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Video: MHS5200A Function Generator Review and Teardown

I have the signal generator referenced in this video. As the video says, its output impedance is specified into an open load.

It has a floating ground. Its ground is not tied to the mains.

I wish to use this to do some tests with MOSFETs and transistors.

How can I apply a DC offset to raise the voltage to a level in which I want to experiment with, say 10 volts? 2 volt sine wave pk-pk.

Is it as simple as putting a 10 volt battery in series with the generator?

What happens if the battery can deliver 5 A, but the generator can only deliver .2 A? The AC variation of the generator riding on the 10 volts would not seem to have any oomph to drive a load and the 10 volts would dominate. Is that so?

Could I use an op amp to get more power?

I guess with this signal generator, having its output into an open load does that mean that any kind of load will cause its output to fall to nil? Is it only meant to drive an op amp or MOSFET with high input impedance?

Please offer some of the best ways to use this signal generator to do tests that could present a load to a typical 50 ohm signal generator.

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  • \$\begingroup\$ Why don't you adjust the output DC bias offset on the signal generator? \$\endgroup\$
    – Justme
    Commented Jun 28, 2022 at 16:30
  • \$\begingroup\$ It has a floating ground...That 5V wall-wart might float. But is USB's GND and function-generator signal GND common? One should be very careful about raising signal GND with a battery - your USB connection to a PC might possibly be in peril. \$\endgroup\$
    – glen_geek
    Commented Jun 28, 2022 at 16:40
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    \$\begingroup\$ What sort of transistor circuit's are you planning on testing that require a 10 V dc offset sine wave? \$\endgroup\$
    – user173271
    Commented Jun 28, 2022 at 17:32
  • \$\begingroup\$ @Justme The DC offset of the generator is －120%～+120%（The ratio of the bias voltage and signal amplitude）. This is from the manual (if it even works). It is a very cheap generator . I am wondering how to work around its limitations if I wanted to adjust the DC component a signal rides along. I would think adjusting the DC component to experiment with biasing a transistor is a very basic use of signal generators and how I might achieve a higher DC level than +120%. \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Jun 28, 2022 at 20:58
  • \$\begingroup\$ @glen_geek I know what you mean. I did observe a wire getting hot/melting a year ago as current went back to the computer when I was using the pc software and this generator. If I did not use the usb, I suppose everything would have been fine. I am trying better how to understand its ground and how I can use it to add a dc offset, and not smoke anything. \$\endgroup\$
    – Jeffrey Edward Messikian
    Commented Jun 28, 2022 at 21:02

2 Answers 2

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This circuit should give you some idea of how it could be done. The capacitor ac couples (dc blocks) the ac signal. The input resistance of the two 100k resistors is 50k and so the signal loss between the sig gen's 50R output resistance and the biasing resistors 50k input resistance is very small.

To make the dc shift variable you could add a potentiometer to the two 100k biasing resistors.

The capacitor and 50k input resistance form a high pass filter which rolls off the low frequency with a -3dB frequency of about:-

fc(-3dB) = 1/(2 * pi * C * 50k)

The op amp buffers the biasing network (the two 100k resistors) and will provide 10mA or so to drive an added load. If more current than that is required then you could consider adding a current boosting stage within the feedback loop of the op amp consisting of driver transistors with a diode/resistor biasing network.

To perhaps give you a better understanding of how the sig gen's 50R output resistance would interact with a directly driven load, if the output with no load is, say, 2Vpp then connecting a 50R load between the output and ground will halve the signal amplitude to 1Vpp. The 50R source resistance (output resistance) creates a potential divider with whatever load resistance is connected to the sig gen's output.

dc shift

EDIT

The circuit below should provide an optional solution.

VR1 varies the output dc level with the resistors added either side of VR1 to prevent clipping.

Level shifter

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Leave the generator output referenced to ground. Take a 9V battery and wire it in series with the output, and connect that to your circuit under test, which will raise the signal output about 9.6 V for a new battery. Then use the small output offset adjustment of the generator to get the exact DC you want.

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  • \$\begingroup\$ Add a capacitor parallel to the battery to bypass the unknown AC impedance and avoid an overload of this shifted output or the battery will rip up the generator driver stage. \$\endgroup\$
    – Jens
    Commented Jun 29, 2022 at 20:33

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