[Note: I saw Hans' message on comp.sys.hp.misc and replied privately, but figure there might be some other folks here who want to know this stuff.] Hans Pufal <hans1(a)filan00.grenoble.hp.com> wrote: Time for some copying from the service manual. Useful information below is from the HP-97 Programmable Printing Calculator Service Manual, HP part number 00097-90130, dated August 1976. Table 4-9. Keyboard Printed-Circuit Assembly A2A1 (00097-60002) Replaceable Parts Ref HP p/n Description R1 0683-4715 RESISTOR, fxd, 470 ohm R2* 0683-2045 RESISTOR, fxd, 200K R2* 0684-3341 RESISTOR, fxd, 330K R3 0812-0058 RESISTOR, fxd, 8.2 ohm, 2W R4 0811-1674 RESISTOR, fxd, 4.7 ohm, 2W R5 0698-8691 RESISTOR, fxd, 4.0 ohm, 1% R6 0683-1835 RESISTOR, fxd, 18K, 5%, 1/4W R7 0683-3915 RESISTOR, fxd, 390 ohm CR1 1990-0450 LED, low battery indicator Q1 1853-0393 TRANSISTOR, PNP Q2 1853-0401 TRANSISTOR, PNP Q3 1853-0374 TRANSISTOR, PNP Q4 1854-0071 TRANSISTOR, NPN U1 1820-1629 INTEGRATED CIRCUIT, cathode driver U2 1990-0595 DISPLAY, numeric U3, 4 1858-0044 INTEGRATED CIRCUIT, quad transistors U5 1810-0252 INTEGRATED CIRCUIT, resistor network P1, 2 1251-3955 CONNECTOR, 9-pin W1 8120-2206 CABLE, 24-conductor 00097-80002 BOARD, etched R2 is marked with a * because it is selected based on the cathode driver IC U1. I don't completely get this, but there is a chart (Table 4-8, Cathode Driver Resistor Selection Chart): U1 Category: I J R2: 200K 330K Given that you say the burned out resistors are "large" I am going to guess that you are looking for R3 and R4. Here's the layout toward the rear of the board if that helps. Q4 R7 R5 Q2 R6 Q3 R2 ---U5--- R1 ---U4--- ---U3--- -------- -------- ---U1--- -------- -----------J1----------- Q1 R3 R4 12.8VAC (from paragraph 2-42). Actually there is a flowchart that includes a test for the AC adapter. For approximately 115VAC input, the output should be between V(in)/10.55 and V(in)/7.82; for 230VAC input, the output should be between V(in)/21.1 and V(in)/15.65. The next part of the test is to connect a 10 Ohm +/- 5% 5W load across the output of the AC adapter and measure the voltage across the load; it should be between 6.2 and 7.4VAC. Hope this helps! NB- I read your message in comp.sys.hp.misc. It looks like you might have posted it to classiccmp too. I am a bit behind reading that. Feel free to repost this there. -Frank McConnell