Fix AC input mains curcuitry, replace electrolytic capacitors, replace transformer
|Target version:||Keithley 2001 #6|
Initial inspection results:
Cut earth wiring
Blown SMD resistors
Blown front panel current input fuse
Electrolyte on PCB near FE autozero and mains curcuitry
Damaged 555 generator SO8
Test digital board
Test controls and front panel display
#2 Updated by tin over 6 years ago
- Due date changed from 30/12/2012 to 31/12/2012
- % Done changed from 0 to 10
Digital board work fine
Front panel/display work fine
Analog board visible inspection:
Q101, VR101, R100, R123, R250 damaged/burnt
Whole AC area side covered in electrolyte/dust from bad C106
S100, Q528 damaged?
#7 Updated by tin over 6 years ago
- % Done changed from 10 to 30
2.10.2 Power supply circuit theory¶
The following information provides some basic circuit theory that can be used as an aide to troubleshoot the
The pre-regulator circuit regulates power to the transformer.
When power is applied to the instrument, a power transformer secondary voltage (pins 12 and 13)
is rectiÞed (CR622), doubled (C624, C630, CR624 and CR625) and applied to U619 which is a +5V regulator.
This +5V (+5VC) is used for the pre-regulator circuit. The pre-regulator circuit monitors the voltage level on C611 using an integrator (U627). The voltage on C611 (typically around 7.5V) is divided by three through
R712 and R713 and applied to the inverting input (pin 2) of the integrator. The +5V (+5VC) is divided by two
through R706 and R708. This 2.5V reference is applied to the non-inverting input (pin 3) of the integrator.
When the voltage on the inverting input of the integrator is less than the 2.5V reference on the on-inverting input, the integrator output ramps in the positive direction. This positive ramp turns on Q608 which pulls the CONT line low to digital common. With CONT connected to common, current ßows through the photodiode of U100 and generates a positive voltage at the gate of FET Q528. As Q528 turns on, the 470W resistor (R100) becomes shunted and results in less effective resistance to the transformer. The resultant increase in current (power) will increase the voltage on C611. Conversely, when the voltage on the inverting input of the integrator is more than the 2.5V reference, the integrator output ramps in the negative direction and begins to turn Q608 off. This will decrease current through U100, decrease the positive voltage on Q528 and thus, increase the effective resistance to the transformer.
The resultant decrease in current (power) will decrease the voltage of C611. This constant regulation of effective resistance in series with the transformer regulates the power delivered to the instrument.
Line voltage (110V/220V) selection circuit¶
This circuit automatically selects the proper power line voltage setting for the instrument. The line selection circuit derives its power from the AC1 and AC2 lines on the primary side of the transformer. RectiÞer CR101 applies approximately +18V to regulator U103. The output of U103 provides the +8V for the line voltage selection circuit and the HI/LO voltage control circuit. U106 is a comparator that has a +4V reference (via voltage divider R125 and R126) applied to its non-inverting input. The inverting input monitors the voltage on C111.
When the voltage at the inverting input is greater than 4V, the output of U106 goes low and turns on FET
Q103. With Q103 on, +8V will be applied to the +RELAY1 line which energizes relay K101 to select the 110V setting. Conversely, when the voltage at the inverting input is less than 4V, the output of U106 goes high and turns off Q103. With Q103 off, the +8V is removed from K101 and thus, the line voltage setting defaults to 220V.
The AC power line is tied to C111 through CR104, R227 and R114 via control line ACL. When the AC power line voltage is less than approximately 135VAC, sufficient charge remains on C111 to keep the inverting input of U106 above 4V to ultimately energize K101 (110V setting). When the AC power line voltage is greater than approximately 18VAC, charge will be pulled from C111 dropping the voltage at the inverting input of the comparator to less than 4V. This will de-energize K101 (220V setting).
HI/LO voltage control circuit¶
This circuit automatically selects the appropriate HI/LO setting for the available power line voltage. During power-up, the line voltage is rectiÞed (CR100), divided (R103 and R105, or R102 and R105) and applied to the
base of Q101. If the voltage level at the base of Q101 is high (above zener VR101), the transistor will turn on and apply power to the ISO1+ and ISO1- lines. With power applied to ISO1+ and ISO1-, U105 will turn on and allow Q105 to be forward biased. With U105 and Q105 on, TRIG of U110 will be pulled low and allow its output
(OUT) to latch at +8V which will turn on FET Q102. With Q102 on, the -RELAY2 line will be connected to common, and thus energize K100 (HI setting). If the power line voltage decreases to a low level, U105 will turn off, but the output of U110 will remain latched at +8V. However, the LOW line will be driven low turning on U109. With U109 and Q106 on, +8V will be applied to THR of U110 forcing its output (OUT) to reset
to low. With the gate of Q102 low, the FET will turn off and open the relay coil circuit for K100 (LO setting). The LOW line is controlled by comparator U628.
The inverting input of the comparator is connected to the 2.5V reference. The non-inverting input monitors (via divider R709 and R711) C611. As previously explained, the typical power line voltage level will apply around 7.5V to C611. However, if the line voltage decreases such that the voltage on C611 becomes less than 6V, the voltage level on the non-inverting input of the comparator will drop below 2.5V causing its output (LOW line) to go low.
#9 Updated by tin over 6 years ago
Replaced Q101, VR100, VR101, K100, K101, R100, U105, U110, S100, R250, R123
Meter now power on ok, able to do measurements.
Self-diagnostics fail on codes:
401.1 Zero Cal Sw.
412.1 ACA switch
Test with Keithley 2400 SMU
|2400 setting||2001 #7 mode||2001 #7 result|
|V 0.182020, 10.5mA, -9.07777mA||2w ohm, 20 range||20.000074, Isrc=9.0766mA, Vdroop 181.53 mV|
|V 0.192785, 1.05mA, -0.962675mA||2w ohm, 200 range||200.00348, Isrc=0.9626mA, Vdroop 192.53 mV|
|V 0.96280, 1.05mA, -0.962675mA||2w ohm, 2k range||1.0000436k, Isrc=0.9626mA, Vdroop 962.67 mV|
|V 0.87480, 100uA, -87.4499uA||2w ohm, 20k range||10.000137k, Isrc=87.460uA, Vdroop 874.62 mV|
|V 0.69530, 100uA, -6.95041uA||2w ohm, 200k range||100.00101k, Isrc=6.9508uA, Vdroop 695.08 mV|
|V 0.75755, 100uA, -0.75856uA||2w ohm, 2M range||1.0000826M, Isrc=757.29nA, Vdroop 757.35 mV|
|V 4.1950 , 100uA, -0.004388uA||2w ohm, 2G range||1.0000368G, Isrc=4.1929nA, Vdroop 4.1930 V|
R592 0.1% 0.91 ohm = 2001 #7 result: 00.909108 ohm
Replaced C108 1000uF 50V to 470uF 50V, need buy proper one
Replace C101, C104 to new ones, 2200uF 35V
Replace cooling FAN, original is failed