I need enlightenment, i don't quite understand this,i'm supposed to make a switch, using 2N5459, the buzzer's operating voltage is 9V, the led V_F is 3.5V, I_LED(max) is 20mA. I dont quite get I_DSS, its a range, what do i use? will this work as a switch? in my mind, if switch is closed,V_GS will -8V, so 0 current, so led off, and if open switch, will led turn on? so confused

 

Great Scott just released a video on using JFETs. Since their use is not-so-common these days, he explains all the considerations of using them. It's probably a good investment of 10 minutes of your time.

 

hi mr,
Welcome to AAC.
Study the attached datasheet for the 2N5458, note the Vgs range.
E

 

JFETs are normally-on devices.
To turn them off, they require a negative bias to deplete the channel.
Otherwise, watch the video.

EDIT: there is a nasty trend on this forum, where single post members, like you, ask a question and never, ever comment again.
Not even a simple “thank you” or “I have solved the issue”.
I trust that you won’t be one of those.

 

if switch is closed,V_GS will -8V, so 0 current, so led off, and if open switch, will led turn on?

Yes, it should turn off.

I dont quite get I_DSS, its a range, what do i use?

You want a minimum Idss value that's larger than the current you want to switch.
The load resistance should determine the ON current.

Why do you have the buzzer connected to the drain and the LED connected to the source?
For using the FET as a switch, it would be better to put everything is series with the drain and ground the source terminal.

But having the LED in series with the buzzer means they both carry the same current.
What is the buzzers operating current?

 

Yes, it should turn off.
You want a minimum Idss value that's larger than the current you want to switch.
The load resistance should determine the ON current.

Why do you have the buzzer connected to the drain and the LED connected to the source?
For using the FET as a switch, it would be better to put everything is series with the drain and ground the source terminal.

But having the LED in series with the buzzer means they both carry the same current.
What is the buzzers current rating?

so, the circuit is my understanding of the instructions for the exercise as follows:
Design and construct a self-bias applied as a switch using n-channel FET. Use 2n5459 for the transistor. (DC Analysis Only)
The following are the Specifications:

• VDD = 15V
• RD becomes a resistance Buzzer
• The buzzer has an Operating Voltage of 9V.
• RS becomes a GREEN LED with a series resistor.(
• A Switch with Negative Source (VIN) and series resistance (RIN) is located parallel at RG and Gate terminal.
• This negative source is equivalent to the max pinch-off voltage of the transistor.
• RG = 1MOhm and RIN = RG / 10
  max pinch off per datasheet is -8V, green led V_F is i think3.5V and I_LED(max) is 20 mA i think.
  
  i want to do the math, but im stuck with what I_DSS to use.im fairly new at using proteus, and the simulation didn't help(probably wrong setup). I actually made this work, by connecting the positive of V_in to gate, but contradicts the instructions.

 

If you don't know the buzzer operating current, then it's not possible to do the design calculations.

 

If you don't know the buzzer operating current, then it's not possible to do the design calculations.

thanks for the help. guess i'll do more tinkering, hopefully will make things work

 

thanks for the help. guess i'll do more tinkering, hopefully will make things work

So what current/resistance value will you use for the buzzer?

 

So what current/resistance value will you use for the buzzer?
[/QUOTE]
so to make v_gs close to 0, I made v_in positive, simulation kind of works, buzzer buzzes, led glows faintly, will compute proper values later. load resistance for the buzzer here is 1.06k

 

hi mr.
I get similar results in LTSpice as per your actual circuit, very low Drain current???
E