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BRogers(at)fdic.gov
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 Posted: Fri Nov 16, 2007 12:44 pm Post subject: How do I convert 14v DC to one-half volt? |
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I am looking for some help or suggestions on how to reduce the normal
bus voltage of 12 to 14 volts DC in my experimental kitplane to one-half
volt for a circuit that activates the GPS OK flag in my CDI.
I have a portable moving-map GPS in my airplane. By sending the GPS
signal through a SMART COUPLER II by Porcine Associates, I get analog
left-right signals that will drive my Navaid autopilot and/or CDI.
However, in order for the CDI to function, it needs an electrical signal
to the GPS OK flag of more than 250 mV and less than 900 mV. I can use
the GPS OK signal from my Smart Coupler II, but it is a full 12-14 volts
(100 milliamps). THIS IS TOO MUCH VOLTAGE for my Course Deviation
Indicator GPS OK flag. I need to reduce the voltage to something
in-between the minimum and maximum allowable voltage (.25 - .90 volts).
My choice is .5 volts or one-half volt.
Can you tell me how to accomplish this? I found voltage regulators in
several supply catalogs (i.e., Jameco) that reduce voltage, but none of
them reduce it down to the needed level of one-half volt.
With appreciation,
Bob J. Rogers
Mustang II (almost ready to fly)
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matronics(at)rtist.nl
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| Posted: Fri Nov 16, 2007 1:56 pm Post subject: How do I convert 14v DC to one-half volt? |
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If you can settle for 0.7V then you could use a resistor and a diode in
series. Wire a 1K2 resistor (1/4W) to the output of your SC-II, then to a
standard silicon diode (1N4004 will do), then to Ground. The diode needs to
be conducting, so there's 0.7V across it. That is well within your margin.
So:
+12V Output o---[ 1K2 ]------(X)----->|-----o GND
The (X) marks your 0.7V signal to the CDI. The resistor causes a current of
about 10 mA to flow when there's 12V present. The diode is conducting and
causes a 0.7V drop.
Hope this helps,
Rob
---
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khorton01(at)rogers.com
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| Posted: Fri Nov 16, 2007 3:47 pm Post subject: How do I convert 14v DC to one-half volt? |
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On 16-Nov-07, at 12:39 PM, Rogers, Bob J. wrote:
| Quote: |
>
I am looking for some help or suggestions on how to reduce the normal
bus voltage of 12 to 14 volts DC in my experimental kitplane to one-
half
volt for a circuit that activates the GPS OK flag in my CDI.
I have a portable moving-map GPS in my airplane. By sending the GPS
signal through a SMART COUPLER II by Porcine Associates, I get analog
left-right signals that will drive my Navaid autopilot and/or CDI.
However, in order for the CDI to function, it needs an electrical
signal
to the GPS OK flag of more than 250 mV and less than 900 mV. I can
use
the GPS OK signal from my Smart Coupler II, but it is a full 12-14
volts
(100 milliamps). THIS IS TOO MUCH VOLTAGE for my Course Deviation
Indicator GPS OK flag. I need to reduce the voltage to something
in-between the minimum and maximum allowable voltage (.25 - .90
volts).
My choice is .5 volts or one-half volt.
Can you tell me how to accomplish this? I found voltage regulators
in
several supply catalogs (i.e., Jameco) that reduce voltage, but none
of
them reduce it down to the needed level of one-half volt.
|
Assuming that very little current would be needed at this input, you
could simply use two resistors to make a voltage divider. Put two
resistors in series, with one of them approximately 24 times the
resistance of the other one (the range of required voltages is quite
large, so you have a fair bit of leeway on this value of 24 times).
Hook one end of the big resistor to 14v, and one end of the small
resistor to ground. Pick off the voltage you need where the two
resistors join up. Make sure the sum of the two resistors is high
enough to limit the current flow through the divider circuit to an
acceptably low level. The power draw will be very low as long as you
keep the current low by choosing high enough values for the
resistors. E.g, if the total of the two resistors is 1000 ohms, the
current would be 0.014 amps, and the power dissipated in the large
resistor would be about 0.19 watt, so 1/4 watt resistors would work.
I'm not sure what the commonly available resistor values are, so I
won't try to suggest which ones to look for.
--
Kevin Horton RV-8 (finishing kit)
Ottawa, Canada
http://www.kilohotel.com/rv8
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rnewman(at)tcwtech.com
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| Posted: Fri Nov 16, 2007 3:54 pm Post subject: How do I convert 14v DC to one-half volt? |
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Here's a little circuit to act as a voltage detector, resistor should be
at least 1/4 watt rated.
-Bob Newman
www.tcwtech.com
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voltage_detector.pdf |
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kuffel(at)cyberport.net
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| Posted: Fri Nov 16, 2007 4:52 pm Post subject: How do I convert 14v DC to one-half volt? |
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Bob,
<< bus voltage activates the GPS OK flag in my CDI. >>
Rob Turk's method of tapping the voltage between a resistor and a
forward biased diode is an excellent way to get a low current
regulated voltage. For your application an even simpler way will
work.
A sufficiently high resistor in the line between the Smart
Coupler output and the CDI input will do the job. I would use
the highest resistor which turns on the flag with a 9 volt
battery. To be on the safe side, I would start with at least
220k ohms and slowly work my way down until the battery reliably
turns on the flag. Almost any local amateur radio operator will
have a selection of resistors, or small variable resistors, with
which to find the right value.
If the CDI documentation specifies the current drawn by the flag
then you can use Ohm's law to calculate the resistor:
resistor equals voltage divided by current
If the current is given in milliamps (ma) then the resulting
resistor is in kilohms (kohm). But even then I would use the
battery/test resistor method. Setting the resistor value with 9
volts gives you a margin that if the buss voltage is high enough
to run the coupler then the output is high enough to display the
flag.
Tom Kuffel
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mprather(at)spro.net
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| Posted: Fri Nov 16, 2007 7:04 pm Post subject: How do I convert 14v DC to one-half volt? |
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Probably unlikely, but the CDI might have a FET input which would have
extremely high equivalent resistance (or low leakage current - in the nano
Amps), and the IV curve may be non-linear. The output leakage current on
the Smart Coupler may be significantly higher than the input
leakage/operating current on the CDI. If this were the case, no matter
how high the resistance value of the interconnect resistor you're
proposing, you may not be able to control the voltage on the CDI. A
voltage divider is a safer way to go (at least for prototyping) though
your proposal will probably work.
Matt-
| Quote: |
<kuffel(at)cyberport.net>
Bob,
<< bus voltage activates the GPS OK flag in my CDI. >>
Rob Turk's method of tapping the voltage between a resistor and a
forward biased diode is an excellent way to get a low current
regulated voltage. For your application an even simpler way will
work.
A sufficiently high resistor in the line between the Smart
Coupler output and the CDI input will do the job. I would use
the highest resistor which turns on the flag with a 9 volt
battery. To be on the safe side, I would start with at least
220k ohms and slowly work my way down until the battery reliably
turns on the flag. Almost any local amateur radio operator will
have a selection of resistors, or small variable resistors, with
which to find the right value.
If the CDI documentation specifies the current drawn by the flag
then you can use Ohm's law to calculate the resistor:
resistor equals voltage divided by current
If the current is given in milliamps (ma) then the resulting
resistor is in kilohms (kohm). But even then I would use the
battery/test resistor method. Setting the resistor value with 9
volts gives you a margin that if the buss voltage is high enough
to run the coupler then the output is high enough to display the
flag.
Tom Kuffel
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kuffel(at)cyberport.net
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| Posted: Fri Nov 16, 2007 9:38 pm Post subject: How do I convert 14v DC to one-half volt? |
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Matt,
<< high resistor in the line >> << CDI might have a FET input
which would have extremely high equivalent resistance .. If this
were the case .. you may not be able to control the voltage >>
My King KI209 specifies a flag input impedance of 1k. So at
least for this instrument a single resistor should work. But
issues like this are why I suggest he try the 9 volt battery/test
resistors method first. No objection to the 2nd resistor or
forward biased diode method (in fact my first thought) but in
this application simplest might work just fine.
Tom Kuffel
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