Using Zener to drop 13.8V to 5V for input to AVR microcontro

[James W]

I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is to it?

I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.

tia
- jim

 

[James W]

Use a 78L05 for an INPUT? That seems like a bit of overkill. I'm sure it
would work, but do people really do things like that?

Bubba Jones wrote:

Why not just use a 78L05???




On Sat, 27 Mar 2004 08:14:08 -0800, James W
change-to-jw22-at-trlp-dot-com@trlp.com> wrote:


I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is to it?

I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.

tia
- jim

 

[grahamk]

James W wrote:

I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is
to it?

I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.

tia
- jim

Have a look at
http://homepage.ntlworld.com/g.knott/elect437.htm

 

[James W]

Thanks.. I had failed to consider the the zener would need a "minimum"
current. I'll have to read up a bit on this.

- jim

John Fields wrote:

On Sat, 27 Mar 2004 08:14:08 -0800, James W
change-to-jw22-at-trlp-dot-com@trlp.com> wrote:


I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is to it?


---
No. That 4.7V is specified with a certain current (the "Test Current")
flowing through the Zener, which will certainly be more than the 7ľA
you'll be supplying if you use a 1 megohm resistor. The test current
for a 1N5230B (4.7V +/- 5%, 500mW) is 20 mA, so to make that happen
you'd need 365 ohms in series with the Zener. The closest 5% value is
360 ohms, so with that value and 12V in, the current into the Zener
will be ~ 20.3mA, which will be fine. Assuming a high of 14V in will
put the Zener current at about 26mA, so it'll be dissipating about
120mW, which will also be OK. With 14V in the resistor will be
dissipating about 242mW, so you should use a 1/2 watt resistor.
---


I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.


---
I don't know if that would help much, but if you put the LC on the
input of the divider and mounted the divider as close to the processor
as possible that _would_ help. Also, putting the processor (and any
other circuitry associated with it) in a covered, grounded steel
enclosure with all the inputs and outputs to the divider being made
through feedthrough filters mounted on (and grounded to) the wall(s)
of the enclosure sure wouldn't hurt!

 

[Chaos Master]

Somebody hiding from the police, using the fake name of "James W" wrote:

Use a 78L05 for an INPUT? That seems like a bit of overkill. I'm sure it
would work, but do people really do things like that?

And the 7805 regulators are not that good at efficiency either.

[]s
--
_____ ___ Chaos MasterŽ
| | Posting from Brazil
| | MSN: wizard_of_yendor at hotmail.com
___|_____| irc.brasnet.org #XLinuxNews #POA

 

[James W]

Again, thanks... Pulled out my trusty copy of H&H, and read a bit about
Zeners.. and there "dynamic resistance", etc..

- jim

John Fields wrote:

On Sat, 27 Mar 2004 08:14:08 -0800, James W
change-to-jw22-at-trlp-dot-com@trlp.com> wrote:


I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is to it?


---
No. That 4.7V is specified with a certain current (the "Test Current")
flowing through the Zener, which will certainly be more than the 7ľA
you'll be supplying if you use a 1 megohm resistor. The test current
for a 1N5230B (4.7V +/- 5%, 500mW) is 20 mA, so to make that happen
you'd need 365 ohms in series with the Zener. The closest 5% value is
360 ohms, so with that value and 12V in, the current into the Zener
will be ~ 20.3mA, which will be fine. Assuming a high of 14V in will
put the Zener current at about 26mA, so it'll be dissipating about
120mW, which will also be OK. With 14V in the resistor will be
dissipating about 242mW, so you should use a 1/2 watt resistor.
---


I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.


---
I don't know if that would help much, but if you put the LC on the
input of the divider and mounted the divider as close to the processor
as possible that _would_ help. Also, putting the processor (and any
other circuitry associated with it) in a covered, grounded steel
enclosure with all the inputs and outputs to the divider being made
through feedthrough filters mounted on (and grounded to) the wall(s)
of the enclosure sure wouldn't hurt!

 

[petrus bitbyter]

"James W" <change-to-jw22-at-trlp-dot-com@trlp.com> schreef in bericht
news:106ba2ijpj5gqb9@corp.supernews.com...

I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is to it?

I may well add an LC between the Zener and the input pin, as there's a
fair amount of RF (ham radio) being used in my car.

tia
- jim

Jim,

Using that high a resistor will not provide enough current to make a zener
working. They tend to need some mA. If you get spikes on that line, the
zener will not react fast enough to protect the AVR. You can do some or all
of the following:

- Make a low Ohms voltage divider that has let's say 4.7V at the AVR pin at
the lowest voltage on the line you want to monitor.
- Connect a 4.7V zener parallel to the lower resistor to keep the voltage at
the AVR pin below 5V if the nominal 12V on your ignition sense becomes 14V.
- Connect to elco of let's say 1-10uF parallel to it to catch fat spikes.
- And a 10nF ceramic capacitor to catch high spikes and hf.
- A (Schottky) diode to +5V to discharge the capacitors at power off.
- A coil in series with the upper resistor to suppress spikes and hf.
- A diode in series with the upper resistor to block negative spikes.

Can't find more in my mind for the moment.

petrus


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[AC/DCdude17]

X-No-Archive: Yes

Chaos Master wrote:

Somebody hiding from the police, using the fake name of "James W" wrote:
Use a 78L05 for an INPUT? That seems like a bit of overkill. I'm sure it
would work, but do people really do things like that?

And the 7805 regulators are not that good at efficiency either.

Not that zener and power transistor combo is any better.

Any linear regulator will lose delta V * I as heat.

7805 regulates well and is highly convenient. It is used in a lot of stuff.
The SNES console, USB hub just to name a few.

If the dissipation in watts comes to be more than a single figure number, go
switch mode.

If the load is only a few miliamps and doesn't need exactly 5.0V resistive
divider will work also.

Resistive divider is something like this: Y is pull resistor. It's not
necessary if you're absolutely sure load will be drawing same current
constantly, but if there's a slightest doubt it will unload, it prevents the
voltage from shooting up to 12V. It's more complex to design(that is coming
up with x and y values) but it's the most economical.

+12V-----^^^^-------------
x | |

|
y LOAD
|
| |

--GND-----

 

[Watson A.Name - \"Watt Su]

"John Fields" <jfields@austininstruments.com> wrote in message
news:4068cdd5.350298906@news.texas.net...

On Sat, 27 Mar 2004 08:14:08 -0800, James W
change-to-jw22-at-trlp-dot-com@trlp.com> wrote:

I'm designing a circuit based around an AVR for use in an automobile.
One of the inputs is "ignition-sense", i.e. is the ignition/acc "On".
It's easy enough to find a 12V line that switches with the ignition.

However, I need to condition this for an input to the AVR. My current
plan is to use a ~1M resistor and 4.7V Zener. Is that all there is
to it?

---
No. That 4.7V is specified with a certain current (the "Test Current")
flowing through the Zener, which will certainly be more than the 7ľA
you'll be supplying if you use a 1 megohm resistor. The test current
for a 1N5230B (4.7V +/- 5%, 500mW) is 20 mA, so to make that happen
you'd need 365 ohms in series with the Zener. The closest 5% value is
360 ohms, so with that value and 12V in, the current into the Zener
will be ~ 20.3mA, which will be fine. Assuming a high of 14V in will
put the Zener current at about 26mA, so it'll be dissipating about
120mW, which will also be OK. With 14V in the resistor will be
dissipating about 242mW, so you should use a 1/2 watt resistor.
---

I may well add an LC between the Zener and the input pin, as there's
a
fair amount of RF (ham radio) being used in my car.

---
I don't know if that would help much, but if you put the LC on the
input of the divider and mounted the divider as close to the processor
as possible that _would_ help. Also, putting the processor (and any
other circuitry associated with it) in a covered, grounded steel
enclosure with all the inputs and outputs to the divider being made
through feedthrough filters mounted on (and grounded to) the wall(s)
of the enclosure sure wouldn't hurt!

Or else use optoisolators on all the inputs, like the commercial
controllers do.

--
John Fields