In a message dated 9/19/2002, you write:
> So far, I've taken a large resistive load and measured the voltage drop on > the AC line. From that I calculated the total impedance of the AC line. > However, as you may suspect, with a resistive load, the power factor is > 1.0. > So I can't vectorly, calculate the resistive and inductive components. > Hi Dave: I am not familiar with the specific measurement you are trying to make, so the following may not be applicable to your situation. However, some of the basic concepts may be useful. In the telecom business it is sometimes necessary to know the complex impedance presented by a 2-wire port. In theory, if you drive the port with a voltage source applied through a known resistance, you can calculate the complex impedance of the port based on measurements of the following two things: 1) The voltage drop across the source resistor 2) The phase of the current through the resistor, relative to the phase of the source I designed a simple fixture to measure the voltage and phase, then derived the necessary equations and created a spreadsheet to calculate the complex impedance. Soon after completing my measurements this way, I got a good price on a used test instrument that makes this measurement directly. I was very pleased to find that the values measured by the test instrument matched my calculated values almost exactly. This gave me good confidence in my earlier test method, even though I no longer need to use it. It might be possible for you to adapt this method for the test you want to make. If so, I would be happy to send you a copy of the equations and a copy of the spreadsheet I used. Since this was developed internally for my own use, the notes are a little sketchy, but I think I could fill in the gaps with a telephone conversation. While I am not familiar with the details of the test you are trying to perform, I can see that one possible complication would occur if the test is supposed to be performed under a specified load condition. In that case, you would have to find a way to ensure that the AC impedance of your load does not affect the measured impedance of the mains. I think there are ways to accomplish this either with a test fixture or by making the spreadsheet calculations take the load impedance into account. Joe Randolph Telecom Design Consultant Randolph Telecom, Inc. 781-721-2848 http://www.randolph-telecom.com
<HTML><FONT FACE=arial,helvetica><FONT SIZE=2>In a message dated 9/19/2002, you write:<BR> <BR> <BR> <BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">So far, I've taken a large resistive load and measured the voltage drop on<BR> the AC line. From that I calculated the total impedance of the AC line.<BR> However, as you may suspect, with a resistive load, the power factor is 1.0.<BR> So I can't vectorly, calculate the resistive and inductive components. <BR> </BLOCKQUOTE><BR> <BR> <BR> Hi Dave:<BR> <BR> I am not familiar with the specific measurement you are trying to make, so the following may not be applicable to your situation. However, some of the basic concepts may be useful.<BR> <BR> In the telecom business it is sometimes necessary to know the complex impedance presented by a 2-wire port. In theory, if you drive the port with a voltage source applied through a known resistance, you can calculate the complex impedance of the port based on measurements of the following two things:<BR> <BR> 1) The voltage drop across the source resistor<BR> 2) The phase of the current through the resistor, relative to the phase of the source<BR> <BR> I designed a simple fixture to measure the voltage and phase, then derived the necessary equations and created a spreadsheet to calculate the complex impedance.<BR> <BR> Soon after completing my measurements this way, I got a good price on a used test instrument that makes this measurement directly. I was very pleased to find that the values measured by the test instrument matched my calculated values almost exactly. This gave me good confidence in my earlier test method, even though I no longer need to use it.<BR> <BR> It might be possible for you to adapt this method for the test you want to make. If so, I would be happy to send you a copy of the equations and a copy of the spreadsheet I used. Since this was developed internally for my own use, the notes are a little sketchy, but I think I could fill in the gaps with a telephone conversation.<BR> <BR> While I am not familiar with the details of the test you are trying to perform, I can see that one possible complication would occur if the test is supposed to be performed under a specified load condition. In that case, you would have to find a way to ensure that the AC impedance of your load does not affect the measured impedance of the mains. I think there are ways to accomplish this either with a test fixture or by making the spreadsheet calculations take the load impedance into account.<BR> <BR> <BR> Joe Randolph<BR> Telecom Design Consultant<BR> Randolph Telecom, Inc.<BR> 781-721-2848<BR> http://www.randolph-telecom.com<BR> <BR> </FONT></HTML>
