Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

class A transitor rating

Status
Not open for further replies.

electronics_kumar

Advanced Member level 2
Joined
Nov 29, 2004
Messages
659
Helped
34
Reputation
68
Reaction score
9
Trophy points
1,298
Location
Tamilnadu
Activity points
5,552
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
 

Eric Best

Member level 4
Joined
Sep 3, 2001
Messages
77
Helped
25
Reputation
50
Reaction score
12
Trophy points
1,288
Activity points
863
electronics_kumar,

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).
Note:
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
Eric

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.

Thanks,
Eric
 
Last edited by a moderator:

Kral

Advanced Member level 4
Joined
Mar 28, 2005
Messages
1,326
Helped
280
Reputation
558
Reaction score
85
Trophy points
1,328
Location
USA
Activity points
13,418
electronics_kumar,
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.
Regards,
Kral
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top