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.

 

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.

 

See

 

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.

 

Is it possible to use N channel mosfet(2N7000) as switch and put load on low side?

Yes, but you will need more voltage so as to allow for the Vgs of the mosfet plus the voltage required by the LED.

 

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.

 

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!!!

 

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

 

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. . .

 

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.

 

@atferrari that is an interesting circuit. Thanks for sharing.

@Bordodynov Thanks for circuit and part numbers. I will try to get these parts and experiment.
Thanks!!!

 

You can use a voltage multiplier:

 

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?

 

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

Why do you ask? He has V2 = 0V in post 10.

 

Why do you ask? He has V2 = 0V in post 10.

It just quite strange and why not just connected to gnd?

It looks like there is a voltage at the Vgs.

 

In post # 10 I used a control voltage source V2. If there is no control, then why transistor? Then, only one resistor.

 

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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