Help me find the right JFET VCR

Thread Starter

ltkenbo

Joined Sep 11, 2008
30
Have recently been reading a on the forum, and other places about JFET transistors. I have a simple circuit with a variable resistor that can vary from 100 ohms and up, but for my application a range of 100 ohms to 200 ohms is all I care about (micro controller analog application).

So, I am wanting to replace the variable resistor with a P junction JFET, to act as a voltage controlled resistor which I can control with a gate voltage (ideally controlled by a micro also, so 0-5V range).

So as I understand it, to act as a linear voltage controlled resistor (ohmic/triode region), I need to keep Vds below a certain level. Then, I can vary Vgs, which will then linearly alter the value of the "resistor."

http://www.nxp.com/documents/data_sheet/PMBFJ174_175_176_177_CNV.pdf

I found this one because the Rds-on it has is slightly below what I need for my application (85Ω, and I need just 100Ω and up). My question is, where in the datasheet can I find the gate voltage that causes the current to "pinch off" (in the tutorial it is called Vgs cutoff) and the largest Vds I can have and remain in the ohmic region? I can't seem to understand the sheet exactly, especially since sometimes things are named differently.

So I want a transistor which will accurately have a varried DS "resistance" 100Ω-200Ω within the 0-5V range. Did I pick the right one? I hope I have provided enough information.
 

Thread Starter

ltkenbo

Joined Sep 11, 2008
30
So, I am wanting to replace the variable resistor with a P junction JFET, to act as a voltage controlled resistor which I can control with a gate voltage (ideally controlled by a micro also, so 0-5V range).
Voltage controlled resistor. I am trying to simulate a resistance and was hoping to be able to use a microcontroller to control that resistance (with the gate). I chose a JFET because I read they are good for doing that, if they are operating in the right conditions. As I am using a microcontroller, I could do without linearity (because I can account for this with C code), but linear functionality would be ideal. I can't use a digital resistor because those only come in the larger flavors (KΩ) and I need something in the 100Ω - 200Ω range.
 

mik3

Joined Feb 4, 2008
4,843
What is the puspose of simulating a resistor?

What will be its use?

I am asking because you might be able to do it in a different way.
 

mik3

Joined Feb 4, 2008
4,843
You can try a JFET but note that you have to keep Vds low and the Vgs between some limits. Also, the resistance of the channel changes with temperature and variations in Vds.
 

Audioguru

Joined Dec 20, 2007
11,248
Jfets have a wide range of spec's. The resistance of a Jfet becomes non-linear when the voltage across it exceeds about 0.05V.
 

Hi-Z

Joined Jul 31, 2011
158
I think a quick look at the characteristic curve for a jfet would be illuminating - try this here:

http://electronics.indianetzone.com/1/characteristic_curves.htm

The right hand side of the characteristic (i.e. for Vds larger than a few volts) shows Ids varying very little as you change Vds. At the extreme left (i.e. for small Vds), you'll see Ids changing at varying rates (according to the applied gate voltage) when you change Vds - and you'll see that the slopes are fairly straight-line. Somewhere in between you'll see a region which is neither of the above.

Well, the slope of Ids versus Vds is the resistance of the device, and this varies with gate voltage, for low Vds. If you know Vds is going to be small enough for the characteristic to be sufficiently linear for your needs, then you can use it as a voltage-controlled resistor. If you're a hi-fi buff, you'll need to limit Vds fairly severely to avoid distortion, but since I don't know your application, this decision will have to be made by you.

When you mention linearity, I suspect you're referring to Rds versus Vgs. Well, if you're prepared to experiment, you'll be able to find out what you need.

But I will echo what's been said: jfets will vary significantly sample to sample, and with temperature. If there's any significant power dissipation, there could be self-heating to cope with too.
 

Hi-Z

Joined Jul 31, 2011
158
Just to clarify for when you're looking at characteristic curves (assuming we haven't put you off!), the resistance is 1/slope. So, for 10mA Ids as you go from Vds = 0 to 1V, the resistance would be 100 ohms at that gate voltage.
 

Thread Starter

ltkenbo

Joined Sep 11, 2008
30
Thankyou for all the information guys.

You could use a digipot, but only use the lower 1/10th of its range, to get a 100-200 ohm resistor.
I couldn't do that because I would have less than 256 bits precision within that 100-200 ohm range wouldn't I? And I need at least 8-10.


So Rds will fluctuate based on Vds (when on the left side) and Vgs correct? But won't Vds fluctuate based on Rds (like a real resistor) as well?

When the resistor is open (or very high resistance) in my application, the voltage across will be ≈ 5V, but as the resistance decreases it will quickly (due to another resistor in series to create a divider) settle between 0.231 V - .0455 V (100Ω - 200Ω range). I guess I should just buy one of the transistors and experiment for my application.

What I was wondering is how can I interpret the characteristic curve from the datasheet. I have seen the curve you are talking but as you mentioned they differ slightly by transistor.

Also is there any better way to do what I am trying to do?

Don't worry you haven't put me off, this is what I signed up for when going into electrical engineering. :)
 
Last edited:

Hi-Z

Joined Jul 31, 2011
158
This all sounds do-able, as long as you're happy with the fact that each individual fet will have to be calibrated (Rds vs Vgs, keeping Vds low), and that there'll be some temperature sensitivity.

At the left-hand side of the graph, you'll see that the curves are very linear, and, as I say, you can deduce the resistance because it's 1/slope. So, for the example I gave, you'll see that the I vs V graph for a 100 ohm resistor would be identical to that of the fet (at that Vgs) - for low values of Vds, of course. So, at low voltages the 100 ohm resistor and the fet are identical.
 

Audioguru

Joined Dec 20, 2007
11,248
The drain-source resistance of a Jfet changes when the voltage across it changes. That is what causes severe distortion unless the signal level causing the voltage change is very low (less than 0.05V peak). If you feed half the signal to the gate then it is much more linear and the signal level can be 0.1V peak.
 

Thread Starter

ltkenbo

Joined Sep 11, 2008
30
Ok. Sounds like my best bet is to experiment I guess, and see how it works out. Though if you guys have any other ideas that could replace the resistor let me know.
 

tom66

Joined May 9, 2009
2,595
I couldn't do that because I would have less than 256 bits precision within that 100-200 ohm range wouldn't I? And I need at least 8-10.
If you truly need 8-10 bits of precision then you are out of luck on a digipot, but I've seen 10-bit 1k digipots, you'd still have ~6.68 bits if you only used 100 ohms of that. I thought I saw a 12-bit one a while back, too...
 

Thread Starter

ltkenbo

Joined Sep 11, 2008
30
Unfortunately with this setup using 6 bits of precision within that range would give me a temperature precision of 2.75 degrees Celsius and it needs to be much more precise than that. :(
 
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