hi Jony, ebp.
This is LTS sim of the diode temperature range, showing Collector current change with temp.
E

Upper plot trace is -20C

 

Why are little low current and low power transistors being severely overloaded in this circuit? LTspice should warn about it.
Why is the input signal 16V peak causing the output to be severely clipping and not heating as much as when its output is a sinewave?

 

eric, that certainly shows the sensitivity of the idle current to difference in temperature between the diodes in the bias network and the transistors.

Can you specify VC as a function of time so that in a transient analysis you get a single curve covering the range of temperature? For example, with the same 80 °C temperature range could you use an expression something like VC = t - 20 where t is time in milliseconds and it ranges from 0 to 80 ms? If that can be done then you can conveniently add things like diode forward voltage, base current and collector current to the plot. Changing the Y axis to logarithmic would be instructive. [EDIT - fixed the VC = ... equation if t is simply a numeric value where 80 ms is 80 and not 0.08; intent is that VC should evaluate as -20 to +60 as t varies from 0 to 80 ms]

Is there simply a way to set the X axis to some variable other than time?

 

Amplifiers usually have the diodes bonded to the output transistors so that the idle current does not change much as the transistors heat up.

 

Most of the simulation packages seem to be quite content to let you run devices far beyond their nominal limits. This is actually necessary since there may be cases when it is perfectly acceptable. It is up to the user to evaluate when something will detonate or melt itself off a circuit board. LTspice may have warning capability, but since I don't use it, I don't know (I don't use it because I don't do electronics anymore. If I did, I'd most certainly use it because it looks to me like one of the highest value freebies in the industry).

My intent in suggesting experimentation with the diode temperature is to show how sensitive the collector idle current is to diode forward voltage in this particular circuit and how the current varies in anti-logarithmic proportion to the base-emitter voltage in a BJT. Many people tend to think of BJTs as being current controlled, which they are, but there are very important relationships between voltage and current.

 

The base-emitter of a transistor is a diode. When it heats, its forward voltage decreases. If the diodes that provide an idle voltage to the diodes also does not heat then the transistors conduct more current which makes them hotter which makes them conduct more current which makes them hotter which …. it is called Thermal Runaway.
But fastening the diodes to the transistor so the diodes are heated by the transistors cancels current change as the transistors heat up.
Frequently a transistor is used to replace the two diodes and its voltage changes with heat the same as the diodes.

 

Is there simply a way to set the X axis to some variable other than time?

hi @ebp,
This what I have so far.
Put the two diode voltages in series on the top plot, effective Q1b + Q2b, Y axis log inVolts.
Lower plot, Q1 Collector current, Y axis log. 10mA > 1A

X axis = Temperature range. -20C to 60C , 10C steps

E

 

Very nice, eric!

 

Hi Eric,
It would be interesting to see your simulation of how thermal runaway of output transistors heating themselves occurs without the diodes thermally attached, then the cancellation of the thermal effect on idle current when the diodes are attached.

 

hi @Audioguru
Is this what you mean.?
I found it easier to duplicate the circuit for the two conditions, as the Y scaling did not show enough detail.
E

 

Thanks, Eric, fastening the diodes to the transistors makes a huge improvement.
From 0 degrees to 50 degrees the unfastened current went from 15mA to 100mA and the fastened current barely changed from 44.7 degrees to 47.6mA.

 

I suspect you need flashier simulation software and models than LTspice to do the thermal runaway thing. Those poor little TO-92 devices would not be long for this world.

Another thing that would be interesting would be to look at the change in bias current during normal operation of an amplifier that was operating at rapidly changing power level, such as movie soundtracks with sudden extra loud bits. No matter how closely you attempt to couple the biasing semis to the output devices there is going to be a time lag and the bias network will tend to respond to a long-term average. In most amps I've seen the temperature compensation device, usually a Vbe multiplier, is mounted to the heatsink between or among the output transistors. It will respond well to long-term (probably tens of seconds to minutes) average temperature of the transistors and know nothing of short term events.

I think I have a vague recollection of seeing something about audio output transistors with temperature sensing diodes built on the same substrate, but I may be imagining things.

 

Time lag between the transistors suddenly heating then the diodes catching up?
I don't care if the Class-AB transistors slowly go a little more Class-A for a second or two.

 

Simply enter 1N4148 TEMP=47 in a part name box.

Regarding Total DC Drain current. Is this correct?
Assumption: Transistors are complementary pairs and diodes are matched.
ICQ ≈ Ibias

No Signal Condition:
Idrain = Ibias + ICQ

Small Input Signal:
Idrain = Ibias + ICQ + (Vout / (π * RL))

Maximum Input Signal:
Idrain = Ibias + ICQ + (Isat / π)

David

 

Regarding Total DC Drain current. Is this correct?
Assumption: Transistors are complementary pairs and diodes are matched.
ICQ ≈ Ibias

No Signal Condition:
Idrain = Ibias + ICQ

Small Input Signal:
Idrain = Ibias + ICQ + (Vout / (π * RL))

Maximum Input Signal:
Idrain = Ibias + ICQ + (Isat / π)

Looks correct to me.
What is more important. Do you understand why you cannot get more than 8V at the output ?

 

You might be calculating with a badly distorted waveform with severe clipping and the output transistors simply switch on (no voltage then no heating) and switch off (no current then no heating).
Look at the datasheet of an audio amplifier IC like an LM1875 with a 50V supply and an 8 ohm load:
1) Its idle current is typically 70mA. Idle power is 50V x 70mA= 3.5W.
2) It clips at 22W.
3) It heats with 18W when its output is 22W.
Then the total power is 43.5W and the current from the 50V supply is 43.5W/50V= 0.87A.

 

You might be calculating with a badly distorted waveform with severe clipping and the output transistors simply switch on (no voltage then no heating) and switch off (no current then no heating).
Look at the datasheet of an audio amplifier IC like an LM1875 with a 50V supply and an 8 ohm load:
1) Its idle current is typically 70mA. Idle power is 50V x 70mA= 3.5W.
2) It clips at 22W.
3) It heats with 18W when its output is 22W.
Then the total power is 43.5W and the current from the 50V supply is 43.5W/50V= 0.87A.

This is not the answer to the simple question that I asked!

 

Looks correct to me.
What is more important. Do you understand why you cannot get more than 8V at the output ?

Please expalin why.

 

Please expalin why.

With a 12V peak sine in I get 7V peak out. After that the output is clipped. All per LTS.

At first you can read this:
https://forum.allaboutcircuits.com/threads/class-b-push-pull-amplifier.85631/#post-615183
Where you can see that the input source does not provide any current to transistor bases.
And one of a purpose of adding those two diodes is described here:
https://forum.allaboutcircuits.com/...ed-on-push-pull-amplifier.101749/#post-766570

So, as the input signal increases the D1 diode conduct less and less current therefore more current from RB1 can enter Q1 base.
And we get the maximum base current when D1 diode is OFF. In this case the situation looks like this:



And from KVL we have

Vcc = Ib*RB1 + Vbe + Vcap + Ie*RL

Vcc = Ie/(β + 1) *RB1 + Vbe + Vcap + Ie*RL

So we can solve for Ie

Ie = (Vcc - Vbe - Vcap)/(RL + RB1/(β+1))

So if β = 100; and RB1 = 180Ω; RL = 10Ω

The maximum we can get at the output is

Ie = (20V - 0.7 - 10V)/(1.8Ω + 10Ω) = 0.788A

And this correspond to Vo_max = 0.788A*10Ω = 7.88V

 

At first you can read this:
https://forum.allaboutcircuits.com/threads/class-b-push-pull-amplifier.85631/#post-615183
Where you can see that the input source does not provide any current to transistor bases.
And one of a purpose of adding those two diodes is described here:
https://forum.allaboutcircuits.com/...ed-on-push-pull-amplifier.101749/#post-766570

So, as the input signal increases the D1 diode conduct less and less current therefore more current from RB1 can enter Q1 base.
And we get the maximum base current when D1 diode is OFF. In this case the situation looks like this:

View attachment 160841

And from KVL we have

Vcc = Ib*RB1 + Vbe + Vcap + Ie*RL

Vcc = Ie/(β + 1) *RB1 + Vbe + Vcap + Ie*RL

So we can solve for Ie

Ie = (Vcc - Vbe - Vcap)/(RL + RB1/(β+1))

So if β = 100; and RB1 = 180Ω; RL = 10Ω

The maximum we can get at the output is

Ie = (20V - 0.7 - 10V)/(1.8Ω + 10Ω) = 0.788A

And this correspond to Vo_max = 0.788A*10Ω = 7.88V

If I input a 12.5 volt sine into LTS it results in a much different Ie than 788 mA,
like 36 to 38 mA although the ouput is 7.18 volts. I am confused.