Missing SMD component on Apollo 1240

[8bitbubsy]

After messing with my totally damaged Apollo 1240, I managed to do some repairs and bridges. I think it's soon up working again, but there's one important SMD component missing that broke off earlier.


Here's a picture from google that I use as an illustration.

The SMD component with the "?" mark is the one that broke off. I don't have that part anymore, so I can't solder it back on.

The red marked traces are the traces that are connected to the SMD component. The left end of the trace goes under the SMD chip.
The right end of the trace goes to the + pad of the SMD eletrolytic capacitor.
The EVDD also leads to the + pad of the cap. EVSS is ground. So this component is between + and - of the power to the CPU or something?

Two of the CPU pins goes through it, and I checked the 68060 CPU pinout and manage to label them. EVDD and EVSS - what are these and what is the missing component? My guess was a digital ferrite bead, but those are black, aren't they?

Thanks in advance.

[Damion]

Looks like an SMD capacitor.

[Zetr0]

@8bitbubsy

This is a SMD 1206 Monolithic Capacitor that is used in a decoupling manner from (e)VDD to (e)VSS (+ive to -neg)

Now - the value of this is going to be a ed-you-ma-kated-guess - but I would say 0.1uf is what you are after

however we can calculate the value of the Decoupling Capacitor that should be used. It depends on your load the IC/Component has to drive.

The load device (CPU) A [the driver] is device B (ground) [the receiver]. If device A has to drive one input at 3.3 volts then the load depends on the load of both inputs, taking into consideration the rise time of the signal.


Ready for the math? (just an example not for the 060)

We need to calculate the existing capacitence - to do this you need the input capacitance for the given device (this parameter is Ci in the device data sheet).

If the receiver [B] has a load of 12pF [Ci] and the output driver [A] has a rise time of 1nS then the current required is: I = dV / dt or I = 12pF*(3.3v)/1nS.

The current required = 39.6mA. If we keep the voltage droop to 3.0 volts, or a reduction of 300mV.

The capacitor then equals: C = I * dt / dV. C = 39.6mA * 1nS/300mV = 22pF

Choose a capacitor whose resonant frequency is at least as high as the corresponding edge rates of the switching signals, where frequency response @ 1 / (3.5 × Trise, or fall).

Thats the text-book approach - however in basically every data sheet a 0.1uF is indicated as a general rule of thumb.

[alexh]

Anyone with a 1260 & a decent MultiMeter (not all have capacitance measuring) could measure the value for you?

[pandy71]

Originally Posted by Zetr0

@8bitbubsy

This is a SMD 1206 Monolithic Capacitor that is used in a decoupling manner from (e)VDD to (e)VSS (+ive to -neg)

Yes, but probably not 1206 - (resistors are usually 1206 and 805), those caps can be something from 35xx series - a bit different ratio between W and H - value something betwen 0.1 - 1uF - if color is lighter then probably it is 0.1uF - darker - brown or dark red are usually over 1uF

@8bitbubsy
Rip similar looking caps from broken equipment - should fit - search for the caps around power supplies or close to big IC

[Shadowfire]

That's definitely a bypass capacitor for the '040. I checked the '040 user guides and they give no suggestions for bypass caps.

[JACK98]

Hi.i have the same card.

[chiark]

Originally Posted by alexh

Anyone with a 1260 & a decent MultiMeter (not all have capacitance measuring) could measure the value for you?

Problem is that you need to take the capacitor out of circuit to reliably measure its capacitance...

[Dimlow]

Its .1uF, just removed one for ya!

[Zetr0]

DIMSY!!!!!!

good to see you about!!!!


Imma up having a bit of late night SCSI Hackery.... what about you?

[Dimlow]

Well the insomnia has kicked in again!

[Zetr0]

yeah, its always the way my friend..... alas mine is because I had a brilliant idea.... just stifled by poor execution lol!