Current supply.
I WANT TO USE A SIGNAL FROM A PIEZORESISTIVE PRESSURE SENSOR TO CONTROL SIGNALS FROM A BS2.
THE PRESSURE SENSOR WILL BE POWERED BY A VEHICLE'S 12V SYSTEM.
THE SENSOR NEEDS A CONSTANT 1 mA CURRENT.
I WOULD APPRECIATE IT IF SOMEONE COULD TELL ME HOW TO ENSURE THIS CONSTANT 1mA SUPPLY.
KEITH MOORE.
THE PRESSURE SENSOR WILL BE POWERED BY A VEHICLE'S 12V SYSTEM.
THE SENSOR NEEDS A CONSTANT 1 mA CURRENT.
I WOULD APPRECIATE IT IF SOMEONE COULD TELL ME HOW TO ENSURE THIS CONSTANT 1mA SUPPLY.
KEITH MOORE.
Comments
cheapest (and worst) constant current source around. Just like using an LED
and a dropping resistor. However, since the resistance of the sensor
obviously changes, this might not be good in this application (depends on
WHY it needs exactly 1mA).
The next step up is a current mirror. Think about this:
1) If you put a voltage on the base of an NPN transistor, the emitter
voltage will be about .6V less (some people say .7 -- it is mildly related
to temperature and other factors, but .6 is close enough for us). Suppose
you put 1.2V on the base -- the emitter will have .6V on it. No two ways
about it.
2) If you put a resistor between the emitter and ground, you can easily
calculate the emitter current which is -- in the example -- .6/R. So if R is
600 ohms, the emitter current is going to be 1mA.
3) Here's the tricky part. The collector current of a transistor is related
to the emitter current by the transistor's alpha. The alpha is the
beta/(beta+1). At high beta, this works out to alphas of .99 or even closer
to 1. So for practical purposes, the Ic (collector current) is the Ie
(emitter current).
4) If you hook one end of your sensor to +V and the other end to the
collector, 1mA will flow through the sensor until the transistor can no
longer lower the collector voltage (figure about .2 or .3 V).
What does that mean? Forget your sensor. Suppose we have a 10V supply, and a
variable resistor. We want 1mA through the variable resistor no matter what.
If the resistor is set to 100 ohms, you need .1V across the resistor, so
you'd set the "other end" of the resistor to 9.9V. The drop is .1V and the
current is .1/100 = 1mA. Now, what if you set the resistor to 2200 ohms? Now
the drop should be 2.2V (2.2/2200=.001) so the voltage on the other end
should be 7.8V. What about 25K ohms? The drop should be 25000*.001=25V).
Now this is a problem. To get 25V across the resistor, you need -15V. The
transistor can't do that. In fact, at 10K you need 10V and the transistor
can't do that either because it saturates -- Vce (the voltage between
collector and emitter) won't go below .2 or .3V. So say .5 for a safety
factor -- The largest resistor you can force 1mA through is 9.5/.001=9500
ohms.
You can also use a PNP transistor. In this case, the +V goes (through R) to
the emitter and you set the base to .6V LESS than the emitter (this is PNP,
remember). The collector goes to the + lead of the load and the other lead
is grounded. Same equations. Suppose a 12V supply and R is 1K. For 1mA you
need 1V, so the emitter voltage should be 11V. The base voltage should be
10.4V. (Of course, real 12V supplies in a car are 13.8, but you know what I
mean).
So how do you set the voltage on the base? One way is a voltage divider (see
http://www.al-williams.com/wd5gnr/basiccir.htm). This isn't very good
though -- you'd want to use 1% resistors. Trim pots are nice, but they tend
not to be reliable in environment.
Another way is to use a stack of diodes. Two silicon diodes forward biased
will give you your 1.2V (use a resistor from +V to the non-banded end of the
first diode. Banded end goes to the non-banded end of the second diode.
Banded end #2 goes to ground. The voltage comes off the "cold" side of the
resistor.
A zener diode or a commercial voltage reference is another option.
There are other ways to generate a constant current, but you'll find this is
the simplest.
Al Williams
AWC
* Floating point math for the Stamp, PIC, SX or any microcontroller at
http://www.al-williams.com/awce/pak1.htm
Original Message
From: Keith Moore [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=9d2VDis20sLHopg3wNiZboblk4d6lEWBIHGD7D0JgZa-SJBibYM9-xY4IpuscjNRsd6OmQDYk0BiVm6FHgs98w]keithmoore1@n...[/url
Sent: Friday, November 03, 2000 6:53 AM
To: BASICSTAMPS@egroups.com
Subject: [noparse][[/noparse]basicstamps] CURRENT SUPPLY.
I WANT TO USE A SIGNAL FROM A PIEZORESISTIVE PRESSURE SENSOR TO CONTROL
SIGNALS FROM A BS2.
THE PRESSURE SENSOR WILL BE POWERED BY A VEHICLE'S 12V SYSTEM.
THE SENSOR NEEDS A CONSTANT 1 mA CURRENT.
I WOULD APPRECIATE IT IF SOMEONE COULD TELL ME HOW TO ENSURE THIS CONSTANT
1mA SUPPLY.
KEITH MOORE.
SIGNALS FROM A BS2.
THE PRESSURE SENSOR WILL BE POWERED BY A VEHICLE'S 12V SYSTEM.
THE SENSOR NEEDS A CONSTANT 1 mA CURRENT.
I WOULD APPRECIATE IT IF SOMEONE COULD TELL ME HOW TO ENSURE THIS CONSTANT
1mA SUPPLY.
KEITH MOORE.
To supplement Al's contribution over his morning cup of coffee, here are
some alternatives...
One of the easiest ways is to use a voltage regulator in the following
circuit:
<TT>
|
|
| LT | 5490 ohms
+12v
| 1121 |--/\/\--;
| -5 | |
|______| | 900uA
| 100uA |
'
o
| 1 ma
-load-
|
com
'
</TT>
For this 1 ma level of current, you must use a micropower voltage regulator
like the LT1121-5. The circuit feedback holds the voltage across the 5490
ohm resistor constant at 5 volts, (900 microamps) and added to that is the
ground pin current (100 uA from the LT1121 data sheet).
Another way to make a very accurate 1 ma current in the sensor is to put it
in the feedback loop of an op-amp:
<TT>
200 ohms |
;-/\/\-o
load--;
| | 2|\ | |
com '--|-\ 6 |
LM10 | >
'
__|+/
| 3|/|1,8
;____;
</TT>
The load is a bridge pressure transducer with the sense wires going into a
differential amplifier. The LM10 provides a constant voltage of 0.2 volts
on pins 1, connected to pins 8 and 3, and feedback then keeps 1 ma flowing
to the load and through the 200 ohm resistor. Pin 4 of the op-amp connects
to common, and pin 7 to +12 volts.
-- Tracy Allen
electronically monitored ecosystems
http://www.emesystems.com