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Measurement of Generated Power in a Function Generator

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P = I * V, but the datasheet has sufficient information to calculate it.

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
Of course the generator doesn't display
The current.also, The data sheet doesn't specify the internal resistance
 
Data sheet mentions "CH1: 2 mVpp ~ 10 Vpp(50 Ω),4 mV ~ 20 Vpp(High Z)"

The 2:1 ratio (20 & 10) confirms that the source Z is 50 ohms.

Note that it's Vpp. You'd need RMS for power. Complex waveforms can get a bit complicated.

To measure the power you'd need to know the voltage (easier) and current (a bit trickier). Also, if AC and complex impedance, then phase angle enters into it.

Can you provide more details?

 
the generator displays 20 Vpp and I am at 100 KHz frequency. My circuit consists of a diode bridge and then a voltage regulator.
 
BTW, I am kind of confused. what does High Z mean for the 4mV~20 Vpp
 
A 50-ohm source feeding a 50-ohm load will split the voltage half each (internal and external). If the load is open circuit or high Z, then you'll see the total voltage on the port.

Why use all this feeding a bridge rectifier and voltage regulator? Is the bridge compatible with 100 kHz? ...getting towards RF, diode speed. And did you include a filter capacitor? ...otherwise poor regulator.

What are you up to?
 
What are you trying to do? The maximum power possible from the generator is only (10V)^2/50ohm = 2W. This is hardly an apropos application for a box that consumes 40W.

Is this for school?

TTFN
faq731-376
7ofakss

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!
 
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