# I am solving a question i found in a book on signals and amplifiers.

Joined Oct 2, 2020
1
Exercise says:"An amplifier operating from a single 15 V supply provides a 12 V peak to peak sine wave signal to a 1 kohm load and draws negligible input current from signal source.. The DC current drawn from the 15V supply is 8 mA. What is the power dissipated in the amplifier and what is the ampifier efficiency?

I said:

Pdc= Vdc * Idc = 12 * 0.08 = 120 mW
Pload = (Vo /sqrt(2))^2/Rl = 72 mW

Pdiss + Pl = Pdc + Pi

Since no current is drawn from source, Pi=0 =>

Pdiss = Pdc-Pl = 48 mW

efficiency=Pl/Pdc = 60%.

What am I doing wrong?

#### ericgibbs

Joined Jan 29, 2010
11,663
hi tT,
Welcome to AAC.

Note: 12 V peak to peak sine wave

So what would you have to do to calc power.?

E

#### ci139

Joined Jul 11, 2016
1,696
I assume the 8mA is RMS . . . checking . . . sounds compliant with your "Answer says"

#### MrAl

Joined Jun 17, 2014
7,892
What is confusing here is the 8ma DC current.
An amplifier would not draw a constant DC current like that when supplying an AC sine wave output into a resistive load.
Thus it does not make sense unless we assume maybe one of these two:
1. The 8ma is really the AC current (peak or rms, probably peak if the output is in peak too).
2. The 8ma is a constant DC as stated even though the output load current draw is an AC current.

If #1 then we can figure all AC quantities.
If #2 then we have to assume that the input power to the op amp is 15*0.008 for all time although the output is an AC voltage and AC current. Then the power used by the op amp would simply be the subtraction.

#### ericgibbs

Joined Jan 29, 2010
11,663
hi Al,
If it is a Class A amplifier, there would a quiescent current flowing in the amplifier, in the absence of any signal input.
The low eff% suggests a Class A amp.?
E

#### BobTPH

Joined Jun 5, 2013
2,757
Well, two us have now verified that, if the 8mA is RMS, the book answer is correct.

Bob

#### ericgibbs

Joined Jan 29, 2010
11,663
hi Bob
The way I read the TS's question suggests a Class A amplifier with a 8mA quiescent Collector current.

Which gives a quiescent supply Power of 15Vdc * 8mAdc = 120mWatt

A 12Vppk swing across a 1K Collector load gives a Load power of ~18mWatt

Eff%= 18mW/120mW = 15%

These values are the same as the answers as given in the TS's book.

ie: the 8mA is not an RMS value in this case.

E

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#### MrAl

Joined Jun 17, 2014
7,892
hi Bob
The way I read the TS's question suggests a Class A amplifier with a 8mA quiescent Collector current.

Which gives a quiescent supply Power of 15Vdc * 8mAdc = 120mWatt

A 12Vppk swing across a 1K Collector load gives a Load power of ~18mWatt

Eff%= 18mW/120mW = 15%

These values are the same as the answers as given in the TS's book.

ie: the 8mA is not an RMS value in this case.

E
Hi Eric,

Yes that is a very reasonable interpretation of the initial problem posed in the first post.
In that case, the internal losses due to the output AC would be calculated also and added to the 'quiescent' power loss.