Constructing a circuit that would trigger my cordless house phone intercom circuit?

[Nigel Goodwin]

Hm. Not sure why a seemingly simple question begets such sarcasm?

I have a circuit known to work at 2.4v and probably between 2.0v & 2.5v (min/max charge states). I'd like to replace the 2xNiMH with single 3.2v (nominal) LiFePO4 cell; and it was suggested that I use a diode to drop the voltage.

But, I do not know the current drawn by the circuit; nor do I see an simple way to measure it; given that it is likely that this battery powered DECT phone has many operating states. Ie. maximum draw may be when its ringing. Or when the speaker phone is turned on. Or it could be when it looses connection to the base station and is searching for a signal to reconnect.

You're overthinking this - just stick an IN4001 or similar in. Bear in mind a 1A rectifier is rated for much higher peaks than 1A.

 

[Buk]

You're overthinking this - just stick an IN4001 or similar in. Bear in mind a 1A rectifier is rated for much higher peaks than 1A.

One of the answers to a similar question at another place reads:

The diode may be a good choice if your load current and input voltage are fairly stable and known.
Keep in mind that the nominal 0.7V or so drop from the diode will be less at very low current and you'll pretty much get the full input voltage with a very light load. If whatever your load is goes into a sleep mode and the current drops to about zero then you'll see that voltage increase to maybe 3.9V. If that is within specs, then you are okay, but then maybe you don't need the diode at all.
There are various regulators. The LP5912-3.3DRVT is one. It can output 500mA with only 0.25V drop (typically less than 50mV at 200mA) and typically draws only about 30uA for its own purposes. It can handle up to 5.5V input.
If you choose to use an LDO regulator, be sure to carefully read the section on capacitors and follow it to the letter. In this case, 1uF ceramic capacitors are suitable on both input and output.

?

 

[Les Jones]

For measuring the current when the cells are in a battery holder I use a piece of thin double sides printed circuit board. A wire is soldered to each side of the board and the board inserted between the cell and the battery holder connection. This is what I use.

I can't remember where I obtained the very thin (0.2mm thick.) board.

Les.

 

[Nigel Goodwin]

One of the answers to a similar question at another place reads:


?

You're still over thinking it!.

But if you're that bothered, then stick an LDO regulator on it.

 

[Nigel Goodwin]

For measuring the current when the cells are in a battery holder I use a piece of thin double sides printed circuit board. A wire is soldered to each side of the board and the board inserted between the cell and the battery holder connection. This is what I use.
View attachment 132409

I can't remember where I obtained the very thin (0.2mm thick.) board.

Les.

A VERY, VERY long time ago back when I was repairing radios, I simply got two pieces of tin plate (from a tobacco tin) and glued them to either side of a piece of paper.

 

[gophert]

How do I calculate the required power rating for the diode?

Remember, unit for power is watts.

so you received this guidance....

They are typically rated in amps, not watts.

And this...

Determine the peak current through the diode, and use one rated for at least twice that.

So, I don't understand why your confused statement came...

Hm. Not sure why a seemingly simple question begets such sarcasm?

All i can say is...
Facts are facts, sarcasms is sarcasm - but sarcasm can also be facts. One fact for sure, you do not seem to understand the definition of "sarcasm" or "power" or possible both - so you were guided to look for amp ratings of diodes rather than watt ratings. Also, when you are corrected, look for your error, don't assume people are being sarcastic.

edit: if I was using sarcasm, it would have been much more cutting and/or clever than "they are typically rated in amp, not watts". Something like...

 

[AnalogKid]

Hm. Not sure why a seemingly simple question begets such sarcasm?

It didn't, but your reaction is very informative.

For many decades, a very common industry practice (and in many companies an in-house rule) is to overrate components by at least 2:1 to increase long-term reliability. Use a 2 A diode in a 1 A circuit, use a 50 V power MOSFET in a 24 V circuit, use a 25 V electrolytic capacitor in a 12 V circuit, etc. To see how this can affect MTBF (Mean Time Between Failures) calculations, see MIL-HDBK-217.

MIL-HDBK-217 F RELIABILITY PREDICTION ELECTRONIC

Of course this isn't always possible, often due to physical or financial constraints. But for some projects I've done, any exception required written justification and management signoff.

ak

 

[Buk]

All i can say is...

When I first arrived here, I was prewarned about some of the incumbents here, by a "well-known" regular.

Amongst those warning was:

Gophert: Can be helpful, but is a bully. He seems to focus in on somebody and attacks every comment that person makes. If that person responds in kind, THEY will get a warning to "Ignore him. It's just how he is."

You certainly live up to your billing.

 

[Nigel Goodwin]

For many decades, a very common industry practice (and in many companies an in-house rule) is to overrate components by at least 2:1 to increase long-term reliability.

Not in domestic electronics it's not If anything, it's the complete opposite - for example, with Samsung fitting electrolytics where the manufacturers own datasheets gave a MBBF of only 9 months (and then gave a 12 month warranty on the units? - what could possibly go wrong?).

However, in this case current will be fairly low, with higher peaks - no issue for a 1A rectifier.

As far as we know, the unit might work perfectly well with nothing whatsoever, running directly from the battery - with increased range due to the higher voltage.

 

[gophert]

When I first arrived here, I was prewarned about some of the incumbents here, by a "well-known" regular.

Amongst those warning was:


You certainly live up to your billing.

You're still having trouble with the concept of sarcasm. You should have said, "Maybe someday you'll live up to your billing."

 

[Buk]

You should have said,

Wow! You even want to tell me what to say now.

Point of fact: I wasn't going for sarcasm; just a simple observation of your behaviour.

BTW: If you want to continue to play, shall we take this off-forum; or can you only get it up for an audience?

 

[Buk]

As far as we know, the unit might work perfectly well with nothing whatsoever, running directly from the battery - with increased range due to the higher voltage.

That'd be nice; and experiement with my existing handsets tells me that the extreme bottom of the garden and the garage are not spots.

But how to determine if 3.6v would do any damage?

I could take the phone apart again and try to get clear pics of both sides of the circuit board. How easy is it to turn that into a schematic?

 

[gophert]

Point of fact: I wasn't going for sarcasm; just a simple observation of your behaviour.

And AK and I were not using sarcasm in posts 18 and 19 when you accused us in your post 20. But enjoy playing the victim.

 

[Dr_Doggy]

ok, so you got your bell button trigger, take a handset and solder the trigger in parallel to the intercom button, keep handset plugged in.

if you deactivate door bell power & ringer you can solder button direct, otherwise just put relay between parallel with the chime.

if handset can use intercom when still on base then just leave on base, otherwise pull battery and replace with DC power adapter at the same volts as the battery you are pulling

 

[rjenkinsgb]

OK, the idea of the load voltage climbing due to leakage when at "zero current" got me curious enough to do a simple experiment:

A reversed 1N4007 in series with a 0.47uF polyester cap, across a 9V battery.

After about half an hour, the cap was charged to roughly 2.5V, but discharged very quickly due to the 10M load of the multimeter once I connected that.

So, use a diode and add a 10M resistor across the phone power terminals - that will bypass any slight leakage and prevent the voltage increasing unduly.

At roughly a quarter microamp load, the effect on the battery life will be negligible.

 

[Nigel Goodwin]

OK, the idea of the load voltage climbing due to leakage when at "zero current" got me curious enough to do a simple experiment:

A reversed 1N4007 in series with a 0.47uF polyester cap, across a 9V battery.

After about half an hour, the cap was charged to roughly 2.5V, but discharged very quickly due to the 10M load of the multimeter once I connected that.

So, use a diode and add a 10M resistor across the phone power terminals - that will bypass any slight leakage and prevent the voltage increasing unduly.

At roughly a quarter microamp load, the effect on the battery life will be negligible.

Have you considered the normal power drain is almost certainly going to higher than that anyway?.

 

[JimB]

Wow! You even want to tell me what to say now.

Point of fact: I wasn't going for sarcasm; just a simple observation of your behaviour.

BTW: If you want to continue to play, shall we take this off-forum; or can you only get it up for an audience?

Nobody is telling you what to say, simply trying to explain the difference in terminology, current rating vs power rating.
Please consider that people may be trying to help you, instead of taking every comment which you do not understand as some kind of an attack on you.

JimB
One of the Moderators.

 

[KeepItSimpleStupid]

You can;t even tell from a datasheet of an AA ni-mH battery: https://data.energizer.com/pdfs/hr6-2650_eu.pdf
Could be anywhere from 0-40 Amps. You really have to measure. peak currents of diodes are higher. It's basically a low power device that has some illumination and transmits, so that the highest current consumption, A 1A or 3A diode would be fine. I suspect a 1n4001 would be fine.

 

[Buk]

Nobody is telling you what to say...

You should have said,...

QED.

And pedandtry is rarely helpful.

 

[Buk]

You really have to measure.

Working on it. I was a otherwise occupied (to no avail) yesterday evening.

I found an old circuit board that is thin enough and cut out a section with ground plane on both sides and filed it to fit the battery terminal:

Unfortunately, it seems that all the little dents where the laquer remains after I rubbed most of it off:

must be plated through vias ; so it's back to the drawing board scrap box.