How to use 2N7000 as low side switch

Thread Starter

other_santa

Joined Dec 22, 2015
4
Hi All,

I am learning ABC of electronics.

This time I am learning how to use 2N7000 mosfet as switch.
On breadboard I made a circuit "A" as shown in attachment and it will make LED ON when Gate is made high.

However, when I change the position of load, LED is not being ON as I expected.
Can someone please explain why circuit "B" does not make LED ON when G is made high?

Thank you very much for reading this post.
 

Attachments

#12

Joined Nov 30, 2010
18,186
Same as when you use an NPN transistor. The LED in the collector (drain) circuit sees the transistor as a switch. When you place the LED on the emitter (source) side, the voltage to the LED is always lower than the gate (base) voltage.
 

Thread Starter

other_santa

Joined Dec 22, 2015
4
Hi #12 and Bordodynov

Thank you very much for the post.
I did not expected so fast reply : )

All the tutorials I see show similar circuit to what Bordodynov has shown.
They put load on high side.
Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?

Thank you once again for your kind posts.
 

ScottWang

Joined Aug 23, 2012
6,861
The rating Vgs=10V of 2N7000, if the Vgs too low may not works properly.
You can using circuit B to try, and move the led to the Vd to V+(Vbat), Vs connecting to V-(Vbat).
If the led still didn't light up, then you need to change to other mosfet has very low Vgs as FDV301N.
 

Thread Starter

other_santa

Joined Dec 22, 2015
4
I used 9V battery and can see that LED is going ON.
May be it is time for me to buy a P channel MOSFET for using in circuit B.

#12, Bordodynov and ScottWang Thank you very much. It was my first post here and you guys are really helpful.

Thanks!!!
 

ScottWang

Joined Aug 23, 2012
6,861
I used 9V battery and can see that LED is going ON.
May be it is time for me to buy a P channel MOSFET for using in circuit B.

#12, Bordodynov and ScottWang Thank you very much. It was my first post here and you guys are really helpful.

Thanks!!!
Normally you will need a current limiting resistor for led.
R = (Vcc - V_led) / (I_led*80%)

For example:
R = (9V - 3V) / (20mA * 80%)
R = 6V/16mA
R = 375 Ω.
You can using a 360 Ω or 390 Ω
 

atferrari

Joined Jan 6, 2004
3,653
I know this is not an actual answer to your question but it is what I had to use to drive a LED with a signal from an existing circuit (comparator). The disadvantage is that it is always drawing power.

Just in case, here you have it. . .

Drive the LED.png
 

Bordodynov

Joined May 20, 2015
2,469
other_santa said:
......
All the tutorials I see show similar circuit to what Bordodynov has shown.
They put load on high side.
Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?
Yes.
Use p-cannal. Example FDV302P or FDV304P.

FDV304P+LED.png
 

ScottWang

Joined Aug 23, 2012
6,861
other_santa said:
......
All the tutorials I see show similar circuit to what Bordodynov has shown.
They put load on high side.
Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?
Yes.
Use p-cannal. Example FDV302P or FDV304P.

View attachment 97094
Do you think when you adding positive voltage(V2) to the Vgs will the p mosfet works?
 

ScottWang

Joined Aug 23, 2012
6,861
In post # 10 I used a control voltage source V2. If there is no control, then why transistor? Then, only one resistor.
That is just a testing circuit, and if you need a simple control then it should be adding a resistor connecting from Vg to +V (+Vbat), and the control switch will be connecting from Vgs to ground (-Vbat).

The circuit you attached is like a pulse control, but here is no needed, and normally when you using a P mosfet then you will need a npn bjt to drive the p mosfet, and the input can be input the pulse from 0V to 5V and 5V to 0V ...
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