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    pin diode power

    I designed a RF ciucuit in which there were some PIN diodes to decide a operating RF path.The RF signal will pass by a PIN diode when it's path was choiced.It worked well under a -10dbm signal power generator.But when I added up the RF signal power to 20Watts,the performance of the RF circuit collapsed.The PIN diodes I used is MA4P7104 of M/A company which works with a Rs less than 0.5ohm and the power dissipation under a constant 25C environment is 3Watts.I don't know why the circuit can't work and I don't know how to calculate the power admit of the circuit.I wonder if someone would help me.TKS!!!

    •   Alt10th January 2006, 07:15

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    power diode range calculation

    Hi, what is the operating frequency? Are you sure that with a so hogh power there are not some radianting effects? Is the layout carefully designed?

    Bye



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    calculate power for diodes

    MA4P7104 is not working properly for frequencies higher than 500MHz.



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    calculate diode power

    Did you have it biased up properly? 100 mA per diode when forward biased and 100 VDC reverse bias?

    Also, you might have blown out the diodes when switching from one state to the other. Most people turn off the RF high power when switching, and turn the RF back on well after they are sure the diodes have switched. Notice that the thermal resistance is very high for a "high power" diode, meaning they are NOT designed for actually dissipating a lot of heat. Silicon diodes really should not be running over 150 degrees C for even short periods, and 120 degrees C in the steady state. That means that the diodes junction to lead temperature, and lead to ground plane temperature, can not add up to very much!


    1 members found this post helpful.

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    power dissipation pin diode

    Quote Originally Posted by mwmmboy
    Hi, what is the operating frequency? Are you sure that with a so hogh power there are not some radianting effects? Is the layout carefully designed?

    Bye


    I made a metal cover under the PCB board.The cover was connected with the Ground plane by about 12 pins on different directions.Is that enough ???

    Added after 10 minutes:

    Quote Originally Posted by vfone
    MA4P7104 is not working properly for frequencies higher than 500MHz.


    On the PDF document of the PIN diodes,It doesn't say anything about the highest frequency it can work at.But I see the diagrams of curve listed on the document all give us the performance up to 1GHz.My operating frequency is about 900MHz.
    It can work well at the frequency on a low power condition but not a high power condition.

    Added after 24 minutes:

    Quote Originally Posted by biff44
    Did you have it biased up properly? 100 mA per diode when forward biased and 100 VDC reverse bias?

    Also, you might have blown out the diodes when switching from one state to the other. Most people turn off the RF high power when switching, and turn the RF back on well after they are sure the diodes have switched. Notice that the thermal resistance is very high for a "high power" diode, meaning they are NOT designed for actually dissipating a lot of heat. Silicon diodes really should not be running over 150 degrees C for even short periods, and 120 degrees C in the steady state. That means that the diodes junction to lead temperature, and lead to ground plane temperature, can not add up to very much!


    Thank you very much.My operating condition is here:
    The forword bias current is about 80mA and 3V.
    The reverse voltage is -200V.
    I didn't do anything when I add up the RF signal power.No switching path,no turning off or turning on any path.I just gradually add the power from -10dbm after it worked well.It collapsed soon.
    How can I know the thermal resistance of a pin diode at a high temperature?Is there any parameter to tell the performance?
    Now,I list the related parameter down:


    Power Dissipation and Thermal Resistance Ratings

    Package Style Condition MA4P7100
    PD JC
    F (SMQ Surface Mount) 25C Contacts 3 W 50C/W


    Is this meaning the junction temperature will rise 50C every watts ??If this's ture,I believe I will have to search for another pin diodes.Would you like to give me some suggestions? TKS!



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    ma4p7104

    Well, its been a while. let me see what I can remember!

    I am assuming you have a two throw switch here (SPDT, one input and two selectable outputs).

    So lets say one of the diodes is forward biased at 80 mA. We will call this D1.

    The other diode is reverse biases at -200 volts. Lets call this D2.

    I also assume you have a 50 ohm system (50 ohm load and 50 ohm transmission lines, etc).

    I am also assuming you have the two diodes mounted in a Y-series configuration, that is one diode per output arm.

    The diode is the surface mount package, with a thermal resistance junction to case of 50 deg C/watt. The diode can dissipate 3 watts of heat maximum, according to the data sheet.

    To 20 watts is input to the SPDT switch. Power travelling towards the diodes on the transmission line is going to be P = V*I = V*(V/Z0) So, V = √(P*Zo) = √(20 watts * 50 ohms) = √1000=31.6 volts RMS. IF there is no standing wave, then the voltage travelling wave down the transmission line will have a peak voltage of +/-44.7 volts.

    So far so good. You have 200 volts DC across the diode, so the RF superimposed on the one diode D2 that is reverse biased will still stay reversed biased and there will be no rectification of current in that diode. How hot is that one series reverse biased diode? Well, it has a worst case voltage of 44.7 volts RMS across it (assuming no big standing waves due to load mismatches), so the power dissipated will simply be the PdisOC= Vrms / Rparallel = 44.7/100K ohms=0.02 watts. According to the data sheet, your reversed biased diode should be ok! Now, I am pretty leary of the "100Kohm" parallel resistance mentioned, but lets say it was as bad as 10 times worse, or 10Kohms, you still only have 0.2 watts dissipated! Not too much. The diode junction temperature is the Pdissipated*Θtotal. Θtotal is the devices thermal resistance (50 degC/W) and the printed wiring boards thermal resistance (lets assume another 60 degC/W. So this reverse biased diode junction temperature rise from ambient might be around 0.2W * (50+60 degC/W) = 22 deg C. So if the bottom of your printed wiring board was at say 40 deg C (warm to the touch), the junction is operating at only 62 deg C, which is acceptable.

    Now for the forward biased diode D1. The same 20 watts is incident on it, but is passing through the diode in series. So P=I*V=I*(I*Zo)=IZo. So, I=√(P/Zo).
    Irms=√(20 watts/50 ohms)= 0.4 amps rms. How hot is this diode getting? Pdissipated is going to be Pdiss=Irms*Rseries = (0.4)*0.5 ohm=0.08 watts.

    Since 0.5 ohms does seem realistic for such a diode, the junction temp again is going to be 40 + 110*0.08= 49 degrees C. Not a problem also.

    So, for the assumptions that I made, you should be all set!

    I would look into other stuff! First, is the diode any good? You should be able to, at low power, vary the DC current through the forward biased diode and see differing insertion losses. Do the measured Rseries vs DC current show that the diode has a Rseries of around 0.5 ohm?

    I would look for any sings of arcing anywhere in the circuit. Slight discoloration, like little dark lines between the component or trace and ground can show a carbon scoring track.

    Are the capacitors used in the circuit up to snuff? they have to be at least rated to 300 volts! If not, they could be breaking down and starving the PIN diodes of the proper bias.

    Are the loads good? If the load shorts out, arcs, or even just has a poor VSWR, then all bets are off.

    Then there is always the sniff test! If you can smell smoke, or feel a really hot component under 20 watts (touch briefly at first!), that could show you where things are going bad.

    Good luck.



    •   Alt11th January 2006, 03:53

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    how to ensure the diode is in good condition

    Quote Originally Posted by biff44
    Well, its been a while. let me see what I can remember!

    I am assuming you have a two throw switch here (SPDT, one input and two selectable outputs).

    So lets say one of the diodes is forward biased at 80 mA. We will call this D1.

    .........

    Good luck.


    First of all,Thank you very much!
    Almost all of your assumptions are ture.They are very reasonable.
    I have checked my PCB board carefully.The minimum gap of two components is about 0.8mm!I think they will not produce arc with a constant 40V signal.
    Now,I'm going to do two experiments.
    1:Design an experimental PCB board. It only comprises a microstrip of 50ohm characteristic impedance and a PIN diode with the bias circuit. It's very easy and excludes any other components.I will then add the RF power gradually.By this experiment,I hope I can draw a conclusion about whether the diode can work on a high RF power condition.
    2:I will put my current product into the heating box and let it work first at an ordinary circumstance temperature with a constant low RF power input..Then I add up the temperature gradually to verify whether it was the high temperature that destroyed the PIN diodes' performance.
    I would like to tell you the results of the experiments and post them here.
    Thanks!



    •   Alt11th January 2006, 14:01

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  8. #8
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    diode power dissipation in reverse bias

    One other thing to check is self rectification. The diode is big enough (thick I region) so that it should not be doing, but it would be worthwhile to check that that is true. When the high power is applied, does the 80 mA forward bias change, as read on a digital current meter?



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    calculating power dissipated by a diode

    While ago Ive tried MA4P7104 at 940MHz with 30W and was not working properly.

    Probably at 20W will be the same thing because is no big difference between 44.7dBm and 43dBm.

    In the datasheet all the main parameters are measured at 100MHz, and showed on graphs at max 500MHz.



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    dioda ma 4p 110

    Interesting point. I did not see those graphs. The data sheet does have some graphs that show parallel resistance diving quickly up to 600 MHz. If those sheets can be extrapolated, then the parallel resistance is more like 2K ohm at 900 MHz (not the 10K I assumed), so power dissipation in reverse biase may be more like 44.7/2 Kohm = 1 watt. Add some VSWR, etc, and maybe the diode is just overheating, breaking down, and then failing.

    And the data sheet shows no series resistance data other than 100 MHz, so what if it changes a lot at 900 MHz also?



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    power diode calculations

    Quote Originally Posted by biff44
    One other thing to check is self rectification. The diode is big enough (thick I region) so that it should not be doing, but it would be worthwhile to check that that is true. When the high power is applied, does the 80 mA forward bias change, as read on a digital current meter?

    I got the data of 80mA by measuring the voltage between the bias resistance's ends.I can get the performance of forward bias current and series resistance from the chart of "SERIES RESISTANCE AT 100 MHz vs FORWARD CURRENT
    (MA4P7000, MA4P7100 SERIES)" on it's official data sheet! We can also see M/A claim that the PIN diodes have pass "MIL-STD-750 environmental tests".Does this means the circumstance temperature in this case will not change the performance greatly? Whether or no,thank you,and I will test the series resistance again.

    By the way,there are some importants points I forgot to proclaim.The system insertion loss with PIN diodes is about 2.5dB with the low input power of -10dbm.If I replace the PIN diodes by some little metal patches.The insertion loss is about 1.2db.Does it means if the input power is 20W,the PIN diodes will have absorted about 4Watts to 5Watts?In fact,I have 16 branches connected with the main path with PIN diodes.Some branches will generally work cooperatively.Of course,the power via them is different.So,there's a very likely chance that some PIN diodes will absorb abou 2W to 3W power.Then they will get hother and hotter and performance get worse.Is this reasonable???

    Added after 1 hours 1 minutes:

    Quote Originally Posted by biff44
    Interesting point. I did not see those graphs. The data sheet does have some graphs that show parallel resistance diving quickly up to 600 MHz. If those sheets can be extrapolated, then the parallel resistance is more like 2K ohm at 900 MHz (not the 10K I assumed), so power dissipation in reverse biase may be more like 44.7/2 Kohm = 1 watt. Add some VSWR, etc, and maybe the diode is just overheating, breaking down, and then failing.

    And the data sheet shows no series resistance data other than 100 MHz, so what if it changes a lot at 900 MHz also?


    I believe the data sheet you saw is not the same with me.The PIN diodes I used is MA4P7104.It belong to MA4P7100 series.There are some graphs that mean it can work at least 1GHz.You can download the data sheet from this site:http://www.downeastmicrowave.com/PDF/ma4p.PDF

    On the upper post,I listed some important points about insertion.The insertion loss is about 2.5dB with a low -10dbm input power and the VSWR is about 1.2 .Maybe the bad insertion loss is the chief suspect.



  12. #12
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    rf power capacity

    2.5 dB of loss for a high power switch is too high. 16 branches? Resistor bias? You are saying things that do not make much sense. post a schematic.



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    power dissipated by a diode calculation

    Quote Originally Posted by biff44
    2.5 dB of loss for a high power switch is too high. 16 branches? Resistor bias? You are saying things that do not make much sense. post a schematic.


    I'm sorry。I don't know how to post a picture on this forum.
    It's a tunable filter with a fixed cavity in which there is a pole.Our branches are all connected to the end of the pole.Each branched offer a selected or unselected load capacitance.They make a combline filter.With a capacitance branch selected,the load capacitances changed.So the resonant frequency changed.
    I can describe the circuit of each path by electrical net.
    net 0 denote the end of the pole.
    net 1 denote the input drive signal of bias circuit.
    2,3,4,5,..... denote some assistant nets.
    between 0 and 2,it's a High Q capcitance.M/A c100 series.
    between 2 and 3,it's an high resonant frequency RF inductance.Coilcraft CS series.
    between 3 and 1,it's a resistance of 40 ohm.
    between 2 and ground,it's the PIN diode.M/A ma47104 series.
    between 3 and ground,it's a big capcitance.M/A c100 series.
    If the branch is selected,the input drive signal of bias is about 4V.If not selected,the input voltage is -200V.
    2.5dB insertion loss is the tunable filter's insertion loss.We have replaced the PIN diodes with metal patch with no any input bias.The filter's insertion loss is 1.2dB.So I draw a conclusion,the PIN diodes bring us 1.3dB.If the input RF signal power is 20W,1.3dB is about 4W to 5W.
    I'm sorry for my carelessness.



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    power diode calculator

    Paste your schemtic into something like MS word, and upload it while posting.

    Putting Pin diodes into a filter negates everything posted on this thread so far. Please try not to waste people's time next time.



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    calculate diode rated power

    Quote Originally Posted by biff44
    Paste your schemtic into something like MS word, and upload it while posting.

    Putting Pin diodes into a filter negates everything posted on this thread so far. Please try not to waste people's time next time.


    I'm sorry for wasting your time.
    I will be grateful if you help me.But if you are busy,I really don't want to occupy your time.



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