Hi I don't know how to identify gate-source voltage in this example:
Can anybody help?

 

Posting in the homework section requires that you make some attempts on your own first, and show your work. You probably won't get any help until you show a little effort on your part.

 

At what time are you to answer the question? Ask yourself what the gate current is at or after that time? What is the voltage drop across R1? What is the voltage at the top rail?

John

 

Assuming that this is a DC circuit, the key is to ask what the current is that is flowing in R1 and, from that, ask what the voltage drop across R1 is.

Beyond that, you really need to show some work.

 

In your schematic, the gate-source voltage Vgs is simply the voltage on the gate, because the source, S, is grounded.
Vgs = Vgate - Vsource. Is that what you're asking?

 

In your schematic, the gate-source voltage Vgs is simply the voltage on the gate, because the source, S, is grounded.
Vgs = Vgate - Vsource. Is that what you're asking?

I solve this problem, but I'm not sure about the Vgs. Could you check if this is correct?

 

Were you given Vds? If not, how did you calculate it?

Assuming your Vds is correct, then your answer is close, but there is a small error from rounding: Vds + Vr2 = Vdd = Vgs (after sufficient time for the gate capacitance to charge and assuming no gate leakage, i.e., no gate current).

John

 

Were you given Vds? If not, how did you calculate it?

Assuming your Vds is correct, then your answer is close, but there is a small error from rounding: Vds + Vr2 = Vdd = Vgs (after sufficient time for the gate capacitance to charge and assuming no gate leakage, i.e., no gate current).

John

Vds was given in DC steady state as VDSdc = 4.9V and about the gate current.. Can it be associated with isolated gate and hence Ig=0 in steady state?
I'm a beginner in this, so sorry if my question seems stupid, I'm trying to figure out how does it work...

 

Vds was given in DC steady state as VDSdc = 4.9V and about the gate current.

I did not see that value in the question you posted. With many mosfets at a gate Vgs of 7V, one might expect a lower Vds, but the value you used is not unrealistic.

Can it be associated with isolated gate and hence Ig=0 in steady state?

Yes, to a first approximation. Look up a datasheet for a mosfet that you have available. In my case, I have an IRF1010E. At 20V Vgs, gate leakage is specifed as 100nA. That is very small, but your gate resistor is 0.25M, which is large for a gate resistor. Therefore, VR1 would be 0.1*0.25= 0.025V with a Vgs of 20 V. Now, your Vgs is less and your mosfet isn't specified, so I assume it is "ideal" and has no gate leakage.

Regards, John

 

I did not see that value in the question you posted. With many mosfets at a gate Vgs of 7V, one might expect a lower Vds, but the value you used is not unrealistic.



Yes, to a first approximation. Look up a datasheet for a mosfet that you have available. In my case, I have an IRF1010E. At 20V Vgs, gate leakage is specifed as 100nA. That is very small, but your gate resistor is 0.25M, which is large for a gate resistor. Therefore, VR1 would be 0.1*0.25= 0.025V with a Vgs of 20 V. Now, your Vgs is less and your mosfet isn't specified, so I assume it is "ideal" and has no gate leakage.

Regards, John

so assuming that it is ideal,is the equation for Vgs correct?

 

Hi I don't know how to identify gate-source voltage in this example: View attachment 90906
Can anybody help?

The gate is basically a capacitance - for any effect you can measure, its resistance is pretty much infinite.

There will be no measurable volt drop on the resistor feeding the gate from Vdd.

With a bipolar transistor, a cheap'n'cheerful bias arrangement would have the bias resistor from collector to base - any tendency for collector voltage to fall, also reduces the base bias current, which in turn backs off the collector current (negative feedback) that ultimately stabilises the collector DC operating point.

With a MOSFET, the gate doesn't draw any DC current, so you have to feed the gate with a voltage divider from the drain to set an arbitrary DC point.

 

Trying to get at Vgs by adding Vds to Vr2 is meaningless because Vds+Vr2 is equal to 7V because the top of R2 is tied to an ideal 7V voltage supply. That same supply is connected to the top of R1. Hence the gate circuit is completely shadowed from the drain/source circuit by the ideal voltage source.

In general, you CANNOT say that

Vgs = Vds + Vr2

because you have not included the voltage drop across R1. Hence my original question about what the current in R1 is. If you are going to assume that there is zero current (due to zero gate leakage current) at DC, then there is no voltage drop across R1. That means that Vgs is 7V because the top of R1 is tied to an ideal 7V voltage supply -- it has absolutely nothing to do with Vr2 or Vds.

 

Iza,
I now understand your problem. A MOSFET is connected as shown with known resistors and known VDD, and Vds is measured. From this measurement, you are supposed to calculate
Drain current, ID
DC current gain, Beta
Small signal transconductance, gm

The mistake you have made in your work is how you calculated VGS. You calculated 6.99V, which is close, but only because your equation for VGS is actually the equation for VDD. And the reason you did not get 7 V exactly is because you rounded the current from 19.090909mA to 19mA. Applying Kirchoff's voltage law to the gate path yields:
VDD = IG*R1 + VGS
Therefore the equation for VGS is
VGS = VDD - IG*R1

I'll leave it for you to calculate Beta, the current gain, since you haven't asked about that, but a big hint is that the gate current is zero.
Also I'll leave it for you to calculate gm, the transconductance. Remember, it is the (change in ID) /(change in VGS), and remember ID is zero when VGS = VTh.

Regards

 

Now I understand Thank you all for help