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I have an old PS/2 keyboard, I doing a future project with it, but for now I just want to measure the voltage in the cable. However I don't know how to do it, so let me explain.

It's a PS/2 keyboard, and I have a P2/2 adaptor that transforms it into a micro USB cable that get's connected to the PC.

It has 4 wires:

  • Red: VCC power
  • Black: Ground
  • White: Clock
  • Green: Data

The circuit looks basically like this (apology for my drawing)

enter image description here

The mini USB is connected to the computer, then it's transformed into a PS/2 cable via an adaptor. The adaptor acts as some kind of resistance. Then the wires are connected to the keyboard via a PS/2 port.

I have basically drawn the 4 wires as they are there, I have removed the external insulator cable on that segment in the drawing so that I can measure the voltage.

I have measured the voltages at the points illustrated on the picture.

A1->A2 = 0.6 mV DC

B1->B2 = 0.6 mV DC

C1->C2 = 2.2 V DC

D1->D2 = 5.22 V DC

I have also measured the voltages between the wires, after the adaptor:

A2->B2 = 5.24 V DC

C2->B2 = 5.24 V DC

D2->B2 = 5.24 V DC

So I guess the (black) ground cable is like the return path of the circuit since A2,C2 and D2 have 0 Volts between them, they are only connected with B2 separately.

Meanwhile I have also measured the voltages before the adaptor:

D1 had 0 Volts between all other points

A1->B1 = 5.25 V DC

C1->B1 = 3.05 V DC

A1->C1 = 2.2 V DC

Question

The question is what is the voltage in the "Yellow Area", meaning in the wires that go out of the adaptor and go into the keyboard.

So my goal is to determine how much watts of power the keyboard uses, so I would need to multiply the voltage in each cable "going out" with the amperes going through each wire and add them together.

How to measure exactly the voltage in the yellow area?

So which measurement is the correct measurement from above, in order to determine how many volts there are in the wires in the yellow area. And use that volt to multiply with with the amperes flowing through each cable to determine the watts of current that the keyboard uses.

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    \$\begingroup\$ In practice, you only need to measure the current through A1 and the potential difference between A1 and B1. There is an IC in the keyboard adapter: it is not simply "some kind of resistance". \$\endgroup\$ – Andrew Morton Apr 9 '17 at 15:24
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    \$\begingroup\$ The green and white wires are for signals, leave them alone for power measurements. Voltage is measured not in a wire but between wires. Between the red and black cable there will be 5 V where the red will be positive. You only need to break the connection of the red wire and measure the current. And you really need to do some learning and studying of electronics before you continue with more complex things. \$\endgroup\$ – Bimpelrekkie Apr 9 '17 at 15:26
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    \$\begingroup\$ I am not sure, but I doubt that your statement "The adaptor acts as some kind of resistance" is correct. Does PS/2 keyboard use the same protocol as USB? I guess there should be some active components in there. \$\endgroup\$ – Anonymous Apr 9 '17 at 15:30
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    \$\begingroup\$ Why do you need to know its consumption? You want to power it from some source other than PC PS/2 or USB port? It takes 5 V voltage. Inserting resistor into power line may render keyboard device inoperable, it is not a right way to decrease its power consumption. That's why I asked what is your goal behind asking the question. \$\endgroup\$ – Anonymous Apr 9 '17 at 16:09
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    \$\begingroup\$ no I would not like to do that, because I am both interested in the power in idle state, as in a typing state. you can still do that with the measurement I suggested. You should realise that you have a lot to learn because you make wrong assumptions because you do not understand how things work so make claims that are simply untrue. \$\endgroup\$ – Bimpelrekkie Apr 9 '17 at 18:32
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I don't really care about the battery though, what I am interested in is what is the minimum watts needed to use the keyboard. What other way there is to decrease power consuption?

First thing you need to think about that traditional keyboards were not designed for low-power applications, thus if you think you are critical for power probably you need to use some special input device especially designed for these purposes.

Look here. Power consumption will depend on what keyboard is currently performing. If it is idle there will be small current, if you press 10 buttons on it and have all LEDs turned on, it will be another current and thus another, bigger power consumption.

In my opinion, the only way to properly decrease power consumption is to turn keyboard device off when it is not expected to be used. When operator or application will need to use it, you should refer to maximal device's current per its datasheet.

I just took PS/2 keyboard I have, and on its rear side there's label saying that it requires 5 V DC. No information about current though.

There might be information on the web about maximal PS/2 ratings, I found this one which says max current is 275 mA.

Also please keep in mind that you are currently thinking or talking about your particular case with particular keyboard. Imagine you are designer of consumer product, which can be used virtually with any keyboard, thus you can not know which exact current attached device takes. And thus you should refer to minimal and maximal definitions within the standard, planning your device operating cycle accordingly.

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  • \$\begingroup\$ I understand but it still consumes more power than it's minimum. So for example I have insterted a 100 ohm resistor on the wire and the current dropped by 3% and it still works, so those datasheets are too broad. So yes that energy gets converted into heat so it doesnt really reduce power consumption, however I am still interested how low it can go, before it doesnt work anymore. \$\endgroup\$ – user138887 Apr 9 '17 at 16:55
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    \$\begingroup\$ How did you know keyboard's minimum power to be consumed? And that's why it is called minimum, you will probably not get less than that minimum having it connected to compliant power source (4.5 V to 5.5 V). When you add resistor on the power line, and when keyboard's input voltage drops below 4.5 V, this configuration will become not supported and it will formally not work however in reality it may work for this particular keyboard for this particular state of this keyboard. \$\endgroup\$ – Anonymous Apr 9 '17 at 16:58
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I'm not sure you understand your keyboard or your PS/2 to USB adapter well yet.

Let's start with the PS/2 to USB adapter ....there are two types:

  1. A Passive convertor with nothing but wires in it. In this application the PS/2 keyboard itself can output the serial TTL or USB signals. As soon as 5 V VCC is applied the keyboard tests to see if it is receiving a clock, and if not it goes to USB mode.

  2. An active PS/2 to USB adapter. This is actually a small MCU that talks serial TTL to the keyboard and USB to the host computer. These adapters typically have both a serial PS/2 and Serial Mouse connector (Green and Purple) to USB 1.0.

You might look at the questions on this site ...it might help you.

You could also Google for PS/2 to USB adapter schematics and get a clue as to their operation.

To get to your question.
While you could measure the current in each wire, the results would be adequate if you simply measure the 5 V current consumption of the keyboard. It will vary from just a couple of mA (more modern keyboards) through to 100 mA or ( the classical 8031 processor based keyboards) so no matter whether the keyboard is feeding serial TTL or USB.

In addition, some keyboard processors are actually in a deep sleep mode until a key is pressed or data sent to the keyboard (you see this particularly in battery powered wireless keyboards, but I've seen it in many USB keyboards). In these cases average current consumption will vary from sleep states around 10 uA to 40 mA bursts while scanning the keyboard array.

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My goal is to determine how much watts of power the keyboard uses,

If you have a DMM capable of 1mV accuracy, break/insert an R=10 Ohm resistor on Red wire (+5V) from PC side (how is up to you) The voltage drop for this R displays current, as I=V/R so 50mV/10 = 5mA. We normally choose a current shunt in this range so the voltage drop is negligible yet readable on a DMM.

Without a current probe, this is the only way.

To see what ROOT USB HUB current is allocated, goto Device Mgr and try each USB ROOT HUB > Properties>Power and see what current is allocated to target.

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  • \$\begingroup\$ I already know the amperes in the red wire, it's 0.79mA DC . The voltage between red and black (before the adapter) is 5.25 V. Does this mean that the keyboard (in idle) consumes basically 0.0041475 Watts? \$\endgroup\$ – user138887 Apr 9 '17 at 16:36
  • \$\begingroup\$ .............yes.............. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 9 '17 at 17:21
  • \$\begingroup\$ This is not necessarily accurate. One has to be careful when measuring small consumption, especially with an instrument having a high burden voltage as the device may well end up drawing some of its power through the data lines as well as through the supply ones. \$\endgroup\$ – Chris Stratton Apr 10 '17 at 16:49
  • \$\begingroup\$ that is why 50mV to 75mV is a standard for SMPS and many other low voltages so sense current voltage is <=1% and Pd is low \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 10 '17 at 19:00
  • \$\begingroup\$ @TonyStewart.EEsince'75 Should that be LVDS rather than SMPS in that comment? \$\endgroup\$ – Andrew Morton Apr 11 '17 at 8:03
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Ps/2 and usb are both 5V bus protocols. The data lines use minimal power. Simply measuring the current through the power lines is enough to accurately measure the power usage of the keyboard. It will most likely be under 500mA. Most keyboards are about 100mA or less.

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  • \$\begingroup\$ But do I need measure the power in the data lines as well? Or is that included in the power line already? If the keyboard is idle the data lines have nearly zero voltage, but if I start typing it rises, and also the power in the red lines rises too. So I believe the power consumption is already included in the red line, and it distributes if from there. I am not sure. \$\endgroup\$ – user138887 Apr 9 '17 at 16:43
  • \$\begingroup\$ No, USB is not a 5v protocol. It has 5v supply but the signalling is differential with a 3.something volt or so peak relative to ground. And no, measuring only the power lines is not enough - in low power cases you really cannot neglect the current on the signal lines. In theory in USB that would be balanced, in practice, not necessarily. \$\endgroup\$ – Chris Stratton Apr 10 '17 at 16:51
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@JackCreasey has the correct explanation of how the USB/PS2 keyboards work.

If you just want to measure the voltage/current, I have a handy little device that does that- it has a USB male at one end and a female at the other.

enter image description here

I don't remember the cost, but it was very cheap- I bought a few of them to have around. There are a few variations with different display technologies.

It has an LCD display inside and an MCU to measure the voltage and current, and it will integrate the current to give mAh. The 80mA shown is from a small FPGA board- I tried measuring a no-frills Dell keyboard (no LEDs, not even for caps lock), but it was too low to register with 10mA resolution.

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  • \$\begingroup\$ 1 to 10 bucks depending on the features. \$\endgroup\$ – Passerby Apr 9 '17 at 16:54
  • \$\begingroup\$ @Passerby I think it was more like $1.50-$2 (USD) for this one, or I would not have bought that many. \$\endgroup\$ – Spehro Pefhany Apr 9 '17 at 17:00
  • \$\begingroup\$ I already have a multimeter, I just don't know at what points you need to measure the voltage and ampere,most answers here are pretty confusing. In fact I have already measured the voltage across certain paths in the OP, I just need a confirmation, which one is correct. I also know the amperes of each wire. \$\endgroup\$ – user138887 Apr 9 '17 at 17:05
  • \$\begingroup\$ @user138887 The power supply voltage is just Vcc to ground. It should be a bit more than 5V, as in my photo. \$\endgroup\$ – Spehro Pefhany Apr 9 '17 at 18:23

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