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  #1  
Old 06-30-2019, 05:17 PM
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150W RF PA 0.4MHz - 3MHz

Good day,
Here i attached the schematic i had drawn in ''Paint''.
I asked on stackexchange about this design but they placed my topic on hold. They can't explain nicely to someone new to this, they have to make use of bureaucracy....Really hard thing to be a gentlemen these days and it gets harder and harder with each year. Anyway, i was asking them for advice concerning the value of those capacitors, the polarized and non polarized ones. They could have thought me how to compute the values of those components...
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Last edited by Buddhafollower; 07-01-2019 at 07:46 PM.
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  #2  
Old 06-30-2019, 06:12 PM
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What is the question ? All elements except one resistor seems labeled on schematic
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Old 06-30-2019, 07:31 PM
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I don't know if all values are correct and i don't have a way to know until i build my own amplifier. I will build it. I just want to know from more experienced members if everything is in order with the values of the capacitors. And since there is a resistor in series with the IC and the quiescent current per transistor is set at 25mA, then the value of the resistor should be 200ohm because 65V-35V= 30V and 30V/0.15A=200 Ohm. Also, 65V-48V= 17V so we decrease the voltage with 17V, then 35V-17V=18Vdc so i will still have 18Vdc at the input of the IC even if i decrease the voltage to 48V. But then i will have to retune the amplifier.
My problem are with the capacitors though. Do they have correct values for this range of frequencies? Excuse my ignorance/stupidity....
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Old 06-30-2019, 08:32 PM
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I will go calculate the resistor values as well, but that will take me a bit.

now I can tell you that the unlabeled capacitors are for bypass reasons.
likely 0.1 or 0.01 micro farad
I would look to the data sheet for the LM7805 for the recommended values and then just use that value for all of them.

will post again when I have look all that up and have numbers
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Old 06-30-2019, 09:43 PM
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pcb

I have here the PCB of the same RF PA but with 8 transistors for a power of 200W. Haven't tested it, don't ask.
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Last edited by Buddhafollower; 06-30-2019 at 10:14 PM.
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Old 06-30-2019, 09:50 PM
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when calculating things,
look to where your current is going,
the input (gate) of the mosfets is going to be about zero current,
the idle current of the transistors goes through the source and drain, so it has nothing to do with your value of "R" that is not labeled.

to calculate R, you need to figure out how much the 7905 and your 4.7K (at 5V) is taking.
your 4.7K is about 0.001A
and the 7805 is going to take 0.0043A to 0.008A (according to the data sheet).
max input of 7805 is 35V and you should not run it past 25V
min input for the 7805 is about 7V
your input voltage of the circuit is 48V to 65V
so if we assume worst case (one direction), you will have 48V in with 0.009A needed. we will run the 7805 at 7V (trying to give ourselves room for when the power supply voltage goes up)
so, 48 - 7 = 41V on our resistor.
41V/0.009A=4555ohms
so, I will use 4.55K for the moment.
so with that part in the other worst case you have 65V with 5.3mA,
this gives 24.115V across the 4.55K resistor, 65 minus 24 gives 41V across the 7805, and that would fry it, and likely would then take out all your mosfets a moment later.
so, either you figure out your particular 7805 more careful, or narrow down your power supply voltage range a bit.
that is if I did the math correct, and I did it 3 times just to be sure, but still could have a logic error in there on my part.

and this means there is a reason they did not give the "R" a value
the 7805 is not really an ideal part here (but thinking about it, it is not that bad either, you just need to pin down your power supply voltage better than the circuit is labeled with.

and 0.1 micro farad looks to be good for all your bypass capacitors.
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Old 06-30-2019, 09:55 PM
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I think I might have made a logic error.
going to re think this a bit.
I am not clear headed right now, not sure why,
but will take a nap and try again in a few min.
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Old 06-30-2019, 10:38 PM
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They say that there is no DC current drop between the gate and the source of a MOSFET. There are three 100mOhm resistors which i considered them as paralelled in DC without signal at the input but with bias and supply voltage. The voltage at the gate is above the treshold voltage, 2,7-2,9V as it says in the schematic. Then we have 0.1Ohm+0.0333Ohm=0.1333Ohm. 2.7V/0.1333Ohm=20.255mA and 20.55mA*2=40.51mA. 65V-25V=40V, 40V/0.04051A=1kOhm. Or 65V-20V=45V, 45/0.04051A=1,1kOhm. This i think is correct now. It sure seems to be...
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Old 07-01-2019, 03:41 AM
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you have series VS parallel mixed up
the gate conducts no current at DC, the resistors at the gate are to add stability in the radio frequencies, but could be considered to be a wire at DC
a wire into a capacitor conducts nothing at direct currents
so your calculations are not based on useful data.

you need to set your calculations on the requirements for the 5V regulator chip.
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Old 07-01-2019, 08:01 AM
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Basically the only thing left for me to do is build the actual device and by trial and error find out the value of that multiplication resistor. At 65V i'll try 5k and get lower and lower and measure the voltage at the input of the IC.
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Old 07-01-2019, 10:24 AM
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I took your advice and included in my order, resistors of 4.7k/3W.
I did it like this: 65V-20V=45V | 45V/0.009A=5k
As it says in the pdf of 7805, polarisation current will be from 4.3mA to 8mA and increases 1mA from 7V to 25V at the input. This is the only logical way to compute this resistor, you were right.
Thank you for your help and please excuse me for my stupidity. _/|\_
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Old 07-01-2019, 07:32 PM
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you are doing good, it is why you ask other people about your design,
I have others go over things before I build them if I can.
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Old 07-01-2019, 07:43 PM
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Thank you sir. I appreciate kind words especially at this time. _/|\_


The first diode turned out to be a Zener. Now i don't know what happens with those 7 volts when you supply the installation with 48V and the 24V zener is there but that diode is there solely for when you raise the voltage by 17V meaning 65Vdc.
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Old 07-01-2019, 11:17 PM
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so, you revised the schematic...
are you copying this from hardware you are looking at ? that is always hard.

it being a 24V zener changes the math there
but it also makes it possible to find a workable value for R

this means the 5V regulator is getting 24V at the input at most.
so now the goal is not to overheat the zener and have a lower limit of 7V.
so, 24V minus 65V is 41V
41 * (0.009-0.0053) = 0.15W (I know that math assumes things that are not quite true, but it should never be more than that power if you have the correct resistor)
what is your zener rated for ?
a standard 1N4736 is 1W, so, lots of design room for parts.

I think your 5K choice is going to be a pretty good guess, but it is maybe a tad high. (and will likely still work just fine)
at 4.5K (my calculated high value that is sure to work(actually is 4555, but I rounded down)) it will still work given worst case of 7V at the regulator and 48V in and the highest current required.
and given that the supply is allowed to go to 65V with the lower current possible,
that would be 65 -25 = 40V
40v / 4500 = 0.00888A
0.00888A * 25V = 0.222W, so you are still under the limits of most zeners.
but you should be able to go lower resistance and have it all work fine
2500 ohms would be 0.4W (I only look at this value because you said you ordered 5K, and 2 of them in parallel would still work.)
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Old 07-02-2019, 04:49 PM
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To be honest, the voltage after the resistor is 25V without the Zener when your input voltage is 65Vdc. The Zener is nothing but a safety measure for me personally. It is rated for 24V for it to lower the voltage by 1V to keep the IC in a safe area and 24V of the PL24V not to overheat the diode. You can put a zener that is rated from 1.3W up to even 5W doesn't matter. I ordered 5W ones, they tend to last longer
And the lower limit of the supply voltage after the resistor is 7V just as you taught me and resulted from the calculations.

_/|\_ Namo Buddhaya
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Old 07-03-2019, 09:09 PM
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one last thing,
if you have not powered up a new amplifier before,
make sure to set your gate bias to zero volts before powering up the circuit for the first time.
then watch your power supply current as you up the voltage on the gate to the required idle current.
this is because on the max bias setting you potentially damage your amplifier with over current.
A fuse on your power in can also make sure that error does not break things.
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Old 07-04-2019, 09:16 AM
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Yes. Thank you for your advice
In the recent past i burned up two IRFP250 because of over current.
The quiescent current was set at about 1,5A which was enough to light up a 60W 24V incandescent light bulb at 1MHz. And the two MOSFETs were destroyed. I remembered to set the q current and then everything was well.
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