AboutDavid Freitag Expertise I can answer questions about all kinds of computers, operating systems, software and hardware in clear and easily understandable language. I have dealt with sorts of software and hardware issues in PCs from before windows to the present. I can explain basic concepts and deal with complex detailed issues in the guts of operating systems, including drivers, bios, hardware issues and software problems.
Experience I have over 3 decades of experience ranging from mainframes and minis and microcomputers to all flavors of personal computers. I have programmed in assembler language, Fortran, C, VB, C# and I have worked with the major relational databases. I have been a CIO of an energy company and I have provided hands on technical management for Windows software development. In the course of my career, I have had to deal with all sorts of configurations and nitty gritty system problems which exposed me to all aspects of personal computer systems.
Publications I am an expert author in Ezinearticles and have had articles published in Buzzle, IdeaMarketer, ArticleCube and other article sites.
Education/Credentials I have a BS in Physics from CCNY. I have received credits from courses dealing with OLE automation and computer systems management from the AMA
Awards and Honors Phi Beta Kappa
Achievement awards for software development efforts in a leading Voip provider company.
Expert: David Freitag Date: 5/21/2008 Subject: AC adapter for HP Compaq nx7300
Question QUESTION: Hi,
A short question:
AC adaptors for HP Compaq nx 7300 have two output voltages:
The first voltage is about 18.8V/3.5 A (between the outer and inner cylinders of the output (DC) connector; and
The second voltage, a voltage between the central pin of the connector and the outer cylinder of the output connector, which was about 18.4 V (about 0.4 V less than the first one).
Is this central pin voltage (the second one) a dc voltage, or some thing else? Why is nothing written about this voltage on the AC adapter’s label?
Thanks in advance,
Bashkim
ANSWER: Hi Bashkim,
I need to look into this. I will get back to you shortly.
Best,
David
Does the cord coming from the A/C power brick have 2 wires or 4 in it? If it only has two, then the adapter is only outputting one voltage, and the adapter on the notebook can take either of two different power adapters (possibly one for the U.S. and one for Europe?)
If it has 4 wires, then the A/C brick is supplying two voltage lines because some other component inside the notebook needs a different voltage rail (although I can't imagine what).
Another expert that I contacted suggested the following;
I'm not familiar with that specific adapter, but here's my <guess>, based on other charging systems:
The first voltage you measured is the main output used for charging the battery and powering the unit.
The second voltage is part of a monitoring/regulation circuit that keeps the battery from being overcharged. The voltage there is coming 'backwards' through a signal conditioning network that, when the connector is unplugged, is disconnected from its 'other half.' The network could be as simple as a low resistance element that is used to measure the current draw according to ohm's law (V = I * R). The difference in voltage could be due to the presence of a diode in the circuit to make sure the system isn't back fed when the battery is fully charged and the power supply is unplugged.
To answer your second question, I <assume> the voltage is DC - if your meter measures both AC and DV voltages, switch it to AC mode and check again - it should read a low value in AC mode if the 'real' voltage is DC.
Best,
David
---------- FOLLOW-UP ----------
QUESTION: Hi David,
Thanks a lot for your quick reply.
Let me answer/specify the point/assumptions you rightly outlined.
1-The cable that comes from the AC “power brick” has three wires: one to the outside cylinder of the output connector (o.c.), the other to the inside cylinder of o.c., and an other to the central pin of the o.c.
2-Both voltages, the first one to the outside cylinder of the o.c., and the second one to the central pin of the o.c., both with respect to the “common” electrode (the outside cylinder) are present also without connecting the AC adapter to the laptop (idle voltages). I measured the voltages with a dc voltmeter, not with an oscilloscope. It seems that the input resistance of the laptop for the first voltage, is relatively low (one can measure some current there), while the input resistance of the laptop (central pin jack) seems to be pretty high (could not measure any current).
3-I tried to substitute the original HP AC adapter, with a general purpose AC adapter, applying a resistor divider between input cylinder- central pin-output cylinder, in order to get the second voltage. But the laptop did not function normally: it was very slow, and it did not charge the battery, although it signaled “recharging the battery”.
I tried to contact many people who claimed to be experts on ac adapters, and their answer was: “We do not know. This thing with “two voltages” is only to these HP Compaq nx7300 laptops”
I am just curious what is it.
Thank you again for all you time.
Bashkim
ANSWER: Hi Bashkim,
One of the experts that I know said;
When you say:
'It seems that the input resistance of the laptop for the first voltage, is relatively low (one can measure some current there), while the input resistance of the laptop (central pin jack) seems to be pretty high (could not measure any current).'
Do you mean that you measured these with an ohmmeter or with an ammeter? An ammeter circuit would have required breaking into the circuit to insert the leads in series - it sounds that way from the 'could not measure any current.'
From the low current draw of the central pin I would think it is a monitoring circuit looking at the voltage coming back from somewhere inside the computer - probably the actual battery terminals. If you think of it as a 'voltmeter' this makes sense, as voltmeters have high input impedance.
The slow function of the system with the alternative power source may be due to the system's picking up a low voltage on the 'monitoring' pin. This would indicate a low battery or weak charger and the system responded by cutting back on CPU/mainboard frequency to conserve power.
If you set up a circuit to monitor the current and voltage on both legs while the computer is running with the original and alternative power supplies you could deduce the function of the connections - I would guess you would see high current on the main connection and very low current on the secondary, while the voltage on the secondary should be lower than the voltage on the primary when the battery is low, rising to be closer to the primary as the battery charges.
Interesting investigation!
Best,
David
---------- FOLLOW-UP ----------
QUESTION: Hi David,
I set up a circuit to monitor the current and voltage on both legs while the computer is running, and certainly also without the computer (laptop).
I was alble to measure the currents, with a bad dc ammeter, on the first voltage load. But I had to “calculate the input resistance” facing the second voltage.
I “calculated the input resistance” as follows: I measured the idle voltage (laptop not connected) on the middle point of the resistive voltage divider I have applied (1 kOhm/47 kOhm). Than I measured the “voltage on load” (laptop connected and running). From the voltages idle/running I calculated the “resistance connected in paralel to the 47 kOhm resistance of the divider that could cause the same central pin voltage change. It resulted about 200 kOhm. In the mean time, the “first voltage” (inner cylinder of o.c.) did not change more than 0.1 V, from idle to load (the general purpose AC adapter was a “big one”, about 120 VA).
My opinion is that the central pin of the AC adapter is feeding some week current to the laptop, and not the laptop to the AC adapter.
I short circuited the central pin with the inner cylinder of the o.c. (that means I gave the same voltage on both output voltages, about 18.8V), but still the laptop did not function normally: slow and not charging the battery. I dared to do this because of the following:
I have measured the idle voltages on two types of AC adapters, one original and one “made in China”. The second voltage (the central pin one) was not the same. The original had about 18.4 V and the Chinese one had 6.6 what made me to believe that the value of the second voltage should not be so important.
In fact, the laptop functioned not equally with both AC adapters, the original one and the Chinese on. With the Chinese adapter, the mouse jumped, was not easy to use it, when moved through the touchpad. But if one used an external mouse, it seemed everything the same: no problem.
I tied to get some information about the original HP AC adapter, but I could not find any thing more than the following:
AC adapter for HP notebook Compaq nx7300
Series PPP009L HP
65W
Input: 100-240 V 1.6 A 50-60 Hz
Output: 18.5 V 3.5 A
Efficiency level: IV
Rev. A01
PA 1650-02HC
Part.No 384019-001
HP Spare 391172-001
6Z20837701
Sorry to enter you in all this particularities, but it seemed to me that you too became curious on “the happenings” with this Hewlett Packard AC adapters.
Best regards,
Bashkim
Answer Hi Bakshim,
WB says;
Your inference about the direction of current flow sounds correct - it could still be part of a regulator in the external power supply since the sense lead of a regulator circuit could be designed to operate as either a source or a sink. The only way to find out for sure would be to open the power supply and see what's inside.
However, it would make sense for the regulation circuit to be in the computer as it has to control the battery charge, and keeping laptop batteries 'happy' is not just a matter of looking at the voltage - there is a thermal sensor built into them and that has to be factored in, too. (look at the battery terminals - there are usually more than two, with the others used to monitor temperature, I believe)
You are doing a good job of figuring out what's going on here, but I'm a little puzzled about your voltage divider. When I want to measure current without a good ammeter, I insert a low resistance into the circuit - say, 0.1 ohm if I'm anticipating currents on the order of an amp or two. Then I measure the voltage drop across the resistor - if it's 0.1 volt then 1 amp is flowing through the resistor. (If I use a higher resistance, say 10 ohms, 1 amp would result in 10 volts, a nice signal, but the load would only see 8 V in your case - 10 volts of the source source would be dropped across the resistor - and the device would not be operating with its normal voltage.) In the case of the inner lead, I would try a higher resistance - maybe 1,000 ohms - and, again, measure the voltage developed across the resistor when the unit is operating. If I read 1 volt then 1 mA is flowing in the circuit. I can also determine the direction of current flow from the sign of the voltage drop - positive from source to load means current is flowing 'into' the load.
Given the fictitious numbers above, I can calculate the input resistance (resistance since it's DC) of each part of the circuit. For the main circuit, if the input voltage is 18.6 V and it's drawing 1 A, then the input resistance is 18.6 ohms. For the secondary circuit, if the input voltage is 18.4 V, and it's drawing 1 ma, the input resistance is 18.4 kohms.
So, I don't understand where the 1k/47k divider is connected.
Perhaps I'm totally misunderstanding things here - more likely, actually!
By the way, what is the objective of all this fun stuff?
