Ricoh Aficio 1515 Service Manual page 104

SERVICE PROGRAM
2220 TD Sensor Output Value Display
2220 1 Displays:
a) Vt: the current TD sensor output value and
b) Vref: the target TD output value Vts (SP2-926) + correction for ID sensor
output.
The TD sensor output value changes every copy. If a > b, toner is supplied to the
development unit.
2221 ID Sensor Error Analysis (☛ 5.1.4)
2221 1 Vsg
2221 2 Vsp
2221 3 PWM
2221 4 Vsdp
2221 5 Vt
2221 6 Vts
2301* Transfer Current Adjustment (☛ 6.6).
2301 1 Normal paper
Adjusts the current applied to the transfer roller when feeding from a paper tray.
Use a high setting if the user normally feeds relatively thick paper (within spec) from
a paper tray
2301 2 Thick/Special paper
Adjusts the current applied to the transfer roller when feeding from the by-pass tray.
Use a high setting (a) if the user normally feeds relatively thick paper from the by-
pass tray, or (b) if waste toner is re-attracted from the drum (which can occur when
using transparencies).
2301 3 Duplex
Adjusts the current applied to the transfer roller when carrying out a duplex job. Use
this SP if there is poor image transfer on the rear side of duplex copies.
2301 4 Cleaning
Adjusts the current applied to the transfer roller for roller cleaning. Increase the
current if toner remains on the roller after cleaning. (Remaining toner may cause
dirty background on the rear side.)
2802
Forced Developer Churning
2802 1 Initializes the developer and checks the TD sensor output (Vt). The machine mixes
the developer for 2 minutes while reading and displaying the Vt value. The machine
does not initialize the TD sensor output. If the machine has not been used for a long
period, prints may have a dirty background. In a case like this, use this SP to mix
the developer. The message "Completed" is displayed when the program ends
normally.
Displays the Vsg value.
Displays the Vsp value.
Displays the PWM value.
Displays the Vsdp value.
Displays the Vt value.
Displays the Vts value.
[–2 = –4 µA / –1 = –2 µA / 0 = 0 µ µ µ µ A / 1 = 2 µA / 2 =
+4 µA]
[–2 = –4 µA / –1 = –2 µA / 0 = 0 µ µ µ µ A / 1 = 2 µA / 2 =
+4 µA]
[–2 = –4 µA / –1 = –2 µ/ 0 = 0 µ µ µ µ A / 1 = 2 µA / 2 = +4
µA]
[–10 ~ 1 / –1 / 1 µA/step]
5-6
9 March 2004