JFET Bias 101

[swinginguitar]

Can someone give me the overview of biasing a JFET as it relates to audio circuits?

I've been doing some reading to try to understand in a little more depth. When you apply an AC signal to the gate of the JFET, what is happening to the voltage and current on drain and source? I read that gate voltage should move between 0v and some negative value, but never positive. How does this relate to a signal on the gate vs what is being output? Is it biased to where the input is moving between 0 and some negative voltage (let's say -9v just for kicks) with the midpoint of -4.5v being the bias point? We're talking voltage amplification right (as opposed to current in a BJT)? What is the effect of the resistors on th circuit? How is bias determined?

THANKS!

[R.G.]

I've been doing some reading to try to understand in a little more depth. When you apply an AC signal to the gate of the JFET, what is happening to the voltage and current on drain and source? I read that gate voltage should move between 0v and some negative value, but never positive.

Oh, sure. Only ask easy questions... 

JFETs are weird. They work by the garden-hose effect. No, really. If you put a water pressure across a garden hose from one end (the supply) to the open end (... ground... sorry, I couldn't help it) then water (current) flows. If you then pinch the hose, it causes a restricted flow area, and the amount of flow (current) decreases. You can even pinch it completely off.

The drain to source is a channel of resistive material. If you leave the gate lead open voltage flows through the channel like it's a resistor, which it is. If you put voltage on the gate in the correct polarity, its electrical field pushes the charge carriers toward the edge of the channel and pinches off some of the current flow. It can eventually pinch it completely off.

In a JFET then, current flows until you do something to stop it. This is called a "depletion mode" device, as you must actively deplete current flow.

The "J" in JFET stands for "junction". The gate is insulated from the channel by a reverse-biased diode junction. To make the gate be insulated from the channel, it has to be held more negative than the channel (for an N-type channel; the reverse is true for a P-channel JFET) so it is insulated. With this polarity, the more negative the gate, the more it pinches off the current flow in the channel.

How does this relate to a signal on the gate vs what is being output?

For signal at the gate to change the current flow in the channel, you need some voltage across the channel. In normal amplifying situations, this means setting up a DC current through the channel, and then some external resistor so that when the gate ->voltage<- change changes the ->channel current<-, the current change through the external resistor, and you get a change in voltage across the resistor.

For a JFET, a change in voltage on the gate changes the current through the channel; this is called a trans (i.e. across) conductance (inverse of resistance). The unit of transconductance is the Siemen, which is the fancy name for the inverse of an Ohm - it's one amp per volt. A JFET with a transconductance (sometimes called Yfs) of one milliSiemen will change its conduction by one milliampere per volt of change on the gate.

The gain of a JFET stage is then close to Yfs*Rd/Rs.

That all assumes signals small enough not to bump either the gate or drain against certain voltage limits.

Is it biased to where the input is moving between 0 and some negative voltage (let's say -9v just for kicks) with the midpoint of -4.5v being the bias point?

No. The JFET works reasonably correctly - with some caveats - with the gate voltage with respect to the source between 0V and -Vgsoff.  0V is where the gate is essentially not affecting current flow at all, and -Vgsoff is the voltage between gate and source where the gate field pinches off all current flow. As a practical matter, the gate is at some bias voltage between 0 and -Vgsoff (for N-channel) to hold the JFET partially off. This lets some voltage flow in the channel and drain resistor, and gives a middle voltage on the drain so the voltage at the drain can vary up and down from there. The gate voltage in the middle of the Vgsoff range lets it vary as well.

The Vgsoff is different for each JFET type, and each JFET of the same type. If you had a JFET with a Vgsoff of -9V (that means, -9V on gate compared to source pin cuts off current flow entirely) then -4.5V *might* be an OK place to DC bias the gate. The next JFET you pull out of your bin of the same type number might have Vgsoff of -7V, the next -8.5, the next -4, ...

But yes, the idea is that you bias a JFET to some point where the gate is held between 0V and -Vgsoff compared to the source pin, not compared to anything in the power supply. It's the source pin voltage that matters to the gate.

We're talking voltage amplification right (as opposed to current in a BJT)?

We're talking trans-conductance: change in gate voltage making a proportional change in channel current; this is then converted to a voltage by running it through a resistor, just like happens with a BJT's collector current.

What is the effect of the resistors on th circuit?

ARG! Huge subject.

How is bias determined?

It's difficult, interactive from gate to channel and with transconductance and Vgsoff. And so varies from device to device, even within the same type number.

Sorry if those last two are vague. They *are* vague, and complex issues. BJT biasing is very much simpler.

[abram]

extremely valuable bit of info there, R.G. thanks!

[Tony Forestiere]

Aron, Andrew:
Can we get a STICKY? Great explanation. I almost understand. More reading is due. 

[swinginguitar]

So glad I asked!!! Another vote for a sticky.

1) so it can be said that the voltage on the gate is opposite of what the base voltage does on a BJT then? (rise in gate voltage = decrease in current ; rise in BJT voltage (ok...current) = increase in current)

2) you mentioned the current being converted to voltage by a resistor - I know that's plain ol' Ohms Law, but believe or not, that is a concept I've yet to get my head around...is there a good analogy to help with that? I get confused b/c in my mind, where there is current there *is* voltage (pressure/potential) and vice versa...or am I off here?

3) what's the diff between transconductance and amplification? i see that Transconductance figures into the gain of the amp, but...

4) Let's bring it a little closer to home...using these:

   

Assuming a 9v power supply, can u walk me thru the biasing of these buffers? the voltage divider would put the gate at 4.5v...is that the "negative" with respect to V+ thing?

Great post..tks! Luv this forum!

[R.G.]

1) so it can be said that the voltage on the gate is opposite of what the base voltage does on a BJT then? (rise in gate voltage = decrease in current ; rise in BJT voltage (ok...current) = increase in current)

mmmmm... not exactly. In a BJT, rise in BJT base-emitter voltage causes an increase in the collector current. That makes the collector ->voltage<- drop because more current through the collector resistor makes the resistor's drop down from the + power supply (assuming NPNs here), or a rise in emitter voltage if there is an emitter resistor because the current through it makes the resistor rise from ground bigger.

In the n-channel JFET, a rising gate voltage with respect to the source, in the direction from -Vgsoff toward 0V, makes current increase in the channel because the gate applies less turn-off to the channel. So it acts in the same direction as the BJT's base-emitter. What's different is that the JFET gate-source has to be DC offset substantially negative.

The BJT, is by the way, an ->enhancement mode<- device. That means that if you leave the base open, or tie it to the emitter, no current flows, and you have to raise the base (... enhance it...) to get something to happen. So the actions are analogous, but have different DC bias requirements. JFETs have some other gotchas/differences, notably that if the base is shorted to the source, the channel limits the current through itself to Idss ( current I through the drain with source and gate shorted).

2) you mentioned the current being converted to voltage by a resistor - I know that's plain ol' Ohms Law, but believe or not, that is a concept I've yet to get my head around...is there a good analogy to help with that? I get confused b/c in my mind, where there is current there *is* voltage (pressure/potential) and vice versa...or am I off here?

No worries. Back at pressure and flow rate. If you want to get 10 gallons per minute through a hose, how much pressure do you need to pump it with? Well, that depends on how big a hose you have. The pressure to get 10 GPM through a 1/2" interior garden hose is quite a bit bigger than the pressure needed to get it through a 4" interior diameter fire hose. The "resistance" of a hose can be stated as pressure per flow-rate, or PSI per gallon-per-minute.

Actually, the analogy to fluid resistance extends; in solid resistors, resistance (which has units of volts per ampere) is equal to rho*l/A, where l is the length of the resistor strip, A is the cross sectional area, and rho is the material resistivity, very much like how much the hose restricts water flow. An open hose has a lower resistivity than one packed with sand.

So a resistor converts current to voltage in the same way that a hose converts flow rate to PSI.

3) what's the diff between transconductance and amplification? i see that Transconductance figures into the gain of the amp, but...

Amplification can be many things. It can be voltage amplification, current amplification, or power amplification. It's always just a ratio of how much comes out to how much went in. Voltage amplification is (usually AC) volts out divided by volts in. Current amplification is amperes out divided by amperes in, and power is watts out over watts in; this last is very important in RF amplification. Audio generally works on voltage amplification if something else is not stated.

Transconductance, and generally 'trans' anything is a ratio of how much out to how much went in, but they're not the same kind of thing. Trans'conductance' is amperes out divided by volts in, and has units of one divide by ohms or amperes per volt. This is called a Siemen as an honor to the guy of that name. Trans'impedance' amplifiers have voltage out divided by current in. Engineers have ratios for everything, and name most of them odd names. Ever see the term "Graatz Number" or the more common "Reynolds Number" for fluid flow? EE is dead simple straightforward compared to fluid mechanics and heat transfer.

On the buffers: remember that for an N-channel, the gate has to be equal to or less than the source voltage. And, these being from muzique.com, I suspect the JFET in the diagram is probably a J201, which is an odd duck even among the flock of odd ducks that are JFETs. The J201 has a very, very low Vgsoff for a JFET. So hang on.

In the first one; if you cut R1 out, the JFET will let its particular Idss flow through R2. This is generally from a partial to tens of milliamperes. And the current will be constant over a wide range of V+, because that's what JFETs do - they are voltage controlled current sources, and the voltage with the gate open is zero (leakage currents pull it to the source). If you put R1 back in and vary Vr, then the voltage on R2 will vary. Remember, if the gate is open, R2's voltage is Idss * 3.3k. So what's Idss?

It varies. A lot. JFETs are grouped into about 3:1 to 5:1 Idss groups. Let's pretend that it's really 2ma. So with R1 open, the current through R2 is 2ma, and the voltage at the top of R2 is 2ma*3.3K = 6.6V. It's 6.6V whether V+ is 9V or more, up to where the JFET burns up from the heat. If you connect R1 and vary Vr, you can vary it from 6.6V to anything more negative until the gate junction breaks over and smoke erupts again. As it goes lower from 6.6V, the voltage between gate and source goes negative, and so less current can flow, so the voltage on R2's top goes down. In fact, the voltage on R2 roughly follows the gate down for small drops. This is, after all, a source follower, and an AC signal at the input would come out of the output a just under unity voltage gain.

If R2 was a current source, the voltage on the source would follow the gate down nearly perfectly until some other gotcha conditions happened. But R2 is not a current source, and as the current through it goes down, its voltage goes down following the gate, and that reduces the gate-to-source voltage that the JFET depends on, so it goes down less than you think it would. When the Vr gets to 0V, same as the ground voltage on R2, there is still some voltage across R2, and the JFET settles down to some current through R2 holding up the source above ground, and the gate at ground. The voltage on the source is perfectly the voltage needed to make that current flow through it with the gate at ground; it's a 100% feedback situation, and it's stable there. This is the "self biased" JFET circuit; triodes and pentodes do this as well; they're depletion mode devices.

This sounds great, and is for tubes because they're very uniform. JFETs vary all over the map. Each JFET, even same type, has a different (and interrelated) Idss, Vgsoff, Yfs and rd (channel resistance). This is one reason JFETs are not used more - they're variable, and difficult-ly so.

The third circuit is in fact what I was talking about as the self-bias circuit. Vr = ground.

The middle circuit hoists the gate up to half the power supply. This means that the source will rise to the gate voltage plus some. Some?? How much?

Can't tell. It's like the self bias circuit, but with a source tied to a negative voltage compared to the gate. It will be higher than the gate voltage by something less than Vgsoff.

... unless this makes the current through R2 by ohm's law exceed Idss, which the channel can't do. In that case, the voltage on R2 stops rising when its current is Idss.

... unless this forward biases the gate-channel diode. If the gate voltage is more than about 0.4V (for silicon; don't know of any germanium JFETs), then current starts pouring into the channel through the gate diode. This raises R2 some. But it's limited by the resistance of R3 and R1 in not letting much current flow.

It gets complicated. You have to pick a JFET based on Vgsoff and Idss, do the equations to pick the resistors, then iterate back to see if it still (kinda) works for the minimum and maximum conditions for the possible Vgsoff and Idss, as well as the transconductance.  Yes, this has made JFETs very unpopular.

The J201 is an odd one, as I mentioned. It has a very, very low Vgsoff, something like -0.5 to -0.1V. This means that the source will never be more than 0.1 to 0.5V higher than the gate as long as the JFET has enough Idss to raise the source that high - or isn't choked off by voltage drop from a drain resistor, which isn't shown in any of these. These are all near-unity-gain source followers.

[diydave]

I tend to like fets more than bjt's. And I tend to find biasing a fet simpler than a bjt, regardless off the big differences between fets.
Biasing a fet, I feel, can be straigt forward if you know the Vp and the Idss with a certain supplyvoltage.
Runoffgroove's fetzervalve has a circuit in which you can measure both.
I've written a calculator for the source and drain-resistors in a self-baising configuration. Here's the link: http://www.diydave.be/tools/fetcalc/index_eng.html
It also shows a graph of 10 values of the Vp-voltage and the matching Idss. So you can bias the fet midway the Vp, or near cut-off, or near saturation.
And as always, the eminent R.G. is the one with the good and super explanations.

[merlinb]

Trans'conductance' is amperes out divided by volts in, and has units of one divide by ohms or amperes per volt. This is called a Siemen as an honor to the guy of that name.

Careful now, the unit of conductance or transconductance is the siemens (with an 's' on both ends!). That's caught me out before now, too.

(SI units are also always spelled with small letters, but now I'm really being pedantic!)

So a resistor converts current to voltage in the same way that a hose converts flow rate to PSI.

Another way to explain it is that if you force current through a resistor, then there must be an equal and opposite force pushing back, otherwise the current would grow indefinitely (Newton's third law). That opposing force is the voltage developed across the resistance.

[R.G.]

Careful now, the unit of conductance or transconductance is the siemens (with an 's' on both ends!). That's caught me out before now, too.

I always like the term "mho". It's got a certain ring to it.

And I can spell it.

[Jdansti]

Careful now, the unit of conductance or transconductance is the siemens (with an 's' on both ends!). That's caught me out before now, too.

I always like the term "mho". It's got a certain ring to it.

And I can spell it.

Many years ago I worked in an environmental lab where one of the analyses we ran on water samples was conductivity. We reported the results in micromhos/cm. I always thought it was cool that conductivity is the inverse of resistance and the unit for conductivity is ohm spelled backwards.

EDIT: Maybe Georg Ohm had a cousin named mho. 

[R.G.]

I'm pretty sure the cousin were named Larry and Curley. 

[PRR]

This book tells much about Ohm, and his dog, but not his cousins.

http://www.ampbooks.com/home/books/ohm/

[Jdansti]

I'm pretty sure the cousin were named Larry and Curley. 

Thanks for taking the set up. I would have been disappointed if no one had. 

[Jdansti]

This book tells much about Ohm, and his dog, but not his cousins.

http://www.ampbooks.com/home/books/ohm/

I'll have to read that - as soon as I finish my competition project. 

[swinginguitar]

2) you mentioned the current being converted to voltage by a resistor - I know that's plain ol' Ohms Law, but believe or not, that is a concept I've yet to get my head around...is there a good analogy to help with that? I get confused b/c in my mind, where there is current there *is* voltage (pressure/potential) and vice versa...or am I off here?

No worries. Back at pressure and flow rate. If you want to get 10 gallons per minute through a hose, how much pressure do you need to pump it with? Well, that depends on how big a hose you have. The pressure to get 10 GPM through a 1/2" interior garden hose is quite a bit bigger than the pressure needed to get it through a 4" interior diameter fire hose. The "resistance" of a hose can be stated as pressure per flow-rate, or PSI per gallon-per-minute.

Could we simplify that by saying for current passing thru a given conductor, the potential between any given 2 points along that conductor is zero ( any 2 points between source and ground), but by inserting a resistor, we now have a voltage drop across its terminals?

[R.G.]

Could we simplify that by saying for current passing thru a given conductor, the potential between any given 2 points along that conductor is zero ( any 2 points between source and ground), but by inserting a resistor, we now have a voltage drop across its terminals?

No. That's one way to model it, by using theoretical zero-resistance wires with lumped resistors inserted between points, and this way works for some problems. But it's not how the real world works.

What you're fighting is a mental misconception that you - and almost everyone dealing with electronics! - have. The **>wire<** itself is resistive to some degree. The only wires which don't have a voltage drop between any two points on the wire are superconductors. The only way any two different points on a wire have zero voltage between them is if the current between those points is zero. All wires are resistors. The resistance between any two points is rho*l/A, where rho is the material resistivity, l is the length, and A is the conductor area. Rho is greater than zero for all materials except the super special case of superconductors.

[PRR]

> conductor, the potential between any given 2 points along that conductor is zero

Yes, almost, often close-enough to zero to "be" zero.

As R.G. says, when it really matters, two points on a practical (non-super) conductor often have some non-zero voltage.

Plug in a lamp. 120V at the wall, 6 feet of cord, 119V at the lamp. 0.5V is dropped in each 6-foot run of wire. Do we care? In this case, hardly at all. Lamps work on a range of voltage. You do pay for the volt of lost electricity, but the cost of a 10X fatter cord to drop only 0.1V is much more than the cost of the wasted juice and the inconvenience of a fatter cord.

> The only way any two different points on a wire have zero voltage between them is if the current between those points is zero.

Nit-pickery: or if the current is zero.

Of course if current is dead-nuts zero, we could replace the conductor with cheap plastic or nothing at all, no change. In fact we run wire because there is (or can be) some current. In most cases we hope (or calculate) that the voltage drop is very-very small. Sometimes we compromise (2V-17V of drop in my house's 400 foot feed from the street, because lower loss means more costly copper and I'm too cheap). Around high-gain amplifiers, sometimes 0.001V of drop matters.

[Bill Mountain]

I'm a hack when it comes to FET biasing but one thing I like to do is use them for clean gain stages.  If you run a high enough voltage then your bias point becomes less critical.  At 24 volts I stick a 10k on the Drain and a 10k pot on the source and adjust it to 1/2 supply on the drain for the particular type of FET.  Then I replace it with a resistor and use those same values for any repeating stages with the same type of FET.  If I switch brands or type then I do it again.  I usually only measure the first stage (or more if the sound is off).  As long as my drain is anywhere from 10 to 14 volts on each stage, I can't tell the difference.  If I need more gain in each stage I'll add a bypass cap.  If I'm looking for FET distortion then I'll add more stages but I'm perfectly happing using clean FET stages to push other types of clippers.

[Gus]

When using higher voltages than say 9VDC you need to be aware of the drain to gate voltage difference.

This is another way I posted about in the past.

[Bill Mountain]

When using higher voltages than say 9VDC you need to be aware of the drain to gate voltage difference.

This is another way I posted about in the past.

The data sheet for the MPF102 says that the max Drain-Gate voltage is 25 volts.  Do I need to be looking for something specific or as long as I don't go over 25 volts I'm fine???