I am using this Function Generator

My circuit :

enter image description here

FG input : 5V peak to peak. 1kHz. 1ms time period. 90% duty.

I want to understand the below questions :

  1. Is it safe to do this connection?

  2. I have read that we need to take care whether the input BNC channel output of the FG should be capable of handling the voltage and current? Which is the voltage it must handle and what is the sink current?

  3. How to check in the datasheet of the FG regarding this?

Actually, the FG output BNC channels only produce output voltage right? How to check how much of input voltage that they can handle when they are impressed on those channels?

  • 5
    \$\begingroup\$ As soon as you bring the gate high enough, you will short 10V to ground through the MOSFET burning it out. The gate of the MOSFET doesn't draw current (other than to satisfy its capacitance). What are you trying to do? \$\endgroup\$ Mar 23, 2020 at 14:15
  • \$\begingroup\$ I want to understand the questions regarding the Function generator. Trying to understand just the function generator. Nothing specific to accomplish here \$\endgroup\$
    – user220456
    Mar 23, 2020 at 14:55
  • \$\begingroup\$ Can someone help me regarding this. \$\endgroup\$
    – user220456
    Mar 23, 2020 at 16:39
  • 2
    \$\begingroup\$ What you've asked is the similar to putting a bunch of holes into the hull of a boat, and then asking everyone how to properly operate it from the bridge. Then when people point out there are holes in the hull, you are tell them that you are only concerned with what switches and levers you need to hit in the bridge. It doesn't matter what levers and switches you hit from the bridge: your ship is sinking. In the same way, it doesn't matter if you use your function generator properly with this circuit because if you did then your circuit will fry. \$\endgroup\$
    – DKNguyen
    Mar 23, 2020 at 20:48
  • 1
    \$\begingroup\$ You can add a proper schematic using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar and "Save and Insert" on the editor an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$
    – Transistor
    Mar 23, 2020 at 21:08

1 Answer 1

  1. No. This will burn out your MOSFET as there is nothing to limit the current from drain to source. This should not harm your function generator, since the MOSFET's gate will not draw current other than to satisfy its capacitance.

  2. This unit has a 50 Ω output capable of ±20 V into high impedance or open circuit (±10 V if the frequency is > 25 MHz). Into a 50 Ω load it can do ±10 V (±5 V if the frequency is > 25 MHz). You can use Ohm's Law to derive the currents if that is of interest to you.

  3. Output impedance and voltage range is listed on page 4 of your datasheet under "General Characteristics - Amplitude".

  • \$\begingroup\$ Thank you for your answer. I have a few questions. When you say "MOSFET's gate will not draw current other than to satisfy its capacitance." - Could you explain a little bit more on this. And Could you please tell me, if I connect the circuit in this fashion, what is the amount of voltage that will be impressed onto the BNC output of the FG and how to calculate that voltage and sink current. Please \$\endgroup\$
    – user220456
    Mar 24, 2020 at 3:45
  • \$\begingroup\$ The gate on a MOSFET looks like a capacitor. Like a capacitor, it blocks DC. In your drawing, no voltage is impressed onto the output of the FG because in MOSFETs the gate is not connected to the source or drain directly. You might imagine a MOSFET like this: The gate is a capacitor and the amount of conduction between the source and drain is affected by how 'full' that capacitor is. Almost no current flows into or out of the gate, and its voltage is not affected by the source or drain. \$\endgroup\$ Mar 24, 2020 at 14:40

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