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In HMC453ST89's datasheet, it gives the electrical specifications when Vs = +5V:

enter image description here

The max. output power is around 32.5dBm, about 1.6W. But for a 50 Ohm load, 1.6W means the peak voltage is about 12V. How is it possible with Vs = 5V?

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1.6W mean's the peak voltage is about 12V. How is it possible with Vs = 5V?

Look at the data sheet application circuit diagram below. I've added a red box and a blue box: -

enter image description here

The red box is around L1 (an inductor) and when you have a collector of a transistor being pulled up to Vs, the peak-to-peak voltage that can be attained is nearly twice Vs. So, if Vs is 5 volts then you can probably achieve something like 9 volts p-p at that node.

But to achieve 25 volts p-p (about 1.6 watts into 50 ohms) you need some help from the tuned circuit inside the blue box. You need some form of resonance that lifts the 9 volt p-p to 25 volts p-p; an increase of about 9 dB: -

enter image description here

The interactive 2nd order low-pass filter tool above is loaded with 5.1 nH, 50 ohm and 22 pF and this resonates at around 400 MHz but, importantly it lifts the voltage amplitude by over 10 dB to give the required signal level into a 50 ohm load to achieve 1.6 watts.

So, it's a combination of class B/C output stage with an inductor and using a filter to resonate the signal to a higher amplitude.

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  • \$\begingroup\$ Andy, I noticed the table is based on the example application circuit. But didn't notice the resonatice. So, the 1.6W is just based on their application circuit, not only the chip? \$\endgroup\$ – diverger Dec 12 '18 at 15:11
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    \$\begingroup\$ If you just consider the chip with ONLY a load resistor (and no inductor or anything else) connected to +Vs, the output might produce 4.5 volts p-p. Consider it to be about 1.59 volts RMS (\$\frac{4.5}{2\cdot \sqrt2}\$). Then ask yourself what impedance that load resistor must be to produce 1.6 watts, you would surely agree it has to be 1.58 ohms. So the chip is theoretically capable of delivering 1.6 watts into a 1.58 ohm resistor from a 5 volt supply AND that would be the load seen by the chip in the application circuit. \$\endgroup\$ – Andy aka Dec 12 '18 at 15:20
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    \$\begingroup\$ So whether you attach a 1.58 ohm resistor or attach a 50 ohm load (via all the extra components) the chip produces 1.6 watts and it sees a load at its output terminal of 1.58 ohms. \$\endgroup\$ – Andy aka Dec 12 '18 at 15:22
  • \$\begingroup\$ Yes, I always considering a 50 ohms matched system, but missed that it can be used with any impedance. \$\endgroup\$ – diverger Dec 12 '18 at 15:29
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This power is never achieved.

The 1db compression point is the theoretical point where the input/output curve of your amplifier deviates from a perfect linear amplifier by 1dB.

You can just make an exponential curve fit at much smaller differences, and then find the point where that extrapolated curve deviates by 1dB.

Having the 1dB compression point significantly above the maximum output power means that you're pretty linear for all actually achievable output powers.

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  • \$\begingroup\$ I realized that i missed one line when I take the diagram, which gives Psat, which just a little higher than P1dB. If this rating is never achieved, why they entitle the sheet as "1.6W…". \$\endgroup\$ – diverger Dec 12 '18 at 14:47
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    \$\begingroup\$ same idea: You take measurements, you fit the curve, you take your definition and calculate where on that curve that happens. \$\endgroup\$ – Marcus Müller Dec 12 '18 at 14:50
  • \$\begingroup\$ @Marcus, I think you might have misread the question. \$\endgroup\$ – Andy aka Dec 12 '18 at 15:31
  • \$\begingroup\$ @MarcusMüller: Thanks for the explanation of 'P1dB'. But what confused me is where the 1.6W come from, so, I change the question title back. \$\endgroup\$ – diverger Dec 13 '18 at 1:32

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