the 57mF/50V capacitor:
TimeVoltage TimeVoltage TimeVoltage TimeVoltage
10:18:000.5 10:18:302.3 10:19:004.0 10:19:305.45
10:20:006.63 10:20:307.55 10:21:008.30 10:21:309.00
10:22:009.62 10:22:3010.19 10:23:0010.74 10:23:3011.26
10:24:0011.75 10:24:3012.21 10:25:0012.65 10:25:3013.06
10:26:0013.46 10:27:0014.19 10:28:0014.86 10:29:0015.48
10:30:0016.04 10:35:0018.22 10:40:0019.72 10:45:0020.84
10:50:0021.73 10:55:0022.45 11:00:0023.08 11:30:0025.27
12:00:0026.47 12:30:0027.12

Note that the measurements were initially done at 30-second intervals, then 1-minute intervals, then 5-minute intervals, and finally at 30-minute intervals. This is because capacitor charging is an inverse exponential function. The voltage applied to the capacitor is 31.13V (in later reformation phases, it's 40V -- I had to adjust my reformation jig by adding the 8k2 resistor; also the voltage should be closer to the 50V limit, but in the PDP-12 the capacitor is only used for a 30V power supply).


Calculating the Capacitance

Graph of 57mF/50V capacitor being charged from 31V source S M L XL

This is the graph of the charging function of the 57mF/50V capacitor. We'll use these values to calculate the capacitance. In order to do the math, you need to first calculate the time constant (known as "tau") of the capacitor. This is simply the product of the capacitance (in Farads) and the resistance (in Ohms). For my reformation jig, the resistor used is 6k8 and the capacitance is 0.057F. Multiplying these two numbers yields a time constant of 0.057 × 6800, or 388 seconds. This time constant is the amount of time that it will take the capacitor to reach 63.2% of the voltage differential between its current voltage and the charging voltage. Using this time constant, we can verify that the capacitor has the expected capacitance. Taking a sample value, let's say the value at 10:18:00 (0.5 volts), the difference between the charging voltage (31.13 volts) and the current voltage (0.5 volts) is 30.63 volts. Taking 63.2% of that value gives us 19.36 volts. This means that 388 seconds after 10:18:00, we would expect the capacitor to have reached 19.36 volts plus the initial voltage of 0.5 volts, or 19.86 volts. 388 seconds is 00:06:28, so we would expect to see 19.86 volts at 10:18:00 + 00:06:28 (or 10:24:28). However, you'll notice that at 10:24:30 we have a value of 12.21 volts -- far lower than the expected value.

What does this mean? Well, it can mean one of two things; either the capacitance of the capacitor is much larger than the nominal value written on the capacitor itself, or it means that the capacitor is taking longer to charge as a result of reformation. I'm pretty sure it's the latter, because this capacitor hasn't been used for a long time, so I really hope that it's busy reforming :-)

For the sake of argument, let's pretend that this is a perfectly reformed and functioning capacitor. What capacitance would the 12.21 volt value imply? Since the time constant is the product of the capacitance and the resistance, we need to solve for the capacitance:

tau = C × R

therefore:

C = tau ÷ R

We want to solve for C to find the new capacitance having observed a different time constant than what we expected. It's easy to find the time that corresponds to 19.86 volts -- that happens at (guessing a little) 10:41:00. This is 10:41:00 - 10:18:00 or 1380 seconds (rather than 388 seconds as above). Plugging the number 1380 into the formula gives us:

C = tau ÷ R
  = 1380 ÷ 6800
  = 0.2029411764

or about 203 mF.

You can also calculate the capacitance based on the discharge rate of the capacitor. In this case, the graph of the discharge would look like the graph above, except with the Y-axis inverted (i.e., it starts at a high voltage and drops down quickly to a lower voltage, but then smooths out as it approaches zero volts). You'd still calculate the same time constant based on the capacitance and the resistance, but this time, the time constant indicates how long it should take the capacitor to go down to 1 - 63.2% (or 36.8%) of the voltage.

Suppose that our capacitor had been charged to 30 volts. What voltage should it have after one time-constant period has elapsed? The answer is easy -- 36.8% of the voltage, or 11.04 volts.

In my case, I discharge the capacitor through a much lower resistor than the one I use to charge it -- a 100 ohm resistor. This means that the new value for the discharge time constant is:

tau = C × R
    = 0.057 × 100
    = 5.7 seconds
Therefore, I should see a drop from 30 volts to 11.04 volts in 5.7 seconds. I measured 8 seconds when doing this experiment, which means that the capacitance is actually 8 ÷ 5.7 times as great, or 80mF. If you look on the capacitor can itself, sometimes they give you the tolerance of the value. On one particular capacitor in my collection, it states the tolerance as "-10% / +75%" (which is quite a range!). Therefore, an actual value of 80mF on a 57mF capacitor is within tolerance.

You can perform the same measurement with the capacitor when you've discharged it to 11.04 volts as well. After one time-constant, it should go from 11.04 volts to 36.8% of 11.04 volts, or about 4.06 volts. I did this too, and once again it took 8 seconds, confirming the above readings.
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