From: jmuchow@SPAMMENOTcamlight.com (John Muchow)
Subject: Re: Limiting total wattage for a constant-current dummy load?
Date: Mon, 11 Nov 2002 23:04:12 GMT
Organization: MindSpring Enterprises
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
X-Server-Date: 11 Nov 2002 23:09:32 GMT
X-Newsreader: Forte Free Agent 1.21/32.243
Very cool. :-)
And I think I understand most of it too .
Now I just have to put it all together with my constant-current
circuit and generate the "power level is going to be too high" signal.
Hopefully, this is the fun part.
Thanks for the circuit and info! With Tony's circuit ideas and the
other input I've received, I'm hoping I can work up a circuit soon.
Knowing me, I'm going to try all the ideas just so I can learn more
about each approach. Saving money is something I have a lot of
trouble with when designing a new circuit. There's just so much to
>>> 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.