Small-scale BGA soldering

Empiracal formula. No derivation.

Our production PCBs are also ENIG. The nickel is typically something like 200 uinches thick, and the gold is just a few uin. What I don't get is that you can't solder nickel, and the few uin of gold dissolves in solder instantly, but the combo is superbly solderable. Maybe some chemist can explain that.

Most high frequency loss is on the underside of a trace. That's where the current concentrates (in the higher Er of the epoxy-glass), and the underside of the trace usually has the horrible black-oxide treatment to aid adhesion. So I'm thinking the ENIG doesn't matter. It just looks and solders great.

Somebody proposed using nickel plating on inner-layer power planes to reduce resonances, or something. That Black Magic HoJo lunatic maybe.

Yes, beautiful protos look awful a few weeks later.

It would wash off when I clean off the flux.

I *like* my gold boards. $100 a square foot adds a few dollars to the typical prototype, which is down in the noise.

Manhattan is good for power stuff, not so good for tiny parts.

You can use surface-mount adapters, but that means wires, added inductance and length, which isn't good for really fast stuff.

Nowadays, we have a folder on a server, J:\Protos, where we document breadboards and experiments and unofficial PC boards. Each gets logged and gets its own subfolder, like Z366_Jitter_Test. There will be a whiteboard pic or other schematic image, photos of the proto, measurements, scope screen shots, notes, anything useful.

We include test fixtures and interesting curiosities in the Protos folder.

A 350 MHz oscilloscope has a risetime of about 1 ns. If you build a gaussian lowpass filter with -3 dB bandwidth B, the output step will have a 10:90 risetime T, and the B*T product will be about 0.35.

Scopes have (or used to have!) gaussian step response.

Here is the derivation for a first order lowpass:

-------------------------------------------------------------------------- This is how it works: the circuit is assumed or approximated to behave like a simple RC integrator.

The time constant will then be: t=RC The bandwidth is then BW=1/(2*pi*RC)=1/(2*pi*t)

On the other hand, the rise time for this simple RC circuit is calculated starting from the equation of the voltage across the cap when a step impulse is applied at the input:

u(t)=U*(1-exp(-t/T))

For u(t)=0.1U you get 0.1U=U*(1-exp(-t10/T)), where t10 is the time when the voltage reaches 10% of the final value.

After simplification

Applying the natural log to the equation you get:

ln0.9=-t10/T or t10=-T*ln0.9

Similarly, for u(t)=0.9U, we have t90=-T*ln0.1

The rise time is then:

tr=t90-t10=T*(ln0.9-ln0.1)T2.2*? or T=tr/2.2

Substitute this value of T into the expression for bandwidth:

BW=1/(2*T*tr/(2.2))=2.2/(2*T*tr)~0.35/tr

or

BW*tr~0.35

Comments

This is only to repeat the restriction that this formula applies only for a lowpass system of first order.

One additional comment:

The mentioned formula can be used with good accuracy also for higher order systems when there is a dominant pole which is far below the other poles.

This, for example, is true for universal-compensated opamps.

rise time bandwidth product

Yes, generallly for two pole system this factor can be in a range around 0.35 (for example 0.3-0.45), which depends on a Q factor.

--------------------------------------------------------------------------

How this is extended to Gaussian or critically damped circuits is not clear. And it is not clear how it applies to the risetime of a semiconductor switching element.

It's the way our universe works.

Yes, that is a complete mystery.

That applies to signal traces. It doesn't apply to a solid copper ground plane.

They get worse as time goes on.

Are you sure? Krylon is pretty tough. Also, it depends on the solvent you use. Isoprop probably wouldn't affect the Krylon. I'll know more soonish.

That's all the justification you need.

That is just Manhattan with rectangular pads. You can make them any shape you want.

Those are horrible. Why use them for high frequency stuff?

Take an EP51. Bend the leads up. Glue it to the copperclad. Connect power and ground with appropriate bypasses (PECL or straight ECL).

Take 4 mil mylar copper clad on one side. Cut into strips that give 50 ohms. Glue the strips to the copper ground plane to connect high frequency signals. Keep them short.

You now have fast circuits in a 50 ohm environment over a clean ground plane. It won't match Rogers or some of the newer GHz materials, but it is quick and good enough for prototyping.

I copied the derivation for a first order lowpass in my reply to Jan.

It is not clear how this is extended to the semiconductor switching time. How do you measure the bandwidth of a saturated transistor?

The inner planes are black copper on both sides. But HF loss is OK in planes, maybe even a good thing to reduce resonances a bit. But probably doesn't matter.

I dunk my boards in our degreaser, first in boiling solvent and then sprayed with clean stuff. That would probably eat Krylon.

To use fine-pitch surface-mount parts on breadboards. They're not horrible.

Here's an EL07 and some gaasfets and stuff.

That was before I had the gold FR4.

in

chould try making gerbers for a bare board and sending it to

$98 for 50pcs 100x100mm two layer boards with ENIG

More info:

Tektronix thinks we should use 0.45 instead of 0.35:

Anritsu has a nice derivation that won't copy in plain ASCII:

Sobering does it in Laplace:

Tektronix compares Gaussian response with different scopes:

The way I've always done it is to note that for a 1-pole LP reponse,

f_c = 1/(2 pi RC)

And the rise time is

t_r/RC = ln(0.9) - ln(0.1) ~= 2.2.

Thus t_R = RC (2.2/(2 pi)) ~= 0.37/f_c

Cheers

Phil Hobbs

Simple, easy to understand, and effective. All good things.

How does that relate to the switchng time of a transistor?

We do some protos as pc boards, usually 4 layers, often several different circuits on one board. But I can Dremel a board in a few minutes.

n:

kin

arkin

sure, I was thinking of it as a cheap way get the gold plated FR4 cheap

Modern digital scopes aren't usually gaussian; most ring on their step response, some absurdly so. I don't think it's so much an aliasing issue as it is specmanship cheating on bandwidth.

I usually just Spice things like that and measure the risetime and frequency response.

That's ok for the input filters for the ADC's in scopes. Although Tektronix says it should be 0.45

But how do you relate the switching time of a transistor to the bandwidth?

Say a switching transistor has 120 ps risetime. How can you say this is 3 GHz bandwidth?

Oh, OK. 50 pieces is a lot of area. I always shear my boards up into smaller bits than 100x100.

I don't. You'd have to invent a definition before anybody could try to answer that question.

If the transistor saturates (ie, the circuit is significantly nonlinear) that's sort of meaningless. If the transistor is used as a linear amplifier, the numbers are about right.

That's the problem. You are taking a risetime of 120 ps and calculating a bandwidth of 0.35/120e-12 = 2,916,666,666.67 = 3 GHz.

That's what you did at the start of this thread.

I don't mind. I do the same thing. But where is the justification?