Message #12 Posted by Ellis Easley on 23 May 2003, 9:26 a.m.,
in response to message #1 by Renato
Did you connect the low input to the battery negative terminal? If so, you have shorted out the battery fuse with your test lead! The fuse is between the battery negative terminal and ground. There is also a diode connected with its cathode to the battery positive terminal and anode to ground, between this diode and the fuse, if you accidentally connect the battery backward, the diode will conduct and the fuse will open, protecting the system.
In general, it is not a good idea to measure anything in a test instrument with itself unless it was designed to do so for test purposes. In this case, it would probably be acceptable to measure the battery voltage by connecting just the high input to the battery positive terminal, with a small error due to the voltage drop of the fuse. You would fail to sense an open fuse, but that is impossible anyway because the negative terminal of the battery charging circuit is connected to ground which is ultimately connected to the "analog ground" which the low input is connected to. I haven't yet determined from the service manual just where the analog and digital grounds are joined but I'm pretty sure they are connected, at just one specific point to avoid ground loops affecting the accuracy of the readings. These grounds are not connected to the power cord ground, there is only a spark gap between them.
With your mux board, you could measure the positive terminal and the negative terminal of the battery (with just the high inputs) and accomplish two things: (1)correct the reading from the positive terminal by subtracting the voltage on the negative terminal, which is the voltage drop of the fuse and the wire and (2) detect an open fuse, which would show up as a unusually high voltage on the negative terminal: even if the battery is fully charged and measures over 6V without any charging current, the positive battery terminal will be driven to 7.2V by the charging circuit so you would measure up to 1.2V on the negative battery terminal if the fuse was open.
Another measurement you can make regarding the battery charging circuit is the voltage across the current sense resistor, R705, a 0.27 ohm resistor located between the power transistor Q700 and the power transformer T700. One end is connected to ground and the other end is connected to the negative terminal of the bridge rectifier and its filter capacitor, C700. The negative lead of C700 is a good place to connect to it but it is hard to reach if you have the HPIB board installed, the resistor is easy to reach. I have been using my 3421A to charge some different size 6V lead-acid batteries and so I am very familiar with this circuit right now! After you let the battery run the system for a few minutes (my unit draws about 200 mA - this doesn't show up at the current sense resistor if you disconnect the AC and just let the battery power the unit), if you then turn the unit off and connect the AC again, the charging current goes to the limit right away (on my unit), I measure -136 mV on R705 which is about 1/2 Amp. While the current limit circuit is operating, the voltage is not being regulated (they work through the same pass transistor). After a while the current starts dropping from the limit value and at this point the voltage regulator circuit starts working. My unit puts about 7.2 V on the battery at room temperature (there is an NTC thermister in the voltage regulator circuit that reduces the voltage as the temperature inside the cabinet rises). If you have one of the Panasonic batteries, it should have a description of the charging specifications printed on it. Mine says "Constant voltage charge - Cycle use: 7.25~7.45V (25 deg. C) - (Initial current: less than 1.28A [always .4 times the A-H rating on different size batteries] - Standby use: 6.8~6.9V (25 deg. C)" I have some different size Panasonic batteries that have slightly different voltages printed on them (10 A-H: says 7.3~7.5V for Cycle use). It appears that HP has chosen a voltage between the Cycle and Standby ranges. The different batteries that I have measured finally settle to a charging current corresponding to about 4 to 10 mV across the current sense resistor (about 15 to 40 mA).
With all my oscilloscopes, I need to connect one to this charging circuit so I can understand the function of one component: a capacitor connected to the outputs of the voltage regulator and current limit comparators. There is an LM393 dual comparator with open collector outputs doing both jobs. The voltage regulator comparator drives its output low when the battery voltage rises above its limit. The current limit comparator drives its output low when the current rises over its limit. The two outputs are "wire-ORed" together. At the outputs there is a 33.2K pullup resistor to the unregulated supply (12-18V) and a 1 microfarad capacitor to ground and the base of an NPN transistor with its emitter going to ground through a 200 ohm resistor (feedback, current limit?). The collector of the NPN goes to the base of a PNP power transistor whose emitter goes to the unregulated supply and collector to the positive battery terminal through a forward biased diode (I think this diode is to keep the charging circuit from draining the battery when the AC is disconnected). So the NPN is a phase inverter and voltage level shifter for the PNP pass transistor. When a rising voltage on the base of the NPN turns it on, it drops the voltage on the base of the PNP and turns it on. As long as either comparator output is low (voltage or current too high), the transistors are off. When both comparator outputs are high (voltage and current below their limits), the 33.2K pullup resistor starts charging the 1 microfarad capacitor. If the unregulated supply is at 18V, the 33.2K resistor would source .54 mA which would take a little over 1 ms to charge the 1 microfarad capacitor to 0.6V, at which point the NPN would begin to turn on. The capacitor would continue to charge, shunting some of the current that would otherwise turn the transistor on more quickly. This is where the feedback from the 200 ohm emitter resistor comes in, because as the NPN collector current (= PNP base current) rises, so does the NPN emitter current and emitter voltage and base voltage, which allows the capacitor to keep charging. If the NPN emitter was grounded and the 200 ohm resistor was between the NPN collector and PNP base, providing current limiting, the NPN base voltage wouldn't rise much after the transistor started to turn on, so the capacitor wouldn't charge much more and the transistor would turn on more quickly. When either comparator output goes low again (current or voltage rising above limit), its open collector output transistor turns on and shorts the capacitor to ground, quickly discharging it and turning off the NPN transistor and the PNP pass transistor. Now the 200 ohm emitter resistor might make the NPN turn off more quickly, because its voltage drop raises the emitter voltage and causes the base voltage required to keep the transistor turned on to be higher. Drawing a comparison to a music synthesizer, the capacitor (and the 200 ohm emitter resistor?) cause the battery charging current to have a slow attack and a fast decay. I suspect the voltage at the comparator outputs is a sawtooth wave.
Maybe the capacitor is there to slow the circuit down and prevent uncontrolled oscillations that might occur due to the lack of hysteresis in the comparator circuits. IIRC, the 9114 battery charging circuit has a comparator with hysteresis to turn on a sensitive gate SCR (gate driven with DC, load driven with pulsating [unfiltered] DC). You can see the comparator limits by monitoring the battery voltage and watching the LED (which provides a trickle charge current when the SCR is off).