[YamahaDX] DX7 Memory Corruption

[geoff baltan [e-mail] [YamahaDX]]

Hi All -

I recently managed to get a semi-broken down DX7, first make and
model. The battery needed to be replaced, which was not too much
difficulty. The presets were wiped, but I ran the internal diagnostic
check and everything seemed to check out.

I have started programming new presets and storing them in the
internal memory...but the internal memory seems to "forget" after a
few hours of the machine being turned off. I'm wondering what might
cause this problem.

Thanks,
Geoff

[Brian [e-mail] [YamahaDX]]

That sounds like the diode might be blown.

Have a search through the files here for diode replacement or battery change.
Daniel in his infinite wisdom may be able to help. 😊

Brian.


From: geoff baltan ***@gmail.com [YamahaDX]
Sent: Tuesday, 4 September 2018 6:39 AM
To: ***@yahoogroups.com
Subject: [YamahaDX] DX7 Memory Corruption

Hi All -

I recently managed to get a semi-broken down DX7, first make and
model. The battery needed to be replaced, which was not too much
difficulty. The presets were wiped, but I ran the internal diagnostic
check and everything seemed to check out.

I have started programming new presets and storing them in the
internal memory...but the internal memory seems to "forget" after a
few hours of the machine being turned off. I'm wondering what might
cause this problem.

Thanks,
Geoff


------------------------------------
Posted by: geoff baltan <***@gmail.com>
------------------------------------


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Yahoo Groups Links

[danforcz [e-mail] [YamahaDX]]

Brian,

you are too kind, I’m only musician, my knowledge of electronics is limited, despite my ability to do simple repairs. And I’m far from wisdom even when I became 60 this year :-) (Maybe because I don’t feel so old.)

I would measure voltage on RAM chips when instrument is off.

Checking the diode D4 is highly recommended, but when diagnostics showed proper voltage, probably it’s OK, isn’t it?

I would also check or replace tantalum cap 10/16 connected parallel to battery pins


I suppose the battery polarity is OK...

Good luck!

Daniel Forro

Post by Brian ***@gmail.com [YamahaDX]
That sounds like the diode might be blown.
Have a search through the files here for diode replacement or battery change.
Daniel in his infinite wisdom may be able to help. 😊
Brian.
Sent: Tuesday, 4 September 2018 6:39 AM
Subject: [YamahaDX] DX7 Memory Corruption
Hi All -
I recently managed to get a semi-broken down DX7, first make and
model. The battery needed to be replaced, which was not too much
difficulty. The presets were wiped, but I ran the internal diagnostic
check and everything seemed to check out.
I have started programming new presets and storing them in the
internal memory...but the internal memory seems to "forget" after a
few hours of the machine being turned off. I'm wondering what might
cause this problem.
Thanks,
Geoff

[[e-mail] [YamahaDX]]

A diode can fail by going open circuit, or by becoming practically a short circuit.

If the diode is open circuit, the battery voltage would still be read as being correct - it just wouldn't provide any backup power to the RAM.

- Andy

[danforcz [e-mail] [YamahaDX]]

Thanks for the explanation.

By open circuit you mean the diode is interrupted? But in such case battery will be not connected. When the diode is shorted, it works just as a wire jumper and battery will be connected. Or am I wrong?

If I understand well, the function of the diode in this circuit is not to allow voltage to go the battery when the instrument is on, and to allow voltage go from the battery to RAM IC’s. Is it so?

Or does the diode allow to pass some small voltage to the battery when instrument is on, for loading the battery? Probably these type of battery can’t be recharged but I always wondered how it is possible that such backup batteries in instruments can work 25 or more years and still have 3 Volts or little bit less...

Daniel Forro

Post by ***@ntlworld.com [YamahaDX]
A diode can fail by going open circuit, or by becoming practically a short circuit.
If the diode is open circuit, the battery voltage would still be read as being correct - it just wouldn't provide any backup power to the RAM.
- Andy

[[e-mail] [YamahaDX]]

Dan,

A diode essentially stops current flowing in one direction, as you know. So in this case (I haven't looked at the circuit diagrams) its probably as you said, to stop main power flowing back into the battery and trying to charge it, which would be very bad. A shorted diode would act as a jumper. Very bad. An open diode would stop battery voltage from getting to the RAM, meaning contents lost after power-off.
Those 3V batteries last a long time because the amount of current needed to keep the contents of the RAM is very very small.

[danforcz [e-mail] [YamahaDX]]

Hi, Jan,

thanks for the confirmation my foggy assumptions :-)

Daniel

Post by ***@yahoo.com [YamahaDX]
Dan,
A diode essentially stops current flowing in one direction, as you know. So in this case (I haven't looked at the circuit diagrams) its probably as you said, to stop main power flowing back into the battery and trying to charge it, which would be very bad. A shorted diode would act as a jumper. Very bad. An open diode would stop battery voltage from getting to the RAM, meaning contents lost after power-off.
Those 3V batteries last a long time because the amount of current needed to keep the contents of the RAM is very very small.

[[e-mail] [YamahaDX]]

Daniel,

What pnetops said, and yes, open circuit means no connection, like a simple switch that's off / open. Closed circuit means like a closed switch (on), allowing current to pass both ways.

The diode is meant to only allow current from the battery to the RAM when there is no external power supply providing 5V, and ideally no current at all the other way, though there might be a few microamps.

A healthy diode conducts best when the voltage at the anode is about 0.6V higher than the voltage at the cathode - for a silicon one, or about 0.3V for a germanium diode... it varies, depending on the material and type of diode, but it's generally in about that range.

A dead diode either doesn't conduct at all, or conducts like a small value resistor, with little or no difference in direction.

Since the batteries were soldered in, a lot of these synths still have the original ones because it's not trivial to change one, but a lot of those original batteries still work, after all these years.

- Andy

[[e-mail] [YamahaDX]]

I had a very similar problem - I even sourced some replacement RAM chips as I thought they might be faulty but that wasn't the reason. It turned out that the contents of the RAM were being corrupted as the DX7 was powered off. There is a PD signal coming from the "DC" power board that is designed to prevent this but the Zener diode ZD1 (6.2V) had failed meaning the RAM could get written with random data.


Hope this helps.

[[e-mail] [YamahaDX]]

The zener diode thing is interesting.

That 10uF capacitor would power the RAM for a while. As they only need such a tiny current, and can normally retain memory contents down to quite a low voltage, it may even last for a few hours.

Tantalum capacitors tend to last well, unless they're subjected to reverse voltage. It's usually electrolytic capacitors that need replacing due to age.

- Andy

[[e-mail] [YamahaDX]]

I missed out the other bit - any small silicon or germanium diode would be able to do that job. I don't know what the original specification was though. No doubt someone else does.

I would guess it was a silicon one, and that something like 1N4001 - 1N4009 would be fine. If you find one that measures the same, that will do.

[[e-mail] [YamahaDX]]

FYI the original diode D4 is an OA95 germanium diode.

[Gregg Sheehan [e-mail] [YamahaDX]]

They picked a germanium diode as it has less forward voltage drop than a silicon type.
On Tuesday, September 4, 2018, 10:01:51 a.m. PDT, ***@goffart.co.uk [YamahaDX] <***@yahoogroups.com> wrote:

 
FYI the original diode D4 is an OA95 germanium diode.

[[e-mail] [YamahaDX]]

Okay, I found a datasheet for M5M5118P 16k-B1T (2048-WORD BY 8-BIT) CMOS STATIC RAM, which may or may not be the same chip, but the standby current for that one is up to 10uA, and there are three of them, so it seems I was wrong about that - the 10uF capacitor wouldn't power them for more than a few seconds.

Maybe there's some current through the diode but not enough? Either way, I'd start with that. It's an easy component to replace, and a cheap one.

- Andy

[Bruce Wahler [e-mail] [YamahaDX]]

True, but a germanium diode has more reverse leakage current, which is
not a good thing for a Lithium battery.  A low-current silicon diode --
ex: 1N4148 -- might be the better choice.

Regards,

-BW

Bruce Wahler
Ashby Solutions.com™
***@ashbysolutions.com
http://music.ashbysolutions.com
http://halfmoon-switch.com
http://ashboys.com
978.597.7008

Post by Gregg Sheehan ***@yahoo.com [YamahaDX]
They picked a germanium diode as it has less forward voltage drop than a silicon type.
FYI the original diode D4 is an OA95 germanium diode.

[[e-mail] [YamahaDX]]

A diode like 1N4001 - 1N4009 would be 'much to big'. The very common 1N4148 will do the trick! The diode (D4) will bar the way of the PSU's 5Volts to get to the battery (due not rechargable!). If the synth is OFF the voltage of the battery goes across the diode to Pin24 of the RAMs to keep the content stored. The capacitor (10µF/16V) acts as a kind of 'buffer' to stabilize that voltage(s) a bit more.

[geoff baltan [e-mail] [YamahaDX]]

This is very helpful - so thank you, everyone for the input.
A few questions:
1. The device doesn't immediately lose its memory at power-off, it usually
takes about 8 hours. If the diode were simply open or closed, would it
still partially-work?
2. Is there a list of appropriate replacements for the diode and tantalum
capacitor?

Post by ***@ntlworld.com [YamahaDX]
Daniel,
What pnetops said, and yes, open circuit means no connection, like a
simple switch that's off / open. Closed circuit means like a closed switch
(on), allowing current to pass both ways.
The diode is meant to only allow current from the battery to the RAM when
there is no external power supply providing 5V, and ideally no current at
all the other way, though there might be a few microamps.
A healthy diode conducts best when the voltage at the anode is about 0.6V
higher than the voltage at the cathode - for a silicon one, or about 0.3V
for a germanium diode... it varies, depending on the material and type of
diode, but it's generally in about that range.
A dead diode either doesn't conduct at all, or conducts like a small value
resistor, with little or no difference in direction...
Since the batteries were soldered in, a lot of these synths still have the
original ones because it's not trivial to change one, but a lot of those
original batteries still work, after all these years.
- Andy

[danforcz [e-mail] [YamahaDX]]

1. Then maybe tantalum cap is OK and power goes only from it after switching the instrument off, until it lose the loaded voltage ???

2. Yes, it’s in the service manual and schematics. Diode - OA95, tantalum cap 10 uF/16 V.

Daniel Forro

Post by geoff baltan ***@gmail.com [YamahaDX]
This is very helpful - so thank you, everyone for the input.
1. The device doesn't immediately lose its memory at power-off, it usually takes about 8 hours. If the diode were simply open or closed, would it still partially-work?
2. Is there a list of appropriate replacements for the diode and tantalum capacitor?

[Bruce Wahler [e-mail] [YamahaDX]]

1. When in standby mode, the RAMs need *very* little current to
maintain their data.  (That's the reason why the battery can last
many years.)  The RAM holds data down to around 2V of backup.  It's
entirely possible that the 10uF capacitor can maintain that for a
while after the DX7 is powered off.  In fact, that seems to imply
that the capacitor is fine, IMO.

2. The diode is probably a small signal part like a 1N4148 or 1N914A. 
They have < 25nA of reverse leakage at the voltages used here, and
at the tiny currents we are talking about, the voltage drop is only
going to be 0.3-0.4V.

Regards,

-BW

Bruce Wahler
Ashby Solutions.com™
***@ashbysolutions.com
http://music.ashbysolutions.com
http://halfmoon-switch.com
http://ashboys.com
978.597.7008

Post by geoff baltan ***@gmail.com [YamahaDX]
This is very helpful - so thank you, everyone for the input.
1. The device doesn't immediately lose its memory at power-off, it
usually takes about 8 hours. If the diode were simply open or closed,
would it still partially-work?
2. Is there a list of appropriate replacements for the diode and
tantalum capacitor?
Daniel,
What pnetops said, and yes, open circuit means no connection, like
a simple switch that's off / open.  Closed circuit means like a
closed switch (on), allowing current to pass both ways.
The diode is meant to only allow current from the battery to the
RAM when there is no external power supply providing 5V, and
ideally no current at all the other way, though there might be a
few microamps.
A healthy diode conducts best when the voltage at the anode is
about 0.6V higher than the voltage at the cathode - for a silicon
one, or about 0.3V for a germanium diode.... it varies, depending
on the material and type of diode, but it's generally in about
that range.
A dead diode either doesn't conduct at all, or conducts like a
small value resistor, with little or no difference in direction.
Since the batteries were soldered in, a lot of these synths still
have the original ones because it's not trivial to change one, but
a lot of those original batteries still work, after all these years.
- Andy

[danforcz [e-mail] [YamahaDX]]

Now I have found in schematics, that battery voltage is measured by diagnostics directly, not through diode - it’s a wire marked VB (probably Voltage Battery) which goes to IN6 (this looks like one of inputs to IC11 = ADC), you can find it near IC10 2/2.

Daniel Forro

Post by ***@ntlworld.com [YamahaDX]
A diode can fail by going open circuit, or by becoming practically a short circuit.
If the diode is open circuit, the battery voltage would still be read as being correct - it just wouldn't provide any backup power to the RAM.
- Andy

[danforcz [e-mail] [YamahaDX]]

Thanks for the explanation.

By open circuit you mean the diode is interrupted? But in such case battery will be not connected. When the diode is shorted, it works just as a wire jumper and battery will be connected. Or am I wrong?

If I understand well, the function of the diode in this circuit is not to allow voltage to go the battery when the instrument is on, and to allow voltage go from the battery to RAM IC’s when the instrument is off. Is it so?

Or does the diode allow to pass some small voltage to the battery when instrument is on, for loading the battery? Probably these type of battery can’t be recharged but I always wondered how it is possible that such backup batteries in instruments can work 25 or more years and still have 3 Volts or little bit less...

Daniel Forro

Post by ***@ntlworld.com [YamahaDX]
A diode can fail by going open circuit, or by becoming practically a short circuit.
If the diode is open circuit, the battery voltage would still be read as being correct - it just wouldn't provide any backup power to the RAM.
- Andy

[[e-mail] [YamahaDX]]

Daniel,

The idea of the diode (germanium, then) is to stop current flowing from the 5V supply to the 3V (or so) battery, while there is a 5V supply, but to allow current (not voltage) to flow from the battery to the static RAM chips, when there is no 5V supply.

In an idealised view of how the circuit works, it behaves like an automatic switch.

However, no component behaves quite that perfectly. There will always be some reverse or leakage current, from the 5V supply to the battery, and there will always be a finite amount of resistance when it's conducting (forward biased) to power the RAM chips from the battery.

With a healthy diode, that reverse leakage current will be small enough not to matter - hence batteries which are not rechargeable being able to last for years.

One of the ways a diode fails is that the leakage current goes up dramatically. It might not happen all at once though. The two distinct layers of semiconductor start to break down, polluting each other, so the diode behaves less and less like a diode, and more like a resistor.

Another way a diode fails is simply that a gap forms between the two layers, typically due to a zap of high current, for a moment. In that case, the diode stops conducting, and becomes open circuit.

However, it's possible that a gap starts to form, with age, but not all the way across, so the two different pieces of semiconductor don't completely separate - maybe forming little pits or cavities. In that case, the forward resistance of the diode would start to increase - it would still pass current, when it was forward biased (anode voltage exceeds cathode voltage by a sufficient amount), but it wouldn't pass as much.

- Andy

[charles copp [e-mail] [YamahaDX]]

broken cartridge in the cartridge port?
or other misdirected connectivity?

[geoff baltan [e-mail] [YamahaDX]]

Alright, I'm a little new to the repair side of this stuff, so I just
wanted to double-check before I start soldering... please let me know if
I'm understanding everything correctly:

It sounds like there are two potential diodes that could be causing this
problem.

First, there is the diode that the battery is normally attached to - D4,
which is a general purpose germanium diode. Anything general purpose diode
could probably be used as a substitute, since the main function is
protecting from reverse current. However, I should be able to determine
whether this diode is "open" with a simple multimeter check, prior to
replacing it.

The other diode, is labelled ZD1 - and refers to a 6.2 zener diode.
Apparently, this should be located somewhere around the RAM chips (ICs 32
and 4 on the SchematicRotated.pdf) - but I'm having some difficulty finding
it. This diode probably needs to be exactly 6.2v, correct?


Cheers!

Post by charles copp ***@sympatico.ca [YamahaDX]
broken cartridge in the cartridge port?
or other misdirected connectivity?

[[e-mail] [YamahaDX]]

You should use the resistance section of the meter and put it across the diode, one way will be infinite, the other will show a few hundred ohms, depending on the type. Most meters have a selection at the bottom of the resistance scale, with a diode symbol next to it. This is usually the one that beeps when you short circuit (<100Ohms), but basically if your meter doesn't have that, use the 2k range.
Zener diodes have a special properties, which is why they are used in power supplies, they leak current when the reverse voltage is over the specified value, in order to keep the voltage at that value. So, if you replace your 6.2v zener with an 8v zener, you'll get 8v probably where you don't want it.


BTW - Without lifting one leg of the component under test, it's not guaranteed you'll determine that it's faulty. For diodes its usually not a problem, shorts are easy to spot, open (both directions) are easy to spot. but there could be a cap in parallel which could create the short condition, for example.

[[e-mail] [YamahaDX]]

You should use the resistance section of the meter and put it across the diode, one way will be infinite, the other will show a few hundred ohms, depending on the type. Most meters have a selection at the bottom of the resistance scale, with a diode symbol next to it. This is usually the one that beeps when you short circuit (<100Ohms), but basically if your meter doesn't have that, use the 2k range.
Zener diodes have a special properties, which is why they are used in power supplies, they leak current when the reverse voltage is over the specified value, in order to keep the voltage at that value. So, if you replace your 6.2v zener with an 8v zener, you'll get 8v probably where you don't want it.


BTW - Without lifting one leg of the component under test, it's not guaranteed you'll determine that it's faulty. For diodes its usually not a problem, shorts are easy to spot, open (both directions) are easy to spot. but there could be a cap in parallel which could create the short condition, for example.

[[e-mail] [YamahaDX]]

You should use the resistance section of the meter and put it across the diode, one way will be infinite, the other will show a few hundred ohms, depending on the type. Most meters have a selection at the bottom of the resistance scale, with a diode symbol next to it. This is usually the one that beeps when you short circuit (<100Ohms), but basically if your meter doesn't have that, use the 2k range.
Zener diodes have a special properties, which is why they are used in power supplies, they leak current when the reverse voltage is over the specified value, in order to keep the voltage at that value. So, if you replace your 6.2v zener with an 8v zener, you'll get 8v probably where you don't want it.


BTW - Without lifting one leg of the component under test, it's not guaranteed you'll determine that it's faulty. For diodes its usually not a problem, shorts are easy to spot, open (both directions) are easy to spot. but there could be a cap in parallel which could create the short condition, for example.

[[e-mail] [YamahaDX]]

Geoff,

If you're confident that the contents of memory are retained for a while after power off, and don't get corrupted so long as you turn it back on soon enough, then it seems most likely that it's the germanium diode that's faulty.

As Gregg said, they chose germanium for the lower voltage drop - with a silicon one, the voltage to the battery would be about 0.3V lower, so the RAM would lose power sooner - maybe reducing the effective battery life by a few years.

Their plan worked pretty well, so I'd be tempted to stick with that. However, if you're also fitting a battery clip, battery life is not so important - changing it becomes much simpler - so you could go with silicon (e.g. 1N4148 or 1N4001). I don't have any figures, but I'm under the impression that silicon diodes don't wear out nearly as easily as germanium ones.

Since the diode is a known point of failure, you may feel there's a good argument for trying a silicon one instead - or you may be having to do this again in as little as twenty years :)

As pnetops said, checking the zener diode is harder. When a component is cheap and easy to replace, known to be a problem, and difficult to test, I'd just replace it. However, just replacing the germanium diode may well fix it, so you might want to try that first, and decide whether you want to stop there.

I like pnetops's idea of checking diodes on the resistance scale of a meter. I've been spoilt by having one that has a special diode checker, which shows me the forward voltage drop of the diode - I use that for identifying the legs on transistors too.

Ideally, yes, you need to de-solder at least one leg of the diode, to make sure you're measuring it properly, but you can try it in circuit without any risk, and see if that convinces you either way.

Anecdotally, we have zener diode failure damaging RAM contents, but not the RAM itself. If that's the case, I wouldn't necessarily change it just to make sure. So long as you back up your RAM contents (e.g. with JSynthLib, or just a bulk dump to MIDI-OX, or Bome's SendSX), you're not going to lose anything other than time and patience.

My latest tool for desoldering is a pair of cheap ceramic tipped tweezers. They make desoldering things like diodes and resistor really easy -

a) Add a dab of fresh solder - optional, but the flux in the solder tends to help

b) Remove the solder with braid or a solder sucker - still optional, but it helps to leave the hole clear of solder when you remove the component

If step b doesn't leave the leg loose, or you want to go straight to c:

c) Grab one leg, or the whole component with the ceramic tipped tweezers, heat with the soldering iron (ideally from the solder side) and pull.

If there's a lot of leg showing, you might also consider clipping it, then pulling out the stump with the tweezers, but there's usually no need, for a two legged component.

- Andy

[[e-mail] [YamahaDX]]

I've gotta get me some ceramic pliers. One other thing I do, rather than lifting a leg (which puts stress on the PCB pad) I tend to just snip one side of the component, test it, and if it's fine, I just resolder the leg back together, you can hardly see it if it's done right.

[geoff baltan [e-mail] [YamahaDX]]

Alright, I replaced the diode with a 4001, temporarily - and so far, it
works fine. So, thanks everyone!
Is there any benefit to going back and replacing it with a smaller signal
diode, like a 1n4148?

Post by ***@yahoo.com [YamahaDX]
I've gotta get me some ceramic pliers.
One other thing I do, rather than lifting a leg (which puts stress on the
PCB pad) I tend to just snip one side of the component, test it, and if
it's fine, I just resolder the leg back together, you can hardly see it if
it's done right.

[Bruce Wahler [e-mail] [YamahaDX]]

The 1N4148 will likely have a lot less leakage current -- maybe, 15nA
vs. 2uA at the voltages seen here.  Pumping current backwards into the
battery isn't desirable, and will shorten its life.  We probably talking
2-5 years vs. 10-20 years, though, so this may or may not seem important
to you.

Regards,

-BW

Bruce Wahler
Ashby Solutions.com™
***@ashbysolutions.com
http://music.ashbysolutions.com
http://halfmoon-switch.com
http://ashboys.com
978.597.7008

Post by geoff baltan ***@gmail.com [YamahaDX]
Alright, I replaced the diode with a 4001, temporarily - and so far,
it works fine. So, thanks everyone!
Is there any benefit to going back and replacing it with a smaller
signal diode, like a 1n4148?
I've gotta get me some ceramic pliers.
One other thing I do, rather than lifting a leg (which puts stress
on the PCB pad) I tend to just snip one side of the component,
test it, and if it's fine, I just resolder the leg back together,
you can hardly see it if it's done right.

[[e-mail] [YamahaDX]]

Where are you getting the 2uA figure from, Bruce?

The 1N4001-4009 series datasheet I've found only quotes 10uA at maximum reverse voltage, but that's 50V for a 1N4001.

Unfortunately, they don't give comparable figures, because they're meant for different purposes. If you look at the figures for them getting hot, and near their reverse voltage limits, they're in the same ball park.

I couldn't find figures for the 1N4001 for a low reverse voltage. It's probably a little bit higher than the 1N4148, but 2uA at 2-3V doesn't seem right.

- Andy

[Bruce Wahler [e-mail] [YamahaDX]]

I'm assuming a) that the leakage at 5V is less than at 50V; and b) the
actual part received is not at the maximum point; other than that, all
bets are off.  That said, I'd wager that the 1N4001's leakage is at
least 10-20x that of a 1N4148, possibly more.  Also, if the 1N4001
happens to leak say, 4.5uA at 5V, it would still pass its specifications
-- probably won't be that bad, but a manufacturer would not consider the
part 'defective.'

A 1N4001 is designed to be a rectifier diode; most of the time, I see
them in power supplies.  The 1N4148 is designed as a fast signal-level
switch, which is a closer match to this application.

Just my opinion.

Regards,

-BW

Bruce Wahler
Ashby Solutions.com™
***@ashbysolutions.com
http://music.ashbysolutions.com
http://halfmoon-switch.com
http://ashboys.com
978.597.7008

Post by ***@ntlworld.com [YamahaDX]
Where are you getting the 2uA figure from, Bruce?
The 1N4001-4009 series datasheet I've found only quotes 10uA at
maximum reverse voltage, but that's 50V for a 1N4001.
Unfortunately, they don't give comparable figures, because they're
meant for different purposes.  If you look at the figures for them
getting hot, and near their reverse voltage limits, they're in the
same ball park.
I couldn't find figures for the 1N4001 for a low reverse voltage. 
It's probably a little bit higher than the 1N4148, but 2uA at 2-3V
doesn't seem right.
- Andy

[[e-mail] [YamahaDX]]

Okay, I found a more detailed datasheet, from Vishay:

General Purpose Plastic Rectifier 1N4001 thru 1N4007 - Vishay

https://www.vishay.com/docs/88503/1n4001.pdf https://www.vishay.com/docs/88503/1n4001.pdf

This one has a graph, for reverse current. At 25 degrees C, it looks like 10nA for a reverse voltage of about 5 or 6V. See "Fig. 5 - Typical Reverse Characteristics"

In other words, the bigger diode can handle more current, and more reverse voltage (which isn't needed, I agree), but at low currents and a low reverse voltage, it behaves in a very similar way to the smaller diode.

Based on that - no, I wouldn't bother changing it, unless you decide to change back to a germanium one. Personally, I'd leave it in, and fit a battery clip, if you haven't already, Geoff.

- Andy

[[e-mail] [YamahaDX]]

Correcting myself a little (it's late) - I was reading percentage of maximum reverse voltage as volts - so I'm out by a factor of two reading the horizontal axis, taking the maximum reverse voltage as 50V.

It doesn't make a lot of difference to the answer though. It looks like 10nA at about 3-4V reverse voltage, which is more than we'd be getting.

- Andy

[[e-mail] [YamahaDX]]

I doubt if there's any advantage to changing it to a 1N4148, no, Geoff.

The forward voltage drop is the same, at such a low current. The datasheet for the 1N4148 I've looked at gives the reverse leakage current for a low current, which looks good, but the 1N4001 datasheet only quotes it for the maximum reverse voltage - because it's meant as a rectifier diode, so that's seen as more important.

It looks like the maximum leakage current is broadly similar, and I suspect the leakage at low currents is too - or at least, the difference isn't likely to matter.

There's an advantage to changing to a germanium diode, because the forward voltage drop is roughly halved, so you get longer life from the 3V battery, before it falls to the minimum voltage to power the RAM - but probably not from the diode. A small germanium diode should last a long time, but they are known to wear out eventually. The 1N4001 will probably last forever.

- Andy