Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Help with a circuit

Status
Not open for further replies.

307pavlos

Member
Hello my friends, i have this circuit and i'm asked to explain how Vin works here.
I know that FET's NORMALLY ON which means when 0V applied in its base it....works.
And when negative voltage is applied then it start to fall until Vgs(off). When V=Vgs(off) then it's OFF.

Ok i made a first explanation in my mind but it's not correct. :)

Well at first square-wave voltage source has 0V.
Ib(2n2222)=V/R=(-15)/2k= -7,5mA.

The voltage in the 2N2222 base is -15V so the 2N2222 doesnt "work".

So the only voltage we have in the IN4148 is the +15V from the Vcc source.
So the IN4148 works as cutoff, so no voltage in FET. so FET is normally ON.

When Square-wave source has 5V there's voltage in the 2N2927, so it is added in the -15V and the 2N2222 goes to working mode. and "sends" voltage in the 2N2222 collector and there's negative voltage in IN4148 cathode. If this negative voltage is below Vgs(cutoff)<-4V the FET goes OFF.

I know i say nonsense but i had to ask :)
 
What FET are you using?
 
so no voltage in FET. so FET is normally ON.
?? If Vin is 0 and D is 0 (ground) the FET can't be ON (i.e. pass a current from S to D), regardless of its gate voltage. The FET (a depletion mode type) can only be ON when the 2N2222 is OFF and Vin > 0.
 
alec_t it's the first time i see something like this. That's why i dont know anything. I just want to understand how it works. If it helps you, ignore my explanation and just see the schematic. Place your own part details, values, etc but please give me an explanation of its operating mode. In fact he wants us to understand/calculate the variance of Vin of this schematic!
 
and just see the schematic.
I did. I've given you some clues, but I'm not going to do all your homework for you. You need to use the background information you were given by your teacher :).
 
Ok here's the answer.

At the moment when Vsquare=0V then :

Q2=OFF, Q1=OFF. So in the IN4148 cathode we have 15V. So no current pass through the diode, and the Gate is OFF and the current from Vin pass through S to D.

At the moment when Vsquare=5V then :

Q2=ON, Q1= saturation, so in Q1's collector we have 0.3V. So in diode's cathode we have -14,7V. Diode "gets" 0.7V so in the FET's Gate we have -14V.

But the Vgs(off)<-4

So Vin= +10V max.
 
Q1= saturation, so in Q1's collector we have 0.3V.
You were doing well up to that point:). If Q1 is saturated its collector will be 0.3V above its emitter voltage.
So in diode's cathode we have -14,7V. Diode "gets" 0.7V so in the FET's Gate we have -14V. But the Vgs(off)<-4
Agreed. So is the FET now conducting or not?
So Vin= +10V max.
You can't deduce that; all you can say is that the gate voltage is 10V below the cut-off voltage.
he wants us to understand/calculate the variance of Vin of this schematic!
Are you sure that is the exact requirement? It makes no sense to me. I would interpret the schematic as showing Vin as an unspecified input voltage, if the schematic is using conventional symbols.
 
If voltage in Gate is -14V then it's not conducting so Vin current pass through S to D.

When i said Vin=+10V max i meant that Vin can be maximum +10V because Vgs(off)=-4V. Yes this is the exact requirement. It doesnt make sense anyway :)
 
Hi,


It looks like you have the logistics of it right. When the pulse is zero, the FET is 'on' When the pulse is high, the FET is 'off'. But what might vary is fully 'on' or not fully 'on', and also the available turn off voltage is limited to -15v.

Im not sure if i understand the question that well either, but it appears that since the cathode of the diode is biased at +15v it will become forward biased with anything above that on the anode, and that will cause a drop in the output voltage and this means NOT fully 'on' anymore. For some school work this means the max can either be 15v, 15.5v, 15.7v, or 16v, depending on what that course at that particular time in the semester is using for a diode forward voltage drop, 0v, 0.5v, 0.7v, or 1v respectively.

The min voltage for Vin will depend on the available turn off voltage, -15v, and the diode drop and the 2N2222 saturation voltage. We need at least -4v so that means the max Vin will have to be -10v with a 1v diode drop and 0v saturation voltage.
 
Last edited:
If voltage in Gate is -14V then it's not conducting
Agreed.
so Vin current pass through S to D.
How do you figure that? If it's not conducting then, by definition, no current passes from S to D.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top