# Constant Current Source Design

Discussion in 'Electronic Basics' started by Confused Soul, Jan 24, 2005.

1. ### Confused SoulGuest

I need to design a constant current source of 0.1 mA. The load varies
between 1Mohm and 1 Kohm. Using a constant voltage source, is it
possible to design a constant current source?
Any help in this is greatly appreciated.

--Confused Soul

2. ### Jonathan KirwanGuest

Let's see....

At 100uA,

1k requires 1k*100u = 0.1V
1M requires 1M*100u = 100V

So, you require a voltage compliance of three decades and a voltage of 100V
the load is only 1k, the transistors themselves will have to stand off that much
voltage and still provide a reasonably close 100uA?

Jon

3. ### JamieGuest

look for current Regulating diodes.

a couple of Fets and biasing resistors will
do it which is really all the CRD are.

http://www.linearsystems.com/datasheets/J500.pdf

those there don't come in a .1ma size but will give
you and idea of how stable they are.

if the range is not tight enough then i suggest using
an OP-Amp as a comparator as the limiter circuit.
basicly you couple the - and + inputs with a resistor
out put will bias a transistor that is supplying the source.
something to look into.

4. ### Confused SoulGuest

100 V power supply seems impractical in my case. So instead of 100 uA,
I could go for 10 uA or even 1 uA. This would reduce the voltage
required to 10 V or 1 V.
only 1k, the >>transistors themselves will have to stand off that much
voltage and still provide a >>reasonably close 100uA?
-- I couldnt figure out the circuit you are talking about !!! Could you

5. ### Andrew HolmeGuest

Here's a constant(ish) current source using a PNP transistor:

http://www.ece.osu.edu/ee327/Figures/current_source_pnp_schem.gif

Current in load resistor RL = (Vcc * R1 / (R1+R2) - 0.7) / Re

6. ### petrus bitbyterGuest

What do you really need? If you reduce to 0uA you need nothing at all.

petrus bitbyter

7. ### John PopelishGuest

Certainly. Ohm's law tells you how much voltage is required to push a
given current through a resistance.
The concept is to also have the current pass through a second resistor
and a variable resistance device like a mosfet or junction transistor.

Some additional circuitry must subtract the voltage drop across the
second resistor from a reference voltage and amplify that difference,
using that result to control the variable resistance device so that it
burns up all the extra voltage not needed to push the same current
through the first resistor. If the amplifier has high gain, the
current will vary very little as the load resistance changes.

How fancy this circuit gets depends on how accurate and temperature
stable the current must be and how fast the load resistance can
change.

Some examples:
See precision current source on page 7 of:
http://www.national.com/an/AN/AN-20.pdf
See bilateral current source on page 10 of:
http://www.national.com/an/AN/AN-31.pdf
Note that the current is proportional to Vin (that you will have to
provide).

You might also browse Google for [current source] or [constant
current]
or [current regulator]

8. ### Geir KlemetsenGuest

Simple constant current device using tvo PNP transistors and two resistors:

___
.____/ |
| \___|
|
.-.
| |
| |
'-' o Constant current
| |
| |
| |
| |/
o-----------|
| |>
| |
| |
\| |
|-----------o
<| |
| |
| .-.
| | |
| | |
| '-'
| |
| |
=== ===
GND GND
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

The current is about 0.6 divided by value to the lower resistor.

Hope this helps

9. ### ChrisGuest

I guess there are two answers to your question. First, it's relatively
easy to make a fairly stiff current source out of a voltage source with
a series resistor. For the example you specified, though, you require
a very high voltage (view in fixed font or M\$ Notepad):

Stiff Current Source From Voltage Source
` + ___
` o----|___|-------.
` 60 megohms |
` |
` o-----.
` | |
` R(L) .-. |
` 6000V 1K-1meg | |<---'
` | |
` '-'
` |
` I(L) |
` - <------ |
` o----------------'

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

As you can see, a load resistance of zero ohms will produce a current
of 0.1mA. If you have a load resistance of 1 meg, the current will
droop to about 0.098mA. That's pretty good, within a couple of
percent. If you want better, you have to bump up the voltage source
and series resistor more. Not very practical in this case.

If you actually have to make one, you probably want something a little
more practical (and a little safer). In that case, you might try
something like this:

100uA HV Current Source

` + +
` o-----o------------------o------o-----.
` | | |
` | | |
` | R(L) .-. |
` | | |<---'
` | | |
` |20K 1 Watt '-'
` .-. I(L) |
` | | <----- |
` | | .-------o------'
` '-' | -
` 120VDC | |/
` o--------| TIP47
` | |>
` | |
` 1N4740 | .-.
` V(z)=10V /-/ 93K | |
` ^ | |
` | '-'
` | |
` - | |
` o----o----------'

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Use an isolated lab 120VDC supply, and safety first -- if you don't
know what you're doing, ask someone who does.

Here's how it works: The 20K resistor in series with the zener applies
10VDC to the base of the TIP47. As a first cut, you would expect that
would mean 9.3V at the emitter of the TIP47. That makes the emitter
current 100uA (9.3V / 93K, use an 82K and an 11K in series). Since the
TIP47 has a minimum h(FE) of 30, you can be guaranteed that at least
97% of that emitter current is coming from the collector, and 3% or
less from the base. That will give you around 100uA to the load
resistor, and it should be a stiffer current source than the first
example over the range of 1K to 1Meg. You might want to replace the
93K resistor with a 91K resistor in series with a 10K pot to tweak it
in to exactly 100uA.
Again, be sure to play safe around high voltage.

Good luck
Chris