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class A transitor rating

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Advanced Member level 2
Nov 29, 2004
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A class-A transformer coupled, transistor power amplifier is required to deliver a power o/p of 10 watts. then what sholud be the max power rating of the transistor used should be


in an ideal case, neglecting all voltage drops and losses (windings, core,
additional resistors, etc.) the quiescent operating point (without excitation)
in an A-class amplifier with a transformer coupled load (n_primary; n_secondary) will be:

&nbsp &nbsp Vc(q) = V_power_supply, and
&nbsp &nbsp Ic(q) = I_signal_max (amplitude of a transformed load current);

therefore its power consumption without any excitation:

&nbsp &nbsp P_total = Vc(q) x Ic(q) = V_power_supply x I_signal_max

Without any driving signal at the input all this power has to be dissipated in the transistor (heat).
This is the worst case from the efficiency point of view.
When excited (let's suppose a sine waveform), the mean values (V, I) of an ac signal are 0,
thus P_total stays CONSTANT (as above; it does not depend on the excitation) and a part
of the total power from the power supply goes to the load, decreasing the dissipated part.

Max available amplitudes (voltage, current) in case of a fully excited amplifier will be:

&nbsp &nbsp V_signal_max = V_power_supply
&nbsp &nbsp I_signal_max = Ic(q)

and therefore the output power:

&nbsp &nbsp \[P_{out} = V_{SignalRms} . I_{SignalRms} = \frac{V_{SignalMax}}{\sqrt{2}}.\frac{I_{SignalMax}}{\sqrt{2}} = \frac{V_{PowerSupply} . I_{SignalMax}}{2} = \frac{P_{total}}{2}\]

so the maximum possible efficiency \[\eta\] = 50%

In this best case one half of the total power goes to the load as a useful power
and one half of it is dissipated in the transistor (Pc) changing into heat.
In practice the efficiency is much lower (~25-35%).
Thus, for max 10 watts delivered to the output the output transistor max power rating must be higher than 20 watts
(it is not a good practice to exploit fully the max power rating).
transistor Vce_max must be at least twice as high as V_power_supply and also Ic_max = 2Ic(q)
the load value "seen" by the transistor is ~ \[{(\frac{n_{primary}}{n_{secondary}})}^2 . R_{load}\]

Merry Christmas (if you celebrate it)

Best Regards

I'm sorry for making a mistake caused by my inattention. Fortunately, I noticed it (sooner than others ;-)) reviewing my contribution later. The sentence "...for max 10 watts delivered to the output the output transistor max power rating must be higher than 10 watts..." (as it was originally) should have been "...higher than 20 watts...", of course, as it follows from the text itself. I've already corrected it right in the original above.

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To paraphrase Eric (Good job, Eric), the transistor dissipation of a class A amplifier does not vary with load. A more important rating than the transistor power rating is its maximum allowable junction temperature. You could have a 100W transistor with inadequate heat sinking and still exceed this rating and destroy the transistor.

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