I want to control a 12V, high current source (a few amps, propably under 10) with a controlling output that comes out from a device. The issue is that the amplitude of the controlling signal is only 0.5V, making it impossible to use a simple transistor assembly or a relay.
I already read this thread: http://forum.allaboutcircuits.com/showthread.php?t=10473 but didn't get much out of it.

tom66 suggested that I could use a comparator to get a higher voltage signal first. I found it clever.

Has anyone else a different idea?

 

I want to control a 12V, high current source (a few amps, propably under 10) with a controlling output that comes out from a device. The issue is that the amplitude of the controlling signal is only 0.5V, making it impossible to use a simple transistor assembly or a relay.
I already read this thread: http://forum.allaboutcircuits.com/showthread.php?t=10473 but didn't get much out of it.

tom66 suggested that I could use a comparator to get a higher voltage signal first. I found it clever.

Has anyone else a different idea?

How is this different from the last post?

Anyway, one other way to do it would be to use a transistor configured as an amplifier. Maybe bias the signal so it just about switches on the transistor. The signal transistor could then control a power transistor or relay.

 

You could use the .5 volts as an input to a uC and when sensed, trigger whatever you wish.

The line could be run into an ADC input on the uC. The uC could sleep until awaken by interrupt via the .5v signal.

That would really just be another way to get 0-5v logic from the 0-.5v signal.

Comparator would be a touch easier.

 

How is this different from the last post?

Read the previous thread, while it's still up.

 

tom66 suggested that I could use a comparator to get a higher voltage signal first. I found it clever.

I endorse the approach. I'm doing something very similar right now and it's working great. You can use the full 12v to power the comparator and to pull up the output to switch a MOSFET to handle the current you need. The only "tricky" part might be making sure the comparator switches states the way you expect, full on and full off, not oscillating in between.

 

What would be the reasons behind oscillation?

 

What would be the reasons behind oscillation?

Think about it: what if you get exactly 0.5V in? The comparator will switch at a half-way point, and depending on the relay this could cause what is known as relay chatter as the comparator rapidly changes between low and high, as the input signal is unlikely to be stable. This will cause the relay to wear out and the lights to flicker. The solution to this is to ensure that the signal always has enough of a margin, for example, if you want 0.5V to be your set point, compare it to 0.4V.

 

The answer to that problem is a comparator with hysteresis.

http://www.maxim-ic.com/app-notes/index.mvp/id/3616

 

Georacer
Senior Member

A friend of mine has bought this car alarm system:
http://www.clifford.com/Products/Pro...?ProductID=721

He wants to connect the auxiliary output of the alarm to the lights, in order for them to flash I guess.

The problem is that the AUX output is only 0.5V high (It's not written on the site, but I believe him). How can I control a 12V voltage with only 0.5 volts?

He thought of a relay at first, but I don't think that there's a relay that switches on with only half a volt.

My thougt is to use a transistor to activate the relay, or a power transistor to replace the relay altogether. However, I don't have any schematics in mind.

Any thoughts or recommendations?

Has your 'friend' only looked at the output with a DC volt meter? I'm repeating what I said at your 'other' post, from which this quote was taken.

I don't think this AUX port is putting out a steady .5 VDC. I think it is using a PWM or other Digital data signal. I went to the website and read all about the product and all the fancy 'add on' sensors, and devices that can be added to this alarm. The are all addressed digitally by the alarm unit. He needs to read it with a scope and SEE what is really on that line, or talk to an installer that has put one of these units in a car.

 

Kermit2, I understood what you said and will take it into consideration. If this is the case, the comparator will be useless.

P.S. I laughed with the quoted "friend". It reminded me of some very funny situations of people too embarassed to ask help themselves. However, this is not the case. No, sir, I 'd never pay a wad of money to ensure the safety of a 15-year-old Corsa.

 

I do tend to overuse those "quote" brackets.

" "

 

Anyway, one other way to do it would be to use a transistor configured as an amplifier. Maybe bias the signal so it just about switches on the transistor. The signal transistor could then control a power transistor or relay.

If the signal IS a true 0.5VDC output then Toms suggestion is the simplest and probably the cheapest.

PS. I wouldn't pay a heap of money to protect a 15yo Corsa either. I WOULD spend a heap of money advertising the fact that it is unlocked and ready to steal.

 

If the signal IS a true 0.5VDC output then Toms suggestion is the simplest and probably the cheapest.

PS. I wouldn't pay a heap of money to protect a 15yo Corsa either. I WOULD spend a heap of money advertising the fact that it is unlocked and ready to steal.

"AWW CRAP!! Keys are stuck in the ignition again! AND THE DOOR WONT LOCK!

Forget it...Im going to bed...."

 

What would be the reasons behind oscillation?

A slow-moving signal very near the reference voltage will challenge the comparator. For instance I'm dealing with a thermostat situation where the transitions take minutes, not milliseconds. It's not hard to fix, but you just need to realize that the comparator isn't a perfect on-off switch. If your signal is truly either 0 or 0.5v, you won't have any trouble.

 

Here is one circuit using a transistor biased to switch at around 0.5V.

It's probably the lowest cost way of doing it. Q1 (input) should be a small signal NPN transistor such as 2N3904. Q2 (relay) should be a relatively high power transistor, capable of sinking at least 250mA; a 2N2222 would work well.

Here is the code for Falstad circuit sim:

$ 1 5.0E-6 10.20027730826997 37 5.0 50
t 320 224 352 224 0 1 -11.5 0.44539218780850104 100.0
r 304 224 304 160 0 1000.0
w 304 160 352 160 0
w 352 208 352 160 0
w 320 224 304 224 0
r 304 224 304 320 0 10000.0
r 352 240 352 320 0 10000.0
g 304 320 304 336 0
w 304 320 352 320 0
R 304 160 304 128 0 0 40.0 12.0 0.0 0.0 0.5
w 304 224 272 224 0
R 192 224 160 224 0 0 40.0 0.5 0.0 0.0 0.5
s 192 224 272 224 0 0 false
t 400 240 432 240 0 1 -11.945392187368954 0.054607812613434205 100.0
r 400 240 352 240 0 10000.0
w 432 256 432 320 0
w 432 320 352 320 0
178 432 160 480 160 0 1 0.2 8.80627437081291E-13 0.05 1000000.0 0.02 20.0
w 432 208 432 224 0
r 496 176 496 320 0 10.0
w 496 320 432 320 0
w 496 176 496 144 0
w 496 144 480 144 0
w 432 208 400 208 0
d 400 208 400 160 1 0.805904783
w 352 160 400 160 0
w 400 160 432 160 0
w 432 160 432 192 0
o 18 64 0 34 20.0 9.765625E-5 0 -1
o 23 64 0 35 20.0 9.765625E-5 1 -1

 

Thank you for your hard work, tom66!

Just a question: Which is the green and which is the yellow signal, down at the graph?

 

Thank you for your hard work, tom66!

Just a question: Which is the green and which is the yellow signal, down at the graph?

Green = Voltage
Yellow = Current

The scope traces probably don't mean much in this sim.

 

Yes, but which voltage? On the input, or on the controlling coil of the relay?

 

Yes, but which voltage? On the input, or on the controlling coil of the relay?

I look now at the traces and realise they are completely wrong.

When the input is 0.5V or so, the relay coil is disengaged, and the normally closed contact is connected to the 12V input; this turns on the lights.

When the input is less than 0.5V or so, the relay coil is engaged, disconnecting 12V from the normally closed contact and turning off the lights.

 

If the .5 V sig. isn't pure DC, add a RC filter, say 10k & 10 μF, feeding a comparator with a bit of + feedback, 1/10 V or so as prev. suggested.