From: email@example.com (Winfield Hill)
Subject: Re: Limiting total wattage for a constant-current dummy load?
Date: 11 Nov 2002 12:12:36 -0800
References: <firstname.lastname@example.org> <email@example.com>
NNTP-Posting-Date: 11 Nov 2002 20:12:36 GMT
John Muchow wrote ...
> Thanks! I grabbed the data sheets for it from the ADI web site. Is
> there a multiplier you can recommend that operates from a single rail
> (+5V)? It's only going to see positive voltages at the inputs.
Hi John. A multiplier is not a bad idea, but most, like the AD835
from Analog Devices, require split supplies. Instead let me suggest
an approach that's well-suited for single-supplies: log circuits.
Consider, anti-log [log I + log V] = I * V. We'll use Intersil's
CA3096 matched-transistor array IC to implement a single-quadrant
multiplier. The classic Ebers-Moll transistor equations (see AoE
page 80 and) tell us how to execute our design. The equation for
base-emitter voltage is Vbe = kT/q ln Ic/Is, where k is the Boltzman
constant, T is absolute temperature, q is the charge of one electron,
and Is is a very small scaling current called the saturation current.
ln means the natural log. Because the term kT/q will appear often,
we'll simplify it to a voltage Vt = kT/q = 25mV at room temperature.
--> I1 |/ Q3 <--
V1 --/\/\---------+---| ,--- Io
--> I2 | |\V |
V2 --/\/\--, | | |
| |/V | Vo |/ Q4
| | |\ Q1 --> I3 | |\V
| |/V | \ |
'---| | / |
|\ Q2 | \ |
| | | |
We're using PNP transistors Q1 Q2 to add the log of two currents (one
related to current, the other to voltage), and a third NPN transistor
to offset the result. Using the equation above we can write
Vo = Vt ln I1/Is + Vt ln I2/Is - Vt ln I3/Is
= Vt ln (I1 * I2)/I3 M,
where M is a term wrapping up the three Is values. The Ebers-Moll
equation can also be written to express collector current resulting
from a base bias voltage, this is an antilog or exponential function.
Ic = Is e^ Vbe/Vt
We'll take the voltage Vo we obtained above and use it to drive the
fourth anti-log transistor, so the resulting current will be,
I1 I2 Is3 Is4
Io = ----- -------
I3 Is1 Is2
Ideally the last term can be made to cancel out, giving us I1 * I2
plus a scaling factor from I3. Pretty cool, huh? One low-cost IC
plus a few resistors, and voila, fast real-time power calculations.