# MOSFET Basic Biasing Problems

After a lot of theoretical studying of MOSFETs, I decided to try out at least the basics of it in practice.

Here is the first circuit I ever made using MOSFET: simulate this circuit – Schematic created using CircuitLab

https://www.onsemi.com/pub/Collateral/BS170-D.PDF

I haven't used resistor in series with gate, because MOSFET is obviously voltage driven component (input gate current is negligible).

I chosen next parameters for the circuit:

• Id(sat) = 10mA
• Id(bias) = 5mA; where Vds = 5V
• Vr1 = 5V
• Rd = Vr1/Id(bias)

Then I didn't really knew which way to calculate Vgs (since the actual gate current can't be really calculated), so I tried next:

• gm = Id/Vds, then Vgs = Id/gm = 5V,

After I made the circuit on breadboard, this are the values I got with measurement:

• if Vgs = 1.8V or less, then Vds = Vcc
• if Vgs = 2.4V, then Vds = Vcc/2 (I wasn't even close to this value...)
• if Vgs = 3.8V or more, then Vds = 0

Where did I go wrong on this one?

• What were you trying to do? The Vgs threshold is approxmately 2V according to the datasheet, so with Vgs = 3.8V, the Mosfet is conducting, i.e. negligible resistance between the drain and source, so I would expect Vds to be approximately 0 V. – Ben Jul 10 '17 at 19:19
• It sounds like 2.4V is in the middle of the "linear region" of the MOSFET response curve. Check the data sheet for typical response curves, which should explain the response you're getting out of the device. – Jon Watte Jul 10 '17 at 19:51
• @JonWatte: Exactly. How is it possible to be in the middle of ohmic region and also at the middle of the load line?? – Keno Jul 10 '17 at 19:57

You should look more closely at the data sheet. Go to page 2, and about the 3rd item is gate threshold voltage. This is defined as the gate drive necessary to produce 1 ma drain current, and is specified to be in the range of 0.8 to 3 volts, with a typical 2.0.

In your breadboard circuit, in order to get Vcc/2 (5 volts) at the drain, you need 5 mA of drain current. Since a typical current at 2 volts is 1 mA, 5 mA at 2.4 volts seems perfectly reasonable. At voltages much below this the drain current will be right up next to 10 volts, since there is no current being drawn, and for much greater voltage the FET will be turned on hard and Vd will be near zero.

The only thing you really did wrong was computing gm. You seem to lost sight of the fact that this is your desired gm, not what you can actually expect. Go back to the data sheet, and you'll only find one gm value, specifically 0.2. However, notice the operating conditions - Vg is 10 volts and id is 250 mA, which is far, far away from the actual conditions you provided, so you can't trust it for what you were doing.

• So the term Vds(sat) > Vgs - Vt for transistor to operate in saturation/active region isn't relevant when analyzing biasing conditions of MOSFET? – Keno Jul 10 '17 at 20:02
• @Keno - Sorry? That makes little sense to me. Vds(sat) is by definition the voltage across the FET when in saturation. How can it apply to linear operation? And the boundary between linear and saturation is a fuzzy one, anyways. – WhatRoughBeast Jul 10 '17 at 20:08
• You know the relation between Vds and overdive voltage (Vgs - Vt)? The relation which tells us, when the mosfet's channel operates in pinch-off? This Vds described above isn't actually saturated (my bad), but when Vds equals the overdrive voltage, Vds is right on the boundary between linear and saturation region. – Keno Jul 11 '17 at 12:49

Where did I go wrong on this one?

You didn't go wrong anywhere. When you have Vgs below the threshold voltage (typically about 2V, but ) and the FET is off, you should see Vdd across its drain to source. When your Vgs is above the threshold voltage you should see very little voltage across its drain to source. If you are hovering around the threshold voltage you will see something in between 0V and Vdd across its drain to source.

When using a MOSFET as a simple switch, you want to be sure that your "off" Vgs is well below the minimum threshold voltage (0.8V for the BS170), and want to be sure that your "on" Vgs is well above the maximum threshold voltage (3.0V for the BS170).

• But Vds changes so quickly as Vgg is increased for volt or two (when Vgs is above threshold voltage); which seem to me a bit strange, don't you think? – Keno Jul 10 '17 at 19:53
• I know how to use it as a switch, but I don't know how to use it as an amplifier - and of course I already have problems with biasing of transistor. – Keno Jul 10 '17 at 19:55
• @Keno Using a MOS in linear region definitevely calls for strong DC feedback (e.g. source degeneration resistor) – carloc Jul 10 '17 at 20:18

There seems to be several flaws in your understanding of a MOSFET's operation. gm is not equal to Id/Vgs as you noted(you have a typo there saying it is Vds but I think you didn't mean Vds).

The transconductance of a MOSFET is equal to 2*Id/Vov, only when the MOSFET is in saturation. Vov is Vgs-Vth, where Vth is the threshold voltage. In the three cases that you have breadboarded, you might have put the mosfet in any of these three regions of operation, viz. sub threshold, linear and saturation (need to actually solve equations to see which region it is in: check here)

At a high level, when VGS is greater than Vth, the larger the VGS you put in, the more current you make the MOSFET sink. This current flows across the 1kΩ resistor and therefore, the VDS drops as current increases, as VDD is fixed. So what you are seeing is sane.

You seem to be concerned with input current to MOSFET. But in most cases, it is safe to assume this current to be zero(unless the MOSFET is a leaky cheapo). This zero input current is what gives MOSFETS >100 MΩ-s(frequently >1GΩ) of input impedance as compared to BJTs.

I haven't used resistor in series with gate, because MOSFET is obviously voltage driven component (input gate current is negligible).

is actually over-simplified.

In reality, MOSFET devices have significant effective capacitances on the gate as shown below. As such, when turning them on and off there is a finite charge which must be applied or removed during that turn on or off time. This results in a transient currents (and voltages) which, if not limited, can destroy whatever is driving the gate. A gate resistor is therefore almost always required. simulate this circuit – Schematic created using CircuitLab

The size of the resistor is dependent on the maximum current the driving device can source or sink as well as the drain and source voltages applied to the device.

When used as a simple low frequency on off-switch, this resistor can be fairly large.

However, when used in a high frequency switching application you will want the resistor as small as possible so as to limit the effect on the charge time, and consequent switching time, of the MOSFET whilst still protecting the driver.

It should also be noted that in high frequency switching applications this charge/discharge current requirement can produce significant power to be dissipated in the driving gate as well as the gate resistor.

• But obviously you are talking about resistor in parallel with the Gate, right? – Keno Jul 11 '17 at 13:12
• @Keno no.. in series. – Trevor_G Jul 11 '17 at 13:13
• What? Why in series? You don't have to limit the input gate current because it is already limited by oxide resistance, or not? – Keno Jul 11 '17 at 17:44
• @Keno which part of capacitance don't you understand. When you apply a voltage to the gate there is initially close to a short to the drain till it charges. Same when you try to turn it off. Series resistance limits how much current can be drawn from the previous device. – Trevor_G Jul 11 '17 at 17:48
• For example, for a microcontroller with 2 mA max continuous output pin current but 8 mA max surge current, you'd want to make sure you never pull more than 8 mA. To switch Vgs to 3.3V means you'd need a resistor of at least (3.3V / 0.008A) == 412.5 Ohms. Better kick it up to 470 to have some margin. The only time gate resistors aren't needed, is when the output before the gate is self-protecting (has its own source impedance that can be at least briefly shorted.) – Jon Watte Jul 15 '17 at 23:35