PSU schematic

[Mark Hennessy]

The problem with a 2 diode 2 cap circuit for -5V is that I think I'd have to have used a half wave rectifier for +5V, something I would rather not do.

No, not what I had in mind.

Take a look at the attached. Ignore most of it - this was just the first example schematic I came across. it was just something I put together in a hurry for someone - it worked well, but it's not finished by any means. But as a scheme, it's something I've done countless times, and it works well. I claim no originality - it's obviously very similar to the example in the OP...

The main supply is full-wave rectified, the low-current -ve supply is derived by the two diodes and two capacitors.

OK, in your case the AC was fixed so you couldn't have reduced it to reduce dissipation in the 5V regulator. But at least you would have lost ~1W being wasted in the 22 ohm resistor...

[Diabolical Artificer]

Thought I'd replied, must have forgotten to click submit.... The circuit makes sense now, as the 5v rail uses the bulk of the power available. I missed the voltage doubler, which they're doing as 9v rectified won't be enough after voltage drop from diodes, regulator chip and load I guess.

The chassis is definately live in regards to the 9v AC in pins of the power input pins. No biggy, just took the plug apart and disconnected the shield from the metal plug shell.

Here's the service manual, as you see it uses a lithium battery to power the SRAM memory when not in use. How would one replace a battery in such a situation as this? Un solder the battery whilst it's powered up? There is a "how to guide" online - http://se-server.ethz.ch/Staff/af/pdfs/Q...cement.pdf, but this involves loosing any settings etc. Mine is fine BTW, just curious.

Andy.

[Nick]

When replacing a lithium like this (in Tek 'scopes for instance) I came up with the following method:

• Make sure the unit is OFF - leave it for a while to let any DC rail levels settle to zero.
  
• Measure the voltage on the battery precisely.
  
• Charge a 4,700uF electrolyic to the same value (to be honest, 470uF would probably do) - SRAM only consumes a few micro-Watts in standby
  
• Tack the electrolytic across the lithium battery (not on its leads) - no charge transfer as at same potential - obviously make sure it's + to +, - to -...
  
• Remove the lithium battery making sure you don't disconnect the parallel capacitor in the process!
  
• Replace with new battery - this will obviously charge the electrolytic to the new (hopefully higher) voltage - this is why you don't want too big a uF value...
  
• Remove the electrolytic...
  
• Done!
  

Important thing here is that you are at no time trying to force charge into the lithium battery, which would be a BAD idea. You need a capacitor value which is big enough to keep the SRAM happy while you change the battery, but small enough that you don't waste too much charge when the new battery is placed in circuit and raises the potential on the capacitor to the new (higher) voltage - that charge is essentially wasted as you then remove the capacitor, so you don't want a value that is unnecessarily large.

HTH

[Terry]

The chassis is definately live in regards to the 9v AC in pins of the power input pins. No biggy, just took the plug apart and disconnected the shield from the metal plug shell.

I still don't understand your problem here, Andy. If you ignore the ±12V rails, the main +5V rail is fed by a half wave rectifier, the output of which is positive with respect to the chassis - which is connected to one side of the incoming AC supply.

However you look at it, one side of the AC supply must  be connected to the chassis - as shown on the circuit you provided.

So why do you stress the importance of not connecting the shield to the shell? The AC supply is connected to the chassis on the socket - what is wrong with doing the same thing in the plug? 

Perhaps I'm missing something here - please tell me what I am missing.

[Mark Hennessy]

Perhaps he's forgotten that the 9VAC winding is floating w.r.t. mains?

[Mark Hennessy]

That's a complicated method, Nick

If you don't have a schematic, then the easiest thing to do is use a bench PSU set to just below the value of the battery. Perhaps via a 1k "safety" resistor if worried. OK, there are exceptions, but the vast majority of bench power supplies are series regulators that only sink current, so they won't attempt to charge or discharge the battery in that state.

Here we have a schematic. So here, I'd just connect a ~5V PSU to "MEM PWR" - across C16 - as D4 prevents the battery from being charged. Simple

One minor concern with the 4700uF capacitor: the amount of instantaneous current that could flow somewhere it shouldn't... Sure it wouldn't hurt the battery, but I'd be tempted to use a safety resistor in series with it.

[AlanBeckett]

Perhaps it would be best to move that Post to the Battery changing Thread
By the way, most Farnell Bench PSUs don't have any Reverse Polarity protection so connecting a Battery to the Output when it's set to lower than the Battery Voltage destroys a lot of the Transistors.

Alan

[Nick]

That's a complicated method, Nick

You may think so, but I've always hated having trailing leads going from a board I'm working on... you turn the board over and the wretched things drop off.

Charging a small cap is hardly complicated...

[Mark Hennessy]

Perhaps it would be best to move that Post to the Battery changing Thread

Which thread?

By the way, most Farnell Bench PSUs don't have any Reverse Polarity protection so connecting a Battery to the Output when it's set to lower than the Battery Voltage destroys a lot of the Transistors.

Yes, I've reported on that in the past and published a simple mod to fix it.

You have to go over the set voltage by quite a bit, as the failure mechanism is exceeding the ~7V reverse VBE breakdown voltage of the driver transistor (VT5 in the E series, for example). Unlikely to happen here.

It usually happens when you apply an external voltage when the PSU is powered down. A power cut while charging a battery is "fun", as you can imagine what happens to the battery when the mains returns!

[Mark Hennessy]

That's a complicated method, Nick

You may think so, but I've always hated having trailing leads going from a board I'm working on... you turn the board over and the wretched things drop off.

Charging a small cap is hardly complicated...

Each to their own, but if the leads drop off, surely you're using the wrong connectors

And if you're happy to tack on a capacitor, why not tack on a couple of leads?

With a bench power supply, the metering gives you the reassurance of knowing what voltage is applied. With a capacitor, the charge could disappear and you wouldn't know.

Ultimately, any method has pros and cons - it's certainly good to have a range of options on the table...

As far as I can tell from the manual, the battery is only there to store user stuff? So at least it's not calibration data, making it a "low risk" change. The last battery I changed was the one in my HP 3568A - that was somewhat more critical, as re-calibrating that would have been a pain, to say the least. Luckily all went well. Amusingly, the old battery is still perfect, many years on - not bad for a 30 year old battery! Next time it'll be the one in my HP 3314A, but that only stores user data, so I'll only be changing it to mitigate the risk of leakage. Not urgent...