2 state converter, from -24VDC [ON] - +5VDC [OFF] to 5VDC [ON] - 0VDC [OFF]

Thread Starter

lordofentropy

Joined Apr 10, 2010
58
So as the header says, I would like to take a 2 state signal (ON/OFF) where the on state is -24VDC and it's off state is +5VDC and make that/convert to a 0-5VDC 2 state signal, where 0 is the corresponding off state and +5VDC is the corresponding on state. This is more so a triggering event or interlock, so this would not be a high frequency thing.
Thanks.
 

AnalogKid

Joined Aug 1, 2013
8,231
There are lotsa ways to do this.
What is the current load on the converted 0 V to 5 V signal?
Does the output signal have to stop at 0 V, or could it go down to -0.3 V or -0.7 V?

ak
 

Thread Starter

lordofentropy

Joined Apr 10, 2010
58
It's some type of triggering or interlock signal, that's all I know. I don't know the current load either, but being it's some type of triggering/interlock signal it's safe to assume it's small. And it needs to be 0 to 5v. All I have so far is using an opto isolator or finding a inverter with a zener (that was where you were going with your forward Voltage drop of -.3/-.7)
 

crutschow

Joined Mar 14, 2008
23,814
Here's a simple transistor inverter circuit that would seem to do what you want.
The diode prevents the -24V input from exceeding the transistors reverse Vbe rating.
The transistor can be just about any small NPN type.

upload_2016-4-27_0-51-48.png
 

AnalogKid

Joined Aug 1, 2013
8,231
Rather than count on the capacitance of the 4148 vs. the capacitance of the 3904 when both are reverse biased, I'd add a diode across the base-emitter junction to prevent reverse breakdown.

TS: Is signal logic inversion allowed? That is, can the -24 V be mapped to +5 V, or must it be mapped to 0 V?

Also, is there a separate 5 V supply for the output signal as in Wally's schematic, or does the output 5 V have to come from the input when it is at 5 V?

ak
 

crutschow

Joined Mar 14, 2008
23,814
Rather than count on the capacitance of the 4148 vs. the capacitance of the 3904 when both are reverse biased, I'd add a diode across the base-emitter junction to prevent reverse breakdown.
..................
I'm not counting on any capacitance matching for the circuit operation.
Capacitance has little effect on the circuit operation at low speeds.
When the diode is reverse biased and the current stops, any remaining capacitive charge is dissipated through the forward biased base-emitter junction.
If necessary a base-emitter resistor can be added to clear out any charge or absorb leakage currents but I don't see that needed for slow speed switching at typical ambient temperatures.

I prefer the series diode since it draws no current from the source when the input is negative.
 

AnalogKid

Joined Aug 1, 2013
8,231
I agree with the series diode; that's not what I was referring to. In the steady-state -24 V input condition, the reverse-biased 4148 and 3904 junctions look like two capacitors in series. The voltage that develops across each one is proportional to the capacitance. If they are approximately equal, there will be -12 V across the 3904 base-emitter junction. Even at sub-microamp leakage currents, something easy to prevent with a 2nd 4148 across the 3904.

ak
 

Thread Starter

lordofentropy

Joined Apr 10, 2010
58
Rather than count on the capacitance of the 4148 vs. the capacitance of the 3904 when both are reverse biased, I'd add a diode across the base-emitter junction to prevent reverse breakdown.

TS: Is signal logic inversion allowed? That is, can the -24 V be mapped to +5 V, or must it be mapped to 0 V?

Also, is there a separate 5 V supply for the output signal as in Wally's schematic, or does the output 5 V have to come from the input when it is at 5 V?

ak
the -24V is suppose to be mapped to +5V [ON] state, while the +5V of the original signal is mapped to 0V [OFF] state, still trying to get more information; all I have so far is that this -24VDC/+5VDC signal is coming from some commercial electronics unit and is being used as a bias across what is essentially is a diode.

I can add a +5 source to bias it, if that's what your asking, but for the original signal, the +5VDC is the signals off state, to be converted/mapped to 0 volts.
 
Last edited:

crutschow

Joined Mar 14, 2008
23,814
I agree with the series diode; that's not what I was referring to. In the steady-state -24 V input condition, the reverse-biased 4148 and 3904 junctions look like two capacitors in series. The voltage that develops across each one is proportional to the capacitance. If they are approximately equal, there will be -12 V across the 3904 base-emitter junction. Even at sub-microamp leakage currents, something easy to prevent with a 2nd 4148 across the 3904.
I understand what you are saying but, if the base-emitter junction is reversed biased beyond it's break-down point due to the capacitances, it will just zener and conduct. At the small currents and short duration involved, that should not hurt the junction.
But you can add the extra diode as insurance if you prefer.
 

AnalogKid

Joined Aug 1, 2013
8,231
The transition region starts just above +1.3 V. Something like 1/2 of an LM393 could transition right at 0 V and have a little hysteresis. But then you've got more parts, larger circuit area, decoupling, etc.

ak
 

bertus

Joined Apr 5, 2008
20,189
Hello,

As @Picbuster asked, where should the change over from 0 to 5 Volts at the output be, depending on the input voltages.
Should any negative input voltage give the 5 Volts out?
Should any positive input voltage give 0 Volts out?

Bertus
 

crutschow

Joined Mar 14, 2008
23,814
one question what happens to the output when signal = -1V or -15V ?
We are getting the requirements in bits and pieces. :rolleyes:
Exactly what it the transition point that you want at the input to give a change at the output?
How accurate does that point need to be?
 

Thread Starter

lordofentropy

Joined Apr 10, 2010
58
From what I have gathered so far, the -24/+5 signal needs to fit into the TTL envelope; therefore I would guess mapping that range to TTL, hence any number in the range of -24 to -8v should be high; but I'm really not sure on this, another person is slowly feeding me this information and does a very poor job comuicating. I need to get the model number of this contraption so i can pull its datasheet, i will keep you all posted. Hopefully sometime tomorrow i will have it.
 

Thread Starter

lordofentropy

Joined Apr 10, 2010
58
****New information

The signal is a bias shutdown for a detector, it's load current is small (mA's), still don't have anything on the transition of the -24/+5, that is the thresholds (what's high/low). So, as of now the assumption is all negative voltages is a logic high, and all positive voltages is a low.
 

crutschow

Joined Mar 14, 2008
23,814
****New information

The signal is a bias shutdown for a detector, it's load current is small (mA's), still don't have anything on the transition of the -24/+5, that is the thresholds (what's high/low). So, as of now the assumption is all negative voltages is a logic high, and all positive voltages is a low.
If you need a sharp transition at 0V then the circuit I posted won't do, since it starts to change state at about +1.2V.
For that you would need a comparator circuit, such as using an LM339.
 
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