This is a second post for the Project14 Control Systems theme. Probably also the simplest possible entry: a Constant Current Driver with one (1) other active element. In the previous post it was a bipolar transistor. This time a JFET.
The circuit is a little bit simpler than the previous one, A JFET and 1 resistor. But it's a bit trickier to set it exactly to the current you want.
On the other hand, it deals better with changing supply voltages.
This is again one of the circuits from The Art of Electronics. In the 3rd edition, it's circuit 3.20 and 3.23.
What Does this Constant Current Circuit Do?
It regulates the current through a variable load, keeps it constant.
image source: on-semi datasheet for J113
The circuit exploits the very flat part of the drain current vs drain-source voltage graph. And that "the level" of the flat part can be controlled by the gate-source voltage.
The current you can control is lower than the IDSS of your JFET.
IDSS is the current through the drain, when the voltage between gate and source (VGS) is 0 V (gate and source tied together).
You may find it in the datasheet, but the spread is big. That's why we'll measure it below for "the JFET at hand".
With my JFET, a switching transistor that's a bit of an outlier in the family, the minimum IDSS is given as 5mA, no maximum. Mine measured 49 mA.
When you connect the gate to source, you use the IDSS graph. And the current will stay fixed at what happens to be the IDSS of your JFET.
When you put a resistor between gate and source, you can "set" the constant current to a different level.
This is the "feedback" mechanism that's used in this "control circuit".
You have to think along a little bit:
In the above circuit. The gate voltage is lower than the drain voltage. Because the upper part of the resistor is at a positive voltage,and the lower part of the resistor is at ground level.
So if we tie that lower part of the transistor (yes - ground) to the gate, this means that the gate is at a lower voltage than the source (negative relative to source). And that lowers the current through the JFET.
Like before in the transistor blog, it's again a self-regulating process:
- current through drain = current through source
- current through source is defined by VGS / resistor
- the back bias generated by the resistor will take care that if current wants to rise (higher power supply voltage, fluctuation of the load impedance), the VGS becomes more negative and brings the circuit back in balance.
The parameters of the JFET at play here (ID vs VGS) has a wide spread. You can't predict the precise current, so a trim potentiometer may be needed (attention - all the current flows through the trimpot).
The IDSS is also unpredictable. Even though my J113 measures 49 mA, the datasheet defines the minimum at 5 mA. So if you want to be sure that any J113 can do the job, you need to limit yourself to max. 5 mA.
Things can get hot. Watch ID and operational area limits
A good load impedance range where the single JFET can keep the current constant.
A good range where the source voltage can vary while the JFET can keep the current constant.
Current is (a little) more predictable with the resistor in the source circuit than with a raw IDSS solution. A little.
Almost no components. A good candidate if your requirement is "current must be constant, but actual current is not that critical."
What's Needed to Build the Circuit
- an n-channel JFET (I used a ).
- a 330R resistors if you want to do one experiment. 2 of them if you want to check constant current for different settings.
- optionally an LED.
This is a lab experiment, and JFETs vary. Although not likely, It's possible that a component dies in an experiment. Cut your losses, take a new component and carry on.
How to Record the Graphs, What's Needed?
In the spirit of project14, this experiment only needs:
- a 1 - 10 V DC power supply
- a multimeter that can read volts DC and mA DC
IDSS is the current through the drain, when VGS = 0 V.
You force VGS to 0 V by connecting them together.
Then step your power supply in 1 V steps from 1 to 10 V, and write down the current each time.
Then draw the result on a graph.
To measure the current you wire the DMM in mA mode in series between power supply and drain.
Measure fast, in particular the second half of samples. The transistor can get warm.
- unplug the + probe of the power supply,
- set voltage,
- plug the + probe into the power supply, write down measurement, and unplug the probe again.
This measurement can differ wildly between JFET types, but also wildly accross a single JFET type.
I measured 50 mA. There are known graphs for the same J113 with 21 mA measured. And some types have a IDSS lower as 1 mA.
As long as your graph has a similar shape, you are doing fine.
Measure Different Constant Current Set Points
To measure the constant current at different set point, you remove the bridge between gate and source, and replace it with a "current set" resistor.
You have two ways to measure the current. Either by again putting the DMM in current mode in series with the drain (easiest) - or - because the drain current is the same as the source current:
you can put the DMM in volt mode over the resistor(s). You get the VGS is at the measurement points, and you can calculate the current by dividing the measured voltage by the resistance.
No need to be fast here for most JFETs. We're running at lower power.
If, like me, you have a JFET with IDSS being several mA, you can put an LED in series with the drain, just to see the brightness change with the different settings.
Don't exceed the forward current of the LED. Better measure first, then put the LED in if your scenario if the current is within range of that LED.
With a 330R Resistor
column A: supplied DC voltage (in V). You set this on your power supply, in incemental steps of 1 V.
column B: current (in mA)
column C: VGS (in V)
With 2 330R in Series, 660R
With 2 330R in Parallel, 165R
exercise for the watcher: why do I measure absolutely nothing with V = 1 V? Hint: one component is the culprit.
also: for 3 V, drain-source voltage is 242 mV but I wrote 0.42 V...
Here are my measurements with 330 and 660 Ohm set resistor. Add yours with the 2 resistors in parallel, the 165 Ohm scenario.
Note, the x-axis in this case is not the VDS put the power supply voltage. In the video you see an example where VDS is used for the X axis.
This one shows that the current stabilises and becomes constant, but it's not a JFET characteristic. It's a "whole circuit" characteristic.
|Simple Electronic Control Systems - Current Control with 1 transistor|
|Simple Electronic Control Systems - Current Control with 1 JFET|