I connected a signal generator to a 5.1 ohm resistor, which may be too low comparing to the internal resistance of the generator, so I got only several hundreds millivolts amplitude for frequencies below a hundred. However, when the frequency was increased beyond a hundred and something, the voltage could rise to about 4.5Vamp (maximum). Why dose this happen?

(The specified maximum voltage amplitude of the signal generator is 20Vp-p)

Thank you!

  • 1
    \$\begingroup\$ What were you using to measure the amplitude? What signal type was it (sine, square etc.)? Specify units of Hz or kHz. What signal generator is it? \$\endgroup\$
    – Andy aka
    Commented Mar 5, 2016 at 9:25
  • \$\begingroup\$ I was using the oscilloscope to measure the amplitude of the waveform and the signal was sinusoidal.The unit should be kHz. The signal generator is Topward 8120. Sorry for my slow response! \$\endgroup\$
    – K. Man
    Commented Mar 8, 2016 at 4:06

2 Answers 2


There are probably two things going on here.

  • Overloading the output.
  • A decoupling capacitor is having an effect at lower frequencies.


At 20 V\$_{P-P}\$ and 5 Ω resistance the current drawn from the signal generator would be \$\frac {20}{5} = 4~A\$. It seems unlikely that your generator is rated for that so some current limiting is likely to occur along with some distortion of the output.



simulate this circuit – Schematic created using CircuitLab

Figure 1. AC coupled signal generator.

The signal generator output may be decoupled by a capacitor - designed to remove any DC bias from the output leaving only the AC component.

Capacitor impedance is given by \$ Z = \frac {1}{2 \pi fC}\$.

If you try putting a 1 kΩ resistor across the output and measure the voltage across it at several kHz and then again at a low frequency you should be able to work out the impedance of the output and, from that, the actual capacitor value using the formula above.

  • \$\begingroup\$ Thank you very much! Your answer really helps solve my confusions and the above should be the problems of my circuit. Thank you! \$\endgroup\$
    – K. Man
    Commented Mar 8, 2016 at 4:16

Let's assume that your measurement method is OK.

A general purpose wideband signal generator will tend to break its output range up into several frequency bands, and use a different technology for each band, for output amplifiers and even for the generation method.

Typically, below a hundred (you're not clear what here in your question, it could be bananas, but I'm guessing it's MHz), the generator is using a DDS (though older types use a BFO (beat frequency oscillator) which heterodynes the normal RF frequencies down to low frequency), and the output amplification is provided by a wideband 'op-amp' type or discrete bipolar buffer.

Above 50MHz, there are very nice RF heterojunction transistors that provide high output at remarkably low distortion. Unfortunately these do not work below 50MHz.

The output attenuator, if built from high linearity RF MOS switches, will also have headroom problems below several 10s of MHz.

  • \$\begingroup\$ Thank you very much for your answer! I am sorry that I didn't clearly specify the unit of frequency, which should be in the range of kHzs. I will think of the above when working on MHz circuit. \$\endgroup\$
    – K. Man
    Commented Mar 8, 2016 at 4:18

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