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Bipolar amp advice

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Mosaic

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Hi all:
Since RF power is (Vpk)^2/100 for a 50 ohm system.
How does the (88-108Mhz) schematic attached manage 25 W output with just 28VDC supplied?bipolar amp.png
 

JimB

Super Moderator
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Without doing any calculations, my best guess is that the 2T inductor (and microstrip) in conjunction with C16 and C17 make for a nice impedance transformer.
C18 thru C20, and the two 5T inductors make a "half wave" low pass filter, so the 50Ohm of the load (antenna) is presented to C17. The 50Ohm is than transformed to something lower by C17, C16 and the inductors so that the available voltage at the collector of Q1 is operating on a lower impedance load.

JimB
 

ronsimpson

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It's a class C RF power amp.
Since RF power is (Vpk)^2/100 for a 50 ohm system.
I asked about Class C because your power formula is not for class C. I think you used the formula from another type of amplifier.

With your 24V supply you will have a collector voltage of 40 to 60 volts peak, depending on how you bias the amplifier.
 

ronsimpson

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There is a inductor from supply to transistor collector. We can not have DC across an inductor. So the average collector voltage is close to the supply voltage.

Another way to say it: If you turned the transistor on hard for 50% of the time so that 24 volts is across the inductor then for the other half of the time you should have a average voltage of about 48 volts. (think boost PWM power supply)

Picture: Using a 13V supply. The voltage on the collector is not a sign wave but 1/2 sign. The filter fills in the rest of the sign wave. The transistor is on for the flat bottom time in the blue trace. This puts 13 volts across the inductor for more than 50% of the time. So the inductor must fly up above supply for the same amount of volt seconds. This easy causes the collector voltage to get to 2x the supply and 3x is possible.
upload_2016-10-23_21-19-17.png
 

Nigel Goodwin

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Kick back when the tranny switches open?
'Sort of' - it's a class C amp, so it provides pulse to the tuned circuit, which then 'rings' - it's rather like a small hammer hitting a bell.

And as already mentioned, the tuned circuits provide impedance matching, so power output isn't related to supply voltage.
 
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