output impedence


by EE4life
Tags: impedance, impedence, output
EE4life
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#1
Jan19-14, 02:40 PM
P: 41
Hi all,

I am not sure I understand the concept of output impedance.

Say I have a function generator with 50 ohm output impedance, and I am driving a 100 Ohm resistor at 1V at 1kHz. What is the current in the system? does the system behave like 100 ohms in parallel with 50 ohms?



Thanks in advance.
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Averagesupernova
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#2
Jan19-14, 03:31 PM
P: 2,451
Output impedance is an impedance in series with an ideal voltage source. So, you have a generator with a Zout of 50 ohms driving a 100 ohm load. The load is seeing 1 volt AC. The load current is 10 mA. This load current is passing through the Zout. You are losing .5 volt AC across this output impedance. If you unload the output of your generator you should measure 1.5 volts AC. Make sense?
EE4life
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#3
Jan19-14, 05:54 PM
P: 41
So, in this case if I set my function generator at 1.5V amplitude, I will see 1V across my load?

Averagesupernova
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#4
Jan19-14, 07:49 PM
P: 2,451

output impedence


If the output impedance is in fact 50 ohms and you measure the output of the generator unloaded and set it to 1.5 volts, then yes, when you load it with a 100 ohm resistor you will see the output drop to 1 volt across your load. That is generally not how it is done though. Usually you just measure it while hooked to the load. However, it is a lesson in output impedances.
EE4life
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#5
Jan20-14, 10:07 AM
P: 41
The output impedance is place to prevent a complete short circuit right?

Actually, I am driving a low impedance (less than 50ohm) resonant circuit with a function generator. I am measuring current with an inductive current probe and a voltage across the circuit with a voltage probe, both attached to a oscilloscope. I am using the current and the voltage to determine the impedance and the phase.

This means I need to include account for the 50 ohm output impedance to back calculate the impedance of the load, right? Or will the Vrms/Irms=Zload' and Z'load*tan(theta)=Zload" (ie, i dont need to worry about the output resistance)?
uart
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#6
Jan20-14, 10:28 AM
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P: 2,751
Quote Quote by EE4life View Post
The output impedance is place to prevent a complete short circuit right?
Not really, although I suppose it does also serve that purpose. All outputs must have a finite output impedance, and 50 ohms is frequently used on instrumentation like this because it matches the characteristic impedance of common types of coaxial cable. This may not be immediately relevant to you, but impedance matching becomes important to avoid distortion due to reflection in work involving high frequency and pulses.

Actually, I am driving a low impedance (less than 50ohm) resonant circuit with a function generator. I am measuring current with an inductive current probe and a voltage across the circuit with a voltage probe, both attached to a oscilloscope. I am using the current and the voltage to determine the impedance and the phase. This means I need to include account for the 50 ohm output impedance to back calculate the impedance of the load, right?
No, as long as you are directly measuring the actual voltage and current at your load then you can determine things like impedance and phase angle without knowing the source impedance.
EE4life
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#7
Jan20-14, 10:40 AM
P: 41
Thank you for your help

So to be clear, Vrms/Irms=Zload' and Z'load*tan(theta)=Zload" works?

Also, what if i need to measure current by hooking a shunt resistor and using a voltage probe. The ground terminal of the voltage probe must always be at ground, so what will the effect of the shunt resistor be if it has a similar impedance to the load? How would I calculate the impedance of the load in this case?

Sorry for these baby questions. Its been a awhile since I did some AC circuit theory.
uart
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#8
Jan20-14, 10:54 AM
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P: 2,751
Quote Quote by EE4life View Post
Also, what if i need to measure current by hooking a shunt resistor and using a voltage probe.
I'm not sure whether or not you're confusing the terminology "current shunt" and "shunt resistor" with the placement of such a resistor. But be sure it would go in series with the circuit in which you're measuring the current, not in parallel.

Anyway, assuming that you really mean "series resistor" then yes, the presence of the common ground (to scope and sig gen) does complicate the simultaneous measurement of voltage and current in that scenario.

Without going to the trouble of trying to do ground isolation or anything like that, probably the easiest solution is to use one probe to measure just the load voltage and a second to measure the total voltage of the load plus the series resistor. You then use the "invert" and "add" features of the scope (if available) to deduce the resistor voltage and hence the current.
sophiecentaur
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#9
Jan21-14, 07:49 AM
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If you are going to be measuring things around the circuit then why not just measure the volts on the output terminal. They are what they are, whatever the output impedance of the generator. You can relate any phase measurements to PD at the terminals so I don't think you would be at risk of getting things wrong (as long as you don't use the signal generator scale as gospel and you measure the output PD every time).
EE4life
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#10
Jan21-14, 11:05 AM
P: 41
Please verify see the attached 2 page pdf for my impedance calculation of the unknown circuit with a series resistor. I am after the absolute impedance of the circuit (to calculate quality factor) and the phase angle.

From what I understand, the output impedance will not affect my impedance measurement of circuit, correct? What about power reflection?
Attached Files
File Type: pdf impedance calc.pdf (276.0 KB, 8 views)


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