Rf switch using pin diodes

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...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson
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Before getting into inductor and capacitor values, I think you need to get the bias situation better under control. Lets start with the data sheet:

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The diodes are rated for 50 volts reverse, but you appear to be letting them sit at ~zero volts when they are off. Zero volt bias sets them to a bit more than 1 pF capacitance (figure 4), while 20 volts reverse bias lowers this to about .6 pF. Turning them off with zero bias also risks them turning on from signal voltage.

As to forward bias, they have a resistance of only a couple ohms with as little as 1 mA of bias current (figure 3). This falls to a fraction of an ohm at 10 mA, so I can't see any good reason to be biasing them around 20 mA. Actually, since they drop almost a volt at

20 mA and your processor output will certainly sag severely while sourcing 20 mA, your 250 ohm bias resistors will deliver much less than 20 mA, so maybe your circuit will produce a sub 10 mA bias, as it is.

As to inductor and capacitor values, you want no resonances anywhere near the operating frequencies so that the capacitors have impedances much lower than the source impedance and the inductors have impedances much higher than the source impedance, so they look like current sources.

I am also concerned that your ground pour will not look like a single voltage at your operating frequency. I can't be sure of this, since I can't follow the layer structure of your layout from this image. I think that every corner of the top surface pour should be stitched to the underlaying ground layer with at least one via, so there are no risks of resonant peninsulas in the top pour.

Reply to
John Popelish

More comments on your layout (even though you didn't ask):

Extend a thin trace grounded trace (that has a ground via as close as possible) between the two input nodes of the diodes as a shield between the two inputs.

Since you are not trying to pass the signal frequencies through the pair of bias networks on the inputs (or the output inductor), do not use merged pads for these connections, but the thinnest possible traces to supply the bias current to the diodes. This gets the LC network a bit further out of the way, and makes the thin connecting trace part of the isolating inductors.

Put the two bias lines between the LC filters, instead of outside them, to reduce coupling between them and the inputs.

I think you have too much ground pour under the connectors. The big center pin will look like a big lumped capacitor as it is, without crowding the ground pour around it. A square hole that is an extension of the 4 surrounding thermals should be close enough.

You may take a look at shrinking the pads for the diode pack to the absolute minimum consistent with good soldering to reduce the stray capacitance between the two inputs.

Reply to
John Popelish

I think you have made some improvements, (especially around the diodes) but you have not moved the bias components out from the signal lines on thin traces, like I suggested. You still have lots of big pads right against your signal path that add lumped capacitance to those lines that could be isolated by high trace impedances.

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At least you might use a trace width that is the same as the width of the coupling cap pads, ifthat gets you closer to 50 ohms with your ground layer spacing. This eliminates some discontinuities.

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Yikes. Are you sure the diodes are rated for 140 mA RF current? At that amplitude, you might need quite a bit more than -5 volts off bias since you will have unterminated signals arriving at the off diode with some reflection generated voltage multiplication, unless you add more switches to turn on termination resistors when the through path is interrupted. I am not sure the on diode will stay effectively on throughout the cycle with ~10 mA of bias current, either. You will probably get some rectification.

Reply to
jpopelish

Hi,

I made a circuit for an RF switch using pin diodes, and was wondering about what values to use for the capacitors and coils in the circuit. Here is a picture of the circuit:

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D1 is a pin diode, like HSMP-3824 from minikits.com.au the ratings for it are:

Ct = 1pF, Rfs = 0.6ohm, usable to 3GHz

The two bias signals are fed from an AVR microcontroller.

What values would be best to use for the capacitors and coils in this circuit? I am interested in using it for 2.4GHz and 433MHz.

Here is a picture of the layout:

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cheers, Jamie

Reply to
Jamie Morken

Hello Jamie,

As John said the ground trace should ideally go through the diode land pattern so that each pad is somewhat enclosed. T off at the number 3 pin and connect left and right to the ground plane. Via it through to the bottom ground plane (seriously hoping there is one...) at several places.

Make sure the traces have the desired characteristic impedance of 50ohms or whatever you need them to be.

And yes, you typically need negative bias.

Regards, Joerg

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Reply to
Joerg

Thanks for the tips guys, I made most of the changes. (oh and yes there is a ground plane on the bottom!)

new layout: "

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I also put the eagle cad schematic and board file there too: "

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switch 1.sch" "
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switch 1.brd"

50ohms is what I would like. Can I get this with the layout I have and just changing the component values, or do I need to add 51ohm resistors and use 0.1" trace widths? The trace lengths are so short that maybe it doesn't matter? :)

So I should drive the BIAS1 and BIAS2 inputs from -5v (off) to +5v (on)? I guess at 50ohms, and 1watt RF power, that will be about 140mA and

7Volts peak, so I should have at least 7V negative bias for 1watt?

How do these pin diode circuit compare to using an SPDT RF relay instead?

cheers, Jamie

Reply to
Jamie Morken

I can answer your last question. A PIN Diode switch will switch much faster than an ordinary mechanical relay type switch, which will switch in the neighborhood of 10-20 milliseconds or so. The mechanical switch will normally have lower insertion loss and higher isolation at your frequencies.

However, you won't have had the fun and learning experience that building your own gives you.

Hope this helps. Good luck with your project.

Ken WB6QWF

Reply to
kenspam

Hello Jamie,

Whew. Good. Sorry for having you accused of that.

Nice. Now if you could round the traces and structures so there are no sharp corners ... but then again I guess this isn't going to be run at several GHz.

I could read them (not on this PC though) but most people can't. I'd post schematics in some popular graphics file format that web browsers understand.

The trace width, thickness of the FR4 (or whatever material you are using) from trace to ground plane below and it's dielectric coefficient is what matters most for trace impedance. Here is a "cheat sheet" that allows rapid calculation:

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I always bias them off to within a healthy margin below the limits, to make sure there won't be any unwanted partial rectification.

SPDT = Moped PIN diode switch = Ferrari Testarossa

Regards, Joerg

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Reply to
Joerg

Hello Jamie,

For me the most important advantage of PIN diodes is that you don't have to run a solvent soaked paper through the contacts every five years (more often if you run a wood stove). And they also don't need any regular spritzes from a can either.

Regards, Joerg

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Reply to
Joerg

Hello John,

Yes, that would make a lot of sense. The bias traces don't carry any power to speak of. 10mA doesn't require a whole lot of trace.

The trace width should be calculated to have a characteristic impedance in itself of 50 ohms or whatever other value is needed. It should not be based on the width of an attached SMT part. For really, really high frequencies there are other tricks, such as making small holes in the ground plane underneath the SMT solder land to nudge its impedance up a bit.

I didn't check this diode but usually they can. The end of Agilent app note "Application of PIN Diodes" explains power handling capabilities. What matters most is how low you steer the resistance. Most diodes be safely steered well below an ohm and then they can handle a surprising amount of power. I have run pulses in the tens of watts through SOT23.

Since the carrier lifetime of the chosen PIN diode has to be much longer than a period of the signal there won't be much rectification either.

Regards, Joerg

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Reply to
Joerg

I am just going to run it at 100mA RF power to start, and go from there, or make a new design if required. RF is slow learning for me! :)

Here is the new layout with the updates: "

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cheers, Jamie

Reply to
Jamie Morken

That's what I'm talking about! I can't say if your signal path is 50 ohms, but I think this layout concept is quite a bit better than either of the earlier versions.

Reply to
John Popelish

Hello Jamie,

The RF traces look a bit wide. Are they 50 ohms?

Regards, Joerg

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Reply to
Joerg

It has been a while since I did some PIN diode design, but here is a little of what I remember.

You need the negative bias in order to get the diode to turn off fast. Pin diode switches are much faster than an electromechanical relay. You also need a negative bias voltage if the power level that you are passing throught the switch is large (say over 0 to 10 dBm). If you do not use negative bias the diode will take a long time to turn off. If you look at the diode on a scope you will see that the voltage on the diode remains "on" even after the "on current" has been removed. The negative voltage will flush this kind of wierd charge off the PIN diode and make it turn off. You need to have a low impedance from the diode to the negative supply. The lower the impedance and the more negative the voltage, the more quicker the diode will turn off.

Many PIN diode switches I saw years ago had a + 5 V supply and a negative 15 volt supply. The negative supply is used to make the switch fast.

If you do high power through the switch , then more things must be considered.

Big Duck

Reply to
Big Duck

Joerg -

According to my Microwave book, the traces need to be .049 wide for epoxy-fiberglass of 1/32 inch thickness to be 50 ohms (Er = 4.8). For the same material in 1/16 inch thickness the traces need to be .105 inches wide to be 50 ohms.

The traces to X1, X2, X3 are .02 wide. This gives about 75 ohms for 1/32 inch thick material and about 100 ohms for 1/16 inch thick material (glass epoxy with Er = 4.8).

The above is for 1 ounce double clad copper. If the material is Teflon (registered trademark), let me know and I can read those properties from the table.

The traces cant be .105 wide because they won't fit between the pads of X1, X2, X3. However, using 1/32 material, they can be made .049 wide and they will fit.

The pour area effect on the top side is not required and, in fact, is somewhat detrimental to the data I have given. The bottom ground plane does not extend to the connector hot terminal (obviously) so the trace cannot be

50 ohms exactly all the way to the terminal. The distance from D1 to the connector is .22 inches. This is about .019 wavelengths at 1 gHz. All these factors, I think, make this more of a learning experience than a practical matter. However, I agree that everything within reason should be done to comply with the usual guidelines. After all, if you don't pay attention to the details to begin with, you develop bad habits that stay with you for a lifetime. I'm an example.

Good luck Jamie.

John

Reply to
John - KD5YI

Hello John,

That's why I mentioned it. Jamie's traces looked like 0.1" (at first, but I stand corrected now) and on thin stuff that would have been too wide. I just wanted to point it out before he ships it off to fab.

It seems you are right, when comparing it to the SOT23 they might fit.

After my Motorola book disintegrated I use an old Fairchild ECL data book which has the calcs in there. That's about to fall apart as well so I use the Missouri-Rolla web calculator a lot. Feels like cheating though.

There should be a calc in your book with pour. It's standard procedure in microwave and helps a little with crosstalk in the off state. Yes, the wavelength will be large compared to the trace length. But then again why not avoid even the slightest mismatch if it costs nothing?

Regards, Joerg

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Reply to
Joerg

The traces to the sma connectors are 0.04" wide in the current layout:

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I just use the standard FR-4 double sided board from Olimex. It has a thickness of 1.5mm (0,062") and a copper pour of 35um (1 oz).

The pour on the top makes the calculation more difficult for the trace impedance though. :) What are the important effects if the impedance is off?

Are the traces narrower for the same impedance with a pour on the top?

cheers, Jamie

Reply to
Jamie Morken

Hello Jamie,

1.5mm is rather thick. That would come to around 80 ohms for the traces unless I am wrong. A bit less with the fill.

Under 1GHz, not much. I am just being a nitpicker here so don't worry about that too much. What you'd see would be a minute increase in SWR if the trace is off. At several GHz that's a whole other story.

Yes, but only marginally. The capacitance a trace of same width sees per unit length would increase a wee bit and that decreases the line impedance.

Regards, Joerg

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Reply to
Joerg

You can usually find Duroid on eBay -- also some surplus sites have it. That will take you to 10 GHz or so.

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Yes, it's more expensive than FR4, but it works.

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Reply to
artie

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