The AWG series instruments will output some amount of voltage, specified in V p-p into a 50 ohm load, which is a properly terminated configuration. This also results in the best system performance.
The output circuit has a 50 ohm series termination, so when the output is terminated into 50 ohms, forms a voltage divider that drops half of the output voltage across the 50 ohm load and half of the output voltage across the internal 50 ohm series. The V p-p that is specified at the 50 ohm load is only half of what the generator is producing internally.
If you are working at lower frequencies and are not impacted by the very high frequency ringing that can occur on the fast transitions of a squarewave, then using a load that is higher than the recommended 50 ohms can give you as much as double the output voltage.
How high is a higher impedance? Take the case of the 50 ohm external termination.
If we set the output to 2V p-p, we now know that it is producing 4 Volts internally, so:
(50 ohm load / 50 ohm load + 50 ohm series termination) = (50 / 100) = 0.5 * 4 V = 2 V
(1,000 ohm load / 1,000 ohm load + 50 ohm series termination) = (1,000 / 1,050) = 0.9523 * 4 V = 3.8095 V
(1,000,000 ohm load / 1,000,000 ohm load + 50 ohm series termination) = (1,000,000 / 1,000,050) = 0.99995 * 4 V = 3.9998 V
This FAQ Applies to:
Product: AWG7101, AWG7102, AWG7051, AWG7052, AWG5002, AWG5004, AWG5012, AWG5014
FAQ ID : 60876View all FAQs »