Simple monitor for healthy voltages - how?

[stevendt]

Hi,

I have four low current supplies fed from a single 5v supply that splits into 4 with individual fuses on the board.
I would like to have a basic fuse failure led for any fuse failure (I don't have space for individual ones)
I am severely limited in the logic gates that I have available, I can either have 4 parts of a single LS04 (i.e. 4 not gates) or, 2 parts of a single LS00 (i.e., 2 NAND gates), but not both.

I thought about something like a wired AND

The problem with that is that this circuit expects the input to be either logic 0 or 1, if a single input goes low, then the associated diode passes and drags the pull-up down causing the input of the gate to go low.

I thought that something similar could be used for my problem, however, if a fuse fails, the line would not be pulled to a logic 0 and the logic input would continue to float. I wondered whether I could find matching values for the pull-up and additional pull downs that would allow this to work, something like :-


However, I can't find a combination of resistor values that allow this to work, so I'm thinking that what I am trying to do is probably not possible.
I would appreciate any comments please?

If what I am trying to do is fundamentally wrong, can anyone suggest how else this might be done please - using no additional logic than described above and with having only a single indicator?

regards
Dave

 

[NorthGuy]

Instead of 4 pull-up resistors you can use 4 pull-down resistors.

 

[stevendt]

Hi,
thanks for the reply.

The second schematic is not very well drawn, but the 4 resistors ARE pull-downs.

regards
Dave

 

[crutschow]

You could use the two 2-input NAND gates to monitor each of the 4 voltages directly with a pull-down resistor on each input. The problem is that ORing the high output voltage of these gates when any input goes low may not be enough to drive the LED. If you can add a buffer NPN transistor such as a 2N2222 for the LED then the NAND gates can be ORed to the transistor base input with a 3k resistor from each NAND output to the base of the transistor. The transistor emitter is grounded and the collector drives the LED and resistor going to +5V.

 

[MrAl]

Hi,

Then use two LED's, in a single package

Also, if a transistor can be added the diode/resistor logic circuit will work too.

Too bad this wasnt CMOS.

 

[stevendt]

Hi (crutschow),

thanks a lot for the reply and the suggestion, is this what you had in mind?


If I have drawn this right, could you maybe explain 3K resistors please?
I guess the intention is that if only one of the gates is high (single fuse failure) we need to ensure that the combined output is enough to turn on the transistor, but I'm not sure what's happening?

regards
Dave

 

[stevendt]

Hi,

Then use two LED's, in a single package

Also, if a transistor can be added the diode/resistor logic circuit will work too.

Too bad this wasnt CMOS.

Thanks - too bad I've only gotten as far as 74LS

- I already have multiple LEDs in packages of 4 - just a single LED left though

 

[MrAl]

Hi,

Use two diodes on the outputs instead of 2 resistors, then one resistor in series with the base. Might also need another larger resistor like 100k from base to ground to ensure the leakage of the transistor can not turn itself on (a standard design technique).

 

[stevendt]

Thanks Mr Al,
just a query on the pull-downs please . . . .
I see you use used 4K7, which seems to be a pretty standard value, it's probably not significant, but 4 x 4K7 would use about 4mA, using 10K would halve that, would it have a material impact on the operation of the circuit?

regards
Dave

 

[NorthGuy]

It'll probably work with much higher resistors, such as 100K. It'll be very slow, but it's not a problem here.

 

[MrAl]

Hi again,

Actually i think the rule for LS TTL is 0.5v and 0.4ma per gate input, so that means even 4.7k is too high. 1250 ohms is the highest value that should be used for a single gate:
R=0.5/0.0004=1250

Remember that TTL pulldowns are always quite low in value. They have to be able to take the input down to some minimum voltage to ensure the logic sees it as a true logic low. To do that means we have to sink somewhat significant current.

 

[alec_t]

I agree with Mr Al. And as he said earlier, "Too bad this wasnt CMOS"; since CMOS can be used with much higher resistor values.

 

[MrAl]

Hi again,

alec:
Yes that would be nice. TTL can be a pain sometimes. Lucky this isnt standard TTL or we'd need really low values.

Here's a new circuit that just uses one transistor and some diodes. It's basically a home brew four input DTL gate.

 

[stevendt]

Hi Again Mr Al.

thanks for that - that looks like a good solution.

Thanks a lot for all the replies guys, I did not expect my trivial problem to get so many helpful responses,
regards
Dave

 

[MrAl]

Hi again "Dave",

I was surprised to see you sign off as "Dave" when your screen name reads "steven dt" more or less, but that's no big deal.

Did you know that the hardest thing to do in electronics is to light up a light bulb? We came up with this 'theory' a long time ago (circa 1980). The reason is because it always seems like it is going to be so very very easy to do, but then all sorts of problems show up when we actually go to do it.

I can tell you many stories like this one, but the most recent was when i went to install an LED on my glue gun. They never seem to make glue guns with power indicators but my older glue gun had an indicator because i had installed one. It was a neon type indicator and worked pretty well. I could always tell when it was on without feeling it or whatever.
Then i got a new one, and of course that did not have an indicator either, so i went to install one using a super bright LED and very large value resistor. But the new one has two heat settings, high and low, so i decided to use a multicolor LED to show high with green and low with blue. Getting this to work was suprisingly much harder to do than you might think. But more about this in another thread...

 

[stevendt]

Hi again,

yes, "Dave" it is

"stevendt" is the username that I use on most forums, it is rarely taken by anyone else when I have to register and saves me having to invent different usernames.
[It originated from the e-mail address format that my company used many years ago, being the first 6 characters of my surname and two initials, it does often cause confusion though.]

Thanks a lot for the help, as you can tell, this stuff is pretty new to me

regards
Dave (honest!)