# Class AB Amplifier - Determining The Total DC Current Drain (Quiescent)

#### elec_eng_55

Joined May 13, 2018
214
Hi:

I
am still trying to figure out some stuff with these Class AB amplifiers. Its driving me
nuts. I am so stressed that I have lost all my hair! (Seriously, I kid you not). Even
though I now resemble Kojak, minus the lolly pop, I am not giving up.

Attached is a sim.

According to my calculations:
Ibias = Vcc * 1.4 / 2* 180Ω = 51.67 mA

Isat = Vceq / RL = 1.25 A

IC_avg = Isat / π = 1.25 A / π = 1.25 A / 3.14 = 398 mA
Does IC_avg only apply when there is an input signal?

According to a textbook that I read:
Idc = Ibias + IC_avg = 51.67 mA + 398 mA = 449.67 mA

However, LT Spice says Idc = Ibias + Icq = 51.67 mA + 18.35 mA = 70.02 mA with
or without an input signal.

I want to know where the 18.35 mA comes from and how Spice came up with it?

Thanks,

David

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#### Jony130

Joined Feb 17, 2009
5,227
Can you explain to me what IC_avg has anything to do with the "bias network" DC current and the output transistors DC quiescent current?

#### elec_eng_55

Joined May 13, 2018
214
My point is, that every text that I look in, there are different answers for the same thing.

In Malvino's textbook, he says that Total DC Current Drain = Ibias + Ic_avg
in his spreadsheet he says Total DC Current Drain = Ibias + Icq + Ic_avg. I would
agree with the "bold type" description. Do you agree?

I assume that LT Spice is only looking at Ibias and Icq as they are "quiescent"
currents? Do you agree?

Now, I still don't see how LT Spice comes up with Icq = 18 mA?

David

#### crutschow

Joined Mar 14, 2008
26,841

#### Jony130

Joined Feb 17, 2009
5,227
It seems to me is that you do not understand what Ic_avg is, or you do not distinguish between a "static" DC current (Ibias + Icq) that is flowing without the input signal. And "dynamic" average supply current (Ic_avg) that is flowing at "full load" condition.

Now, I still don't see how LT Spice comes up with Icq = 18 mA?
Me neither. In my simulation Icq = 52mA.

#### elec_eng_55

Joined May 13, 2018
214
It seems to me is that you do not understand what Ic_avg is, or you do not distinguish between a "static" DC current (Ibias + Icq) that is flowing without the input signal. And "dynamic" average supply current (Ic_avg) that is flowing at "full load" condition.

The sim is attached, where Icq = 18mA. Not sure what happened. I must have sent you the wrong sim initially. And yes I do understand the difference between no-load and full-load dc current drain. I was trying to illustrate how some of these so-called experts have errors in their work. Malvino is one example. It makes it tough for us
beginners.

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#### elec_eng_55

Joined May 13, 2018
214
Noting is attached.
Post the .asc file.
Correct Sim is re-posted. Thanks Crotchchow.

#### Jony130

Joined Feb 17, 2009
5,227
The sim is attached, where Icq = 18mA. Not sure what happened.
Simply, the diode V-I characteristics do not match the transistor V-I characteristics.

See

So, it not a surprise that in your circuit Ibias do not match Icq.

Icq/Ibias = √((Is_Q1 + Is_Q2)/Is_D^2)

I was trying to illustrate how some of these so-called experts have errors in their work. Malvino is one example. It makes it tought for us
beginners.
I don't see any error.

Simply, the Ibias + Icq is the current that is flowing when Vin = 0V (without the input signal).
And I_avg is an average current drawn from the power supply by the load resistance at "full load" condition.

#### crutschow

Joined Mar 14, 2008
26,841

#### elec_eng_55

Joined May 13, 2018
214
Simply, the diode V-I characteristics do not match the transistor V-I characteristics.

See
View attachment 168708

So, it not a surprise that in your circuit Ibias do not match Icq.

Icq/Ibias = √((Is_Q1 + Is_Q2)/Is_D^2)

I don't see any error.

Simply, the Ibias + Icq is the current that is flowing when Vin = 0V (without the input signal).
And I_avg is an average current drawn from the power supply by the load resistance at "full load" condition.
If I apply this formula in the diode bias example, I get
Icq/Ibias = √((Is_Q1 + Is_Q2)/Is_D^2) = 4.79 mA for the diode bias
circuit. Using IS_Q1 1E-14,IS_Q2 1E-14, IS_D1 2.95E-08 and
IS_D2 2.95E-08. Which does not agree to the sim results.

I have modified the circuit and I am using 2N3904/06 s in diode mode. The Ibias = 51.27mA and Icq = 54.89mA

Using your formula I came up with an Ibias=Icq of 1.41mA.
Still not the same even though the devices are matched.
Not sure why this would be.

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#### elec_eng_55

Joined May 13, 2018
214

#### Jony130

Joined Feb 17, 2009
5,227
If I apply this formula in the diode bias example,
Forgot about this formula, I made a mistake and this is why the formula is wrong. Because the diode Vt voltage (thermal voltage) is not equal to 25mV as for the BJT. And again to solve for Icq will have to use the iterations. But it is pointless to do such calculations. Because no one will use your output stage in real-world design without to worry about thermal-runaway. In the real world, we are using Re resistors and Vbe multiplier to set the Icq current.

And Vbe multiplier is used in 90% of a real-world class AB audio amplifiers.

#### elec_eng_55

Joined May 13, 2018
214
Forgot about this formula, I made a mistake and this is why the formula is wrong. Because the diode Vt voltage (thermal voltage) is not equal to 25mV as for the BJT. And again to solve for Icq will have to use the iterations. But it is pointless to do such calculations. Because no one will use your output stage in real-world design without to worry about thermal-runaway. In the real world, we are using Re resistors and Vbe multiplier to set the Icq current.

And Vbe multiplier is used in 90% of a real-world class AB audio amplifiers.
Thanks Jony130.

I have created a sim with a vbe multiplier. Iref = 50mA and Icq = 214mA. As I decrease the value of the
vbe resistors, Icq decreases and DC Input power decreases but the ac output power remains fixed at 440mW.

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#### Jony130

Joined Feb 17, 2009
5,227
Vbe multiplier will set Icq (the class of a amplifier A or B or AB). But Vbe multiplier would not change the output power. The output power will depend on Vcc and RL but not on Icq.

#### elec_eng_55

Joined May 13, 2018
214
Vbe multiplier will set Icq (the class of a amplifier A or B or AB). But Vbe multiplier would not change the output power. The output power will depend on Vcc and RL but not on Icq.[/QUO

thanks

#### elec_eng_55

Joined May 13, 2018
214
I randomly chose Ibias = 50 mA.

I also randomly chose the vbe multiplier resistors R1 = R2 = 22Ω and the Icq ends up
being 118.3 mA.

There must be a way to calculate the required R1 and R2 values given that you want Icq
to be a certain value.

David

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#### Audioguru

Joined Dec 20, 2007
11,249
Why are you using small low current 2N3904/2N3906 transistors (max allowed 200mA) at the output where the current can be 1000mA??
It is a power amplifier so you should use power transistors.

#### Jony130

Joined Feb 17, 2009
5,227
There must be a way to calculate the required R1 and R2 values given that you want Icq
to be a certain value.
Sure there is a way. We can again try to use the Shockley equation and solve it for Vbe1, Vbe2 etc.
http://www2.engr.arizona.edu/~brew/ece304spr07/Pdf/VBE Design.pdf
http://www.hagtech.com/pdf/vbe.pdf

But it is pointless to do such calculations in real life due to transistor mismatch and big production spread between transistor main parameters (we don't know exact value of a BJT's beta, Is current etc), and it very hard to predict the temperature effect. This is why in almost all class AB amplifiers that we can buy on the market, there is a POT in series with the lower resistor in Vbe multiplier. So that in the factory they can adjust the voltage drop across the output emitter resistors do get desire Icq value.
And the same "procedure" is used by the hobbyist in all DIY constructions.

Because the only thing we need to do is to measure the voltage drop across RE resistors and set via POT the desired Icq value.
And repeat this procedure after amplifier heats-up. And we are also thermally coupled the Vbe multiplier transistor together with the output transistor by mount it on the main heat-sink or screw the transistors together. Hence our job as a designer is to properly choose the component value.

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#### elec_eng_55

Joined May 13, 2018
214
Sure there is a way. We can again try to use the Shockley equation and solve it for Vbe1, Vbe2 etc.
http://www2.engr.arizona.edu/~brew/ece304spr07/Pdf/VBE Design.pdf
http://www.hagtech.com/pdf/vbe.pdf

But it is pointless to do such calculations in real life due to transistor mismatch and big production spread between transistor main parameters (we don't know exact value of a BJT's beta, Is current etc), and it very hard to predict the temperature effect. This is why in almost all class AB amplifiers that we can buy on the market, there is a POT in series with the lower resistor in Vbe multiplier. So that in the factory they can adjust the voltage drop across the output emitter resistors do get desire Icq value.
And the same "procedure" is used by the hobbyist in all DIY constructions.
View attachment 168863

Because the only thing we need to do is to measure the voltage drop across RE resistors and set via POT the desired Icq value.
And repeat this procedure after amplifier heats-up. And we are also thermally coupled the Vbe multiplier transistor together with the output transistor by mount it on the main heat-sink or screw the transistors together. Hence our job as a designer is to properly choose the component value.