(*)Note: I edited this message to correct an error. When I measured the output current of the constant current supply, I connected the milliameter between pin 10 and pin 6, not pin 10 and pin 4. 6/24/2003 3:00 pm
(**)Corrected another error - output voltages measured relative to pin 6 and not pin 4. 6/24/2003 6:40 pm
Now that I have an earlier 34C working, I see what the "2" disease is! I would describe the condition this way: whenever you press the "2" key when there is someplace for a 2 to go in the display, or there is a 2 in the display and you press any key, the minus sign and possibly various commas flash momentarily. If you type ten digits other than 2, filling the display, then pressing "2" doesn't cause flashing. If you type a number in scientific notation without any 2's in the mantissa, you can make the flashing start and stop by shifting a 2 through the exponent; in this case the flashing commas are limited to the exponent digits. My newer unit doesn't do this "2" flashing at all.
The other flashing my older unit does is to flash random commas and the minus sign during execution of built-in functions. Calculating the gamma function (! of non-integers) of negative numbers with large magnitude seems to take the longest. My newer unit is completely dark while executing a built-in function. You can make the calculator work longer with the Integrate function, but then both units flash random digits when the user function subroutine is being evaluated.
I measured the power supply outputs and current drain with no load (power supply module on the bench) and in circuit (calculator assembled). I made the no load measurements with two different input voltages to see the effect of the voltage and current regulation circuits in the power supply module.
First, I connected two NiCads in series to the power supply module, (-) to pin 6 and (+) to pin 4 (see the netlist below to see how I numbered the pins). Both pin 7 (the standby supply that powers the 8 pin IC closest to the power supply module) and pin 8 (which supplies power to the other two 8 pin ICs and the 40 pin IC) measured 7.12V relative to pin 6 (**). Pin 10 of the power supply module, which goes to pin 8 of the 40 pin IC and which is driven by a constant current circuit in the power supply module, measured 4.62V and when connected to pin 6 (*) through a milliameter, it sourced 39 mA. The current drain from the battery with no load was 8 mA and when the constant current supply was loaded, the current drain was 106 mA and the voltage on pins 7 and 8 increased to 7.46V.
Next, I added another NiCad, increasing the input voltage from 2.58V to 3.88V. The no load current drain increased to 9 mA. The voltage on pins 7 and 8 was 7.14V. The voltage on pin 10 was 5.22V. Connecting a milliameter between pin 10 and pin 6 (*), I measured 38 mA, the voltage on pins 7 and 8 increased to 7.28V and the battery current drain increased to 76 mA. So the no load output voltages on pins 7 and 8 and the current sourced by pin 10 change very little with a 50% increase in input voltage.
The constant current supply is similar to the one in the Classic recharger. It works by maintaining a voltage drop equal to the emitter-base junction voltage of a silicon transistor (~.6V) across a 16 ohm resistor: (0.6V)/(16 ohms)=(0.0375A) or ~38 mA. The circuit in the Classic recharger (which provides the current that recharges the battery) uses a 13 ohm resistor and provides ~50 mA.
Next, I assembled the calculator and measured the voltages that are accessible that way. I couldn't get to the pins connected to the constant current supply because of the molded plastic frame but I could measure the two voltage supplies. With power supplied by two NiCads, the voltage on pin 8 of the 8 pin IC closest to the power supply module measured 6.82V and the voltage on pin 8 of the other two 8 pin ICs measured 6.52V, both relative to pin 7 of the 8 pin ICs (which are connected to the negative terminal of the battery).
I also measured the voltages on the other pins of the 8 pin chips with the calculator idle and running a program. Pin 6, which goes to the unknown jumper on the power supply module and to pin 14 of the 40 pin chip, measure 5.33V either way. If the unknown jumper were installed, it would connect the pins to ground. Pin 5 measures 0.65V idle and 0.88V running. Pins 4 and 3 measure 0.86V either way. Pin 2 measures 0.86V idle and 1.35-1.41V running. Pin 1 measures 0.04V idle and 0.3-0.5V running. All were measured with a digital voltmeter with respect to the battery negative terminal (pin 7 of the 8 pin IC's). I'll have to fire up one of my many scopes and see if some of those "quiet" lines have low duty-cycle signal activity on them.
Other than the power supply, the schematic is just connections between the pins of the IC's, LED and keyboard that can be communicated better as a netlist. I've typed that up and include it here:
;HP34C Main Circuit Assembly (solderless version) netlist-;component reference designators and pin numbering
IC1=8 pin chip with standby power (closest to the power supply module)
IC2=second 8 pin chip
IC3=third 8 pin chip
IC4=40 pin chip
PS=power supply module with the flex cable solder pads on its PCB numbered from left
to right as viewed from the rear. Note: pin 1 only goes to the recharger connector,
it doesn't go to the main circuit assembly. The pads where the flex cable contacts
the main circuit assembly flex circuit are PS.2 - PS.10.
LED=LED display with pins numbered from left to right as viewed in normal use
PRSW=Program-Run switch, pin 1=common, pin 2=closed on Run
OOSW=On-Off switch, pin 1=common, pin 2=closed first when switched to "On",
pin 3=closed second when switched to "On"
KBR=keyboard rows numbered in standard keycode format
KBC=keyboard columns numbered in standard keycode format for rows 3-7,
offset by -1 for rows 1-2
AC=AC connection from battery contact assembly to power supply module (return is bat-)
(doesn't go to main circuit assembly);memory chip connections
v+standby = PS.7,IC1.8
v+ = PS.8,IC2.8,IC3.8,IC4.12,PRSW.1
bat- = PS.6,IC1.7,IC2.7,IC3.7,IC4.13,LED.22; LED.22 is minus sign cathode, anode of ???
diode with cathode on LED.1 (goes to psin)
mem6 = PS.5,IC1.6,IC2.6,IC3.6,IC4.14; memory bus, ??? jumper connects to bat- if present
mem5 = IC1.5,IC2.5,IC3.5,IC4.15; memory bus
mem4 = IC1.4,IC2.4,IC3.4,IC4.16; memory bus
mem3 = IC1.3,IC2.3,IC3.3,IC4.17; memory bus
mem2 = IC1.2,IC2.2,IC3.2,IC4.18; memory bus
mem1 = IC1.1,IC2.1,IC3.1,IC4.19; memory bus;control signals
run = PRSW.2,IC4.20; Program:open,Run:switch connects to v+
radix = PS.9,IC4.9; comma:open,period:jumper on power supply module connects to v+;another power supply module output
iled = PS.10,IC4.8; 38 mA constant current supply - for LEDs?;LED anodes
antr = LED.21,IC4.3; top right digit segment
ant = LED.20,IC4.2; top digit segment
antl = LED.19,IC4.1; top left digit segment
anm = LED.18,IC4.4; middle digit segment
anb = LED.17.IC4.5; bottom digit segment
anbl = LED.6,IC4.36; bottom left digit segment
ancom = LED.5,IC4.40; comma
anbr = LED.4,IC4.39; bottom right digit segment
anper = LED.3,IC4.38; period
anms = LED.2,IC4.37; minus sign anode (cathode is on LED.22, goes to bat-);LED cathodes, keyboard rows(numbered in standard keycode format)
ca1 = LED.16,IC4.21,KBR.7; digit 1 (leftmost), keyboard row 7
ca2 = LED.15,IC4.23,KBR.6; digit 2, keyboard row 6
ca3 = LED.14,IC4.26,KBR.5; digit 3, keyboard row 5
ca4 = LED.13,IC4.28,KBR.4; digit 4, keyboard row 4
ca5 = LED.12,IC4.29,KBR.3; digit 5, keyboard row 3
ca6 = LED.11,IC4.30,KBR.2; digit 6, keyboard row 2
ca7 = LED.10,IC4.32,KBR.1; digit 7, keyboard row 1
ca8 = LED.9,IC4.33; digit 8
ca9 = LED.8,IC4.34; digit 9
ca10 = LED.7,IC4.35; digit 10 (rightmost);keyboard columns(numbered in standard keycode format rows 3-7, offset by -1 rows 1-2)
kc0 = IC4.31,KBC.0; "A","X<>Y"(column 1)
kc1 = IC4.27,KBC.1; "B","GTO"(column 2),"ENTER","-","+","*","/"(column 1)
kc2 = IC4.25,KBC.2; "GSB","STO"(column 3),"CHS","7","4","1","0"(column 2)
kc3 = IC4.24,KBC.3; "f","RCL"(column 4),"EEX","8","5","2","."(column 3)
kc4 = IC4.22,KBC.4; "g","h"(column 5),"CLX","9","6","3","R/S"(column 4);power switch
psin = OOSW.1,PS.4,LED.1; + input to PS, LED.1 is cathode of ??? diode
with anode on LED.22 (connected to bat-)
bat+ = OOSW.2,PS.3; makes first when switch turned "On"
chrg = OOSW.3,PS.2; makes second when switch turned "On",
to short out 8.2 ohm current limit resistor on power supply module;AC adapter (doesn't go to main circuit assembly - goes only to power supply module)
acsource = AC.1,PS.1
Now all that's left is a schematic of the power supply module. I will attempt to write a description of the circuits on the power supply module with enough details to draw the schematic.
HP34C Power Supply Module circuit description-
The battery negative terminal and the AC adapter share a common return connection on pin 6 of the power supply module; this is also the return for the main circuit assembly. The other side of the AC adapter is connected to pin 1, which isn't connected to the main circuit assembly. Pin 1 goes to the anode of a rectifier diode whose cathode goes to pin 2. Pin 2 goes to an 8.2 ohm, 1 watt resistor which goes to pin 3. Pin 3 goes to the battery positive terminal. This completes the circuit which charges the battery when the unit is turned off.
Pin 4 is the input to the switching power supply and is connected to the common terminal of the On-Off switch. The switch has two other terminals: one makes contact with the common terminal first when the switch is turned to "On" and it is connected to pin 3, the battery positive terminal. The other switch terminal makes contact with the common terminal second when the switch is turned to "On" and it is connected to pin 2; its purpose is to short out the 8.2 ohm battery charging current limiting resistor when the unit is on. This completes the description of how power from the battery and the recharger is connected to the power supply input.
Pin 5 only goes to a pad which can be connected by a jumper (not installed) to pin 6, the battery negative terminal. On the main circuit assembly, pin 5 is connected to pin 6 of the 8 pin IC's and pin 14 of the 40 pin IC. The purpose of the jumper option is unknown. Pin 7 provides standby power to pin 8 of the 8 pin IC closest to the power supply module. The voltage on pin 7 is supplied by two sources: first, it is connected to the battery positive terminal to provide memory backup when the unit is off - through a diode (anode to the battery positive terminal, cathode to pin 7) on my older unit, or through a 10K resistor on my newer unit. To support this, there is a 22 mf capacitor from the battery positive terminal (cap + lead) to the battery negative terminal to provide backup power while changing the battery. The second source of power for pin 7 is a voltage regulated output of the switching power supply which also provides power to pin 8 for the other chips on the main circuit assembly. Both pins are connected to the supply through separate diodes (anodes to the voltage regulated supply, cathodes to pins 7 and 8). When the unit is on, the voltages on pins 7 and 8 are approximately +7V. Pin 9 only goes to a pad which can be connected by a jumper to pin 8. On the main circuit assembly, pin 9 goes to pin 9 of the 40 pin chip. When the jumper is installed, the calculator uses period as the radix mark; when it is not installed, comma is used. Pin 10 is connected to a constant current supply which sources approximately 38 mA and is powered by the switching power supply. On the main circuit assembly, this supply goes to pin 8 of the 40 pin chip. I believe this current source is used to drive the LED segments. This completes the description of the terminals of the power supply module. What remains to be discussed is the switching power converter, the regulated voltage supply and the regulated current supply.
The power converter transformer is wired as an autotransformer with a primary winding and two secondary windings connected in series with the primary winding. An additional independent winding provides feedback for the oscillator that drives the primary winding. The power supply input (from pin 4) is connected to the top of the primary winding (= the bottom of the secondary windings) and the bottom of the primary winding is connected to the collector of an NPN transistor with its emitter connected to the battery negative terminal (pin 6). The base of the transistor is biased on by a 680 ohm resistor from the power supply input. One end of the feedback winding is connected to the battery negative terminal and the other end is connected to the base of the NPN transistor through a series network of a 2000 pf capacitor and a 47 ohm resistor. The end of the higher secondary winding in series with the primary winding, with a voltage equal to the power supply input voltage (the battery positive terminal voltage) plus the sum of the voltages across the two windings, is connected to a rectifier diode anode. A 22 mf capacitor (33 mf on my newer unit) is connected from the diode cathode (cap + lead) to the battery negative terminal.
This capacitor positive lead is connected to the anodes of the two diodes that supply the regulated +7V to pins 7 and 8 and also to the cathode of a zener diode. The anode of the zener is connected to a 1 mf capacitor (+ lead) with its other lead connected to the battery negative terminal. The anode of the zener is also connected to a 1K resistor which goes to the base of a smaller NPN transistor with its emitter connected to the battery negative terminal. The collector of this transistor is connected to the base of the oscillator transistor. When the output voltage of the supply rises high enough to cause the zener diode to conduct, the smaller transistor turns on and shunts some bias current (provided by the 680 ohm resistor) away from the base of the oscillator transistor, reducing the amplitude of the oscillation and the output voltage. This completes the description of the switching power converter and the voltage regulated supply.
The remaining tap of the transformer - the junction of the two windings in series with the primary winding - is connected to the anode of a rectifier diode. A 22 mf capacitor is connected between the cathode of the diode (cap + lead) and the battery negative terminal. This provides a supply of something less than 7V for a constant current source. The diode cathode and the cap + lead are connected to a 16 ohm resistor which goes to the emitter of a PNP transistor with its collector connected to pin 10, the current regulated output. The base of the transistor is connected through a 2.2K resistor to the battery negative terminal; this biases the transistor on. A smaller PNP transistor has its base connected to the other transistor's emitter, its emitter connected to the other end of the 16 ohm resistor, and its collector connected to the other transistor's base. When the voltage across the 16 ohm resistor exceeds approximately 0.6V, current flows through the smaller transistor's emitter-base junction and it turns on, shunting current away from the larger transistor's base and reducing its collector and emitter current. Since 0.6V divided by 16 ohms = .0375 amps, the circuit reaches equilibrium with about 38 mA flowing out of the larger transistor's collector towards the battery negative terminal.
Edited: 24 June 2003, 7:44 p.m. after one or more responses were posted