I built a CS amplifier today, using a BSS670S2L (N-ch MOSFET).

I got the following results (in RED).

I really dont understand why the MOSFET's VDS was so low.
The voltage drop on RD was 6V, its very weird.
Even when I changed RD from 150Ω to 47KΩ, RD's voltage drop was still 6V.

Why did the MOSFET keep being in the linear region?
I waned to get it into its saturation region.

Could it be because of the diode its got connected between its D-S channel?

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The device is in a saturated condition. Changing the vale of the drain resistor does not chnage this condition.

The linear region would be have a larger vlaue for VDS.

 

You are not correct.
The MOS is saturated if VDS > VGS - VTn, which doesnt happen in this case.

 

Your question is easy: 40mA * 150 = 6V. 6.1V(VDD) - 6V = .1V. If you want active region (satruation) you need either higher VDD, lower drain resistor, or change the gate bias (lower)

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I see, thank you.
I have 2 questions please.

1. At first, I picked RG1=RG2=47Kohm, and when I powered-up the circuit, VGS was not 3.05V but it was less than 0.5V.
I measured the resistance of RG2 when it was still connected to the circuit, and the multimeter read 3Kohm! its like the Gate-source had a resistance of 3Kohm.
Could you please explain this phenomenon to me?

2. If I decreased VDD to 1V, so VGS=0.5V<VTn.
Should I have seen VDS = VDD? (since no current would have flown throught RD).

Thanks.

--
By the way, perhaps "Are our children learning?" is less rare a question

 

The quote in my signature was taken verbatim from one of my country's great leaders.

You can't measure the value of RG2 in your circuit with any accuracy. If you resistance meter uses voltage higher than the transistor's threshold voltage, you turn on the transistor.

If VGS of .5 volts is less than Vt, then you should see VGS=VDD.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I agree

Well, the fact was that when using RG1=RG2=47KΩ, the VGS was equal to VDD/2, but was equal to ~0.4V.
I didnt understand your sentence about the voltage the multimeter uses, could you explain how its got to do with the problem?


You mean VDS=VDD, right?

 

Right. Too easy to get those mixed up.

If you're using a multimeter with resistance measurement selected across R2, you're applying a voltage from the meter which can forward bias Q1, which effectively puts R1 and the DS of Q1 in parallel with R2. I can't begin to explain the 0.4V you get with R1=R1=47K. Something seems wrong there.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Well, I really measured the voltage on RG2 (VGS) and it was ~0.4V when RG1=RG2=47kOhm.
But when I picked up RG1=RG2=100ohm, then VGS was VDD/2.
I dont get it also.

By the way, what is usually the zenner voltage of such D-S diodes?
It wasnt specified in the datasheet.

 

There isn't a zener voltage is the D-S of a mosfet. They are electrically isolated. A voltage that breaks down the barrier would most likely destroy the device.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I believe there is one.
Could you take a look please at the datasheet?

 

Uh-huh, and???

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I was wondering whats its Zener voltage, it isnt specified in the datasheet, so whats the usual zener voltage in such D-S zener diode?

 

There isn't a zener voltage. That's why it's not specified on the data sheet.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

So whats the symbol of the zener diode for?