[sci.electronics] A whole lot of fun

mark@mips.COM (Mark G. Johnson) (07/19/89)

Here's a design problem that I had some with; hopefully you will too.
I enjoyed the tightwad-tradeoff aspects.

    0.  The overriding goal is to have the cheapest possible parts cost.
        Saving $0.01 is Ultimate Victory; cost table is below.
    1.  Given a transducer and a load to drive.  Design a controller
        circuit that varies the voltage applied to the load (approximately)
        linearly with the transducer "reading".
    2.  The transducer is a pressure sensitive resistor.  Its resistance is
        <= 5Kohms when there is too little pressure, and >10K if too much.
    3.  The circuit and the load operate from 12VDC (+/- 10%).
    4.  When there is too little pressure, apply the full 12V (99% of VCC)
        to the load.  Not (VCC-2*Vbe), VCC.
    5.  When there is too much pressure, apply only 6V to the load.
    6.  When the transducer reading is between 5K and 10K ohms, linearly
        control the voltage applied to the load as shown in the curve below.
    7.  The load is basically a 50 ohm resistor (240 mA at 12V).

   Voltage applied to load (% of 12V supply)
                 |
                 |
            >99% |----------+
                 |           \       (approximately linear curve but
                 |          . \       not required to be perfectly precise)
                 |          .  \
                 |          .   \
  !! NOTE:  50%  |          .    +--------------
                 |          .    .                   resistance of pressure-
                 o----------|----|---------------->  sensitive transducer
                           5K    10K ohms


Here are the prices of components; minimize the total parts cost:
   Resistors                      1 cent
   Capacitors <= 0.5uF            3 cents
   Capacitors > 0.5uF            15 cents
   Diodes                         2 cents
   Zener Diodes                   5 cents
   Transistors rated < 0.5 Watt   8 cents
   Transistors rated > 0.5 Watt  20 cents
   Power MOSFET,     > 0.5 Watt  45 cents
   Dual Opamp, 0.5 Watt          20 cents
   Quad Opamp, 0.5 Watt          20 cents
   Transducer (REQUIRED)        100 cents

(My "solution" costs 152 cents; I suspect it can be done for about 141
 cents by a sufficiently clever design).


Transistor and diode VBE is between 0.45 and 0.85 volts (mfg variability
plus temperature sensitivity).  Beta (Hfe) is 40-200 for the small
transistor, 10-80 for the big one.  Vcesat is 0.05 to 0.10 volts.
MOSFET VT is 1.5 to 3.0 volts, and RON is 0.05 to 0.15 ohms.  Resistors,
capacitors, and Zeners are 5% tolerance.  The circuit should not
oscillate or exhibit other instabilities.
-- 
 -- Mark Johnson	
 	MIPS Computer Systems, 930 E. Arques, Sunnyvale, CA 94086
	...!decwrl!mips!mark	(408) 991-0208

tomb@hplsla.HP.COM (Tom Bruhns) (07/22/89)

mark@mips.COM (Mark G. Johnson) writes:
>
>
>Here are the prices of components; minimize the total parts cost:
>   Resistors                      1 cent
>   Capacitors <= 0.5uF            3 cents
>   Capacitors > 0.5uF            15 cents
>   Diodes                         2 cents
>   Zener Diodes                   5 cents
>   Transistors rated < 0.5 Watt   8 cents
>   Transistors rated > 0.5 Watt  20 cents
>   Power MOSFET,     > 0.5 Watt  45 cents
>   Dual Opamp, 0.5 Watt          20 cents
>   Quad Opamp, 0.5 Watt          20 cents
>   Transducer (REQUIRED)        100 cents
>
>(My "solution" costs 152 cents; I suspect it can be done for about 141
> cents by a sufficiently clever design).

-  Are the two opamps really both $.20?
-  What does it mean to call the opamps "0.5 Watt"?
-  What accuracy is required, beyond the low "on" drop specified?
-  If your parts costs are right, and with reasonable accuracy,
   seems like $1.41 might be tough.  (You immediately toss out
   many of the parts as too expensive :-)

BTW, what about the PC board cost?  Connectors?  Mounting/enclosure?
If this were a _real_ problem, an obvious thing to work on is lowering
the cost of the transducer :-).

mark@mips.COM (Mark G. Johnson) (07/30/89)

Recetly I posted a design problem which was claimed to be "a whole
lot of fun" to solve, mostly because it required a minimum-parts-cost
design.  (And thus rewarded "clever" circuits).

The problem was to vary the voltage applied to a 50 ohm load depending
on the value of a variable-resistance sensor, with this xfer function:

   Voltage applied to load (% of 12V supply)
                 |
                 |
            >99% |----------+
                 |           \       (approximately linear curve but
                 |          . \       not required to be perfectly precise)
                 |          .  \
                 |          .   \
  !! NOTE:  50%  |          .    +--------------
                 |          .    .                   resistance of pressure-
                 o----------|----|---------------->  sensitive transducer
                           5K    10K ohms


The two most interesting replies used very different approaches.
One circuit employed two discrete transistors:

	X1   VCC 1       /* (Transducer: 5K - 10K)	$1.00  */
	R1   1   GND     11K			/*	$0.01  */
	R2   1   2       30K			/*	$0.01  */
	Q1   VCC 2   3   NPN-small		/*	$0.08  */
	DZ1  4   3       4.7V   /* Zener diode		$0.05  */
	R3   4   GND     4.7K			/*	$0.01  */
	Q2   5   4   GND NPN-power		/*	$0.20  */
	R4   2   5       150K			/*	$0.01  */
	D2   5   3       diode  		/*	$0.02  */
	RX   VCC 5       50    /* load */
----------------------------------------------------------------------------
        TOTAL COST					$1.39


The second approach used an opamp.  Although one way to account for
this is to charge 1/4 the cost of a quad opamp pkg, I decided to count
it as 1x the full cost of a dual opamp pkg.  Computing parts cost the
other way, the price would be $1.39 (same as the previous design).

	R1   VCC 1       100K			/*	$0.01  */
	R2   GND 1       5K			/*	$0.01  */
	R3   1   2       33K			/*	$0.01  */
	R4   2   3       1.5M			/*	$0.01  */
	A1   3   0   5   2
	   /* OPAMP with   out = 3, in+ = 5, in- = 2	$0.20  */
	R5   3   4       250			/*	$0.01  */
	Q1   6   4   VCC PNP-power		/*	$0.20  */
	R6   6   5       100K			/*	$0.01  */
	X1   5   GND     /* (Transducer: 5K - 10K)	$1.00  */
	DZ1  6   4       4.7V   /* Zener diode		$0.05  */
	C1   6   4       0.1U			/*	$0.03  */
----------------------------------------------------------------------------
        TOTAL COST					$1.54


The discrete circuit is rather non-precision, as it does not compensate
for the temperature-sensitive VBE's or the finite base currents.  However
it certainly *is* inexpensive, mostly because it doesn't need any
capacitors for stability.

The opamp circuit *does* need frequency-compensation elements (R3,R4,C1)
which adds $0.05 to the cost.  However, this topology is much more
likely to be linear than the previous circuit.  Note that the opamp
must handle input signals very near GND, and put out an output signal
near VCC.

Interestingly, both circuits used the same "trick" to get the breakpoint
at (10K, 50%-of-VCC).  A zener diode is placed across the collector-base
junction of the power transistor, to limit the collector-emitter voltage
drop at 6 volts or less.  (DZ1 and D2 in the first ckt, DZ1 in the second).

Special thanks to Charles Flaig of Apple and TomB of HP for the two
most interesting (and lowest cost) solutions.
-- 
 -- Mark Johnson	
 	MIPS Computer Systems, 930 E. Arques, Sunnyvale, CA 94086
	...!decwrl!mips!mark	(408) 991-0208