datasheets Hfe's graph

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julian403

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Hi, for the MJL1302A transistor, the datasheet gives the graph which shows the Hfe for Ic and for temperatura but it's for Vce=5[V]

The transitor will be part of class AB amplifier, so it's going to work in cutoff and saturation (well, it's not going to be in saturation for the current that the tranformers can gives). In cotoff it's going to have a Vce=69 [V] and for the maximun peak signal it's going to have Vce=38 [V].

The graphs for Vce=5 [V] works for every Vce? and Hfe just varies by Ic and temperature? I do not thinks so.

For example, I know that in work the transistor will have (almost) 100ºC and IC=4 [A] but in that case the colector emitter voltaje will be 38 [V] (on peak signal voltages). How can I infer the Hfe with the graph?

https://www.onsemi.com/pub_link/Collateral/MJL3281A-D.PDF

I'm going to put negative feedback to Hfe will not varies. And maybe you can say that I must to measure it. But it dificult works to 70 [V] and make a circuit for it.
 

Aren't the output transistors a complementary NPN and PNP pair of emitter-followers?
Then if the NPN is cutoff with a Vce of 69V the PNP is saturated with a Vce of close to 0V or close to the negative supply voltage, not 38V.

hFE does not matter if the driver transistor can supply the maximum base current needed for the output transistors.

Negative feedback simply makes the DC gain at 1 so that the output idles at half the supply voltage so it can swing symmetrically up and down for the maximum possible output. Negative feedback reduces the very high AC open-loop gain of the amplifier to a much lower gain that you need and it also reduces distortion with the same reduction.
 
Ok, the Vce will be Vce=38[V] + 69[V] = 107[V].

But the graph of Hfe it's for Vce=5[V]. There is a form to infer it when Vce=107[V] using the graph's data?
 

The "Early effect" of increasing Vce causes the current gain of a transistor to increase a little.

Why doesn't your output go down almost to 0V but is limited to 38V? Is your amplifier using an old-fashioned output transformer? Please post a schematic that shows why.
 
Increasing Vce above the measurement conditions does not usually have that much effect (When you consider that Hfe is **Really** poorly specified anyway (100% variation between parts is commonplace)).

You should never design assuming a specific value for Hfe (You can assume a minimum, more or less), and I would note that as Vce drops below the measurement conditions Hfe tends to fall (Early effect), to full saturate a typical power transistor, assume you need to supply Ib = 0.1Ic, even if the device has a much higher Hfe out of saturation.

Any design that assumes more then that Hfe will be greater then the minimum given in the data sheet (Possibly very much greater) is just plain broken, and even that assumption fails near saturation.

Regards, Dan.
 
You are the guy with at least 3 threads running about your Differential input, Darlingtons Output and now these Odd Voltages for your amplifier.
You should have used a single thread. Maybe a moderator will combine them.
 
You are right Audioguru, I didn't think about make just one threat, I thought they are different. Sorry.

Increasing Vce above the measurement conditions does not usually have that much effect (When you consider that Hfe is **Really** poorly specified anyway (100% variation between parts is commonplace)).
Click to expand...

So for example, the graph it's make for the minimum Hfe value (whit 5 [V]) for 107 [V] I must assume that the Hfe will be bigger and the value for a constant Ic and temperature, the minimum it's what graph shows.
 

Yea, generally they give a minimum and the real value is larger in most devices (Except at very low Vce where the early effect causes Hfe to fall).

Regards, Dan.
 
The graphs on a datasheet are usually for a "typical" device but yours might have minimum or maximum spec's. The minimum and maximum spec's are printed on a datasheet.

You did not answer my question about your odd voltages.
How will you control the current in the output transistors when it increases caused by heating?
How will you cool the output transistors?
 

How will you control the current in the output transistors when it increases caused by heating?
Click to expand...

As first protection I use the emitter resistor. And I will study the negative feedback.

How will you cool the output transistors?
Click to expand...
With heat sink
 

julian403 said:
As first protection I use the emitter resistor.
Click to expand...
The emitter resistor reduces the effect of different Vbe and different current gain in transistors, it has only a small help for temperature changes. But the value of your emitter resistors in the other thread was 46,000 times too high.

You should look at Class-AB Amplifier Circuits in Google Images to see hundreds of circuits that use diodes or a transistor to compensate the current when the temperature changes because a diode or a sensing transistor has the same forward voltage changes an output transistor. But watch out, some of the circuits are class-B with the bases tied together without biasing.
 
Ah, Do you remember Audioguru in my last post, the circuit was:



Do you means that instead of putting the resistors R2 and R3 I must put a diode to polarize? (because the resistor is dinamic) In this case I must to put two diodes because they are darlington but it are not the transistor i'm going to use.
 

R2 and R3 are shorted by a piece of wire so they do nothing.
If you look at any of the hundreds of amplifiers in Google Images then you will see that they are diodes. Two series diodes for two ordinary power transistors or four series diodes for darlington power transistors like this:

- - - Updated - - -

The diodes are attached to the heatsink of the output transistors so they all heat at the same time. When the transistors heat then they draw more and more current but when the diodes heat their forward voltage drops which prevents the output transistors from drawing more current when they heat. Then the diodes compensate the temperature effect.
 

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Of course, The semiconductor's conductivity increase with temperature. Using diodes and Re resistors, it's necesary to put negative feedback? or differential amplifier? Maybe as a previous stage I can put just a commun emitter.
 

You can know that the hFE will be higher at higher voltage.
Is that enough? Or do you need some quantification, for
some reason? If you were trying to, say, design for a worst
case driven base bias network for the final stage, you'd be
looking at the other end of the temp range (for drive current)
as well as high temp (overdrive, saturating harder, maybe too
slow in switching off).

If there is an Early voltage spec then you could at least
graphically, if not analytically, assess the ratio of one Vce's
beta to another Vce's.
 
julian403 said:
Using diodes and Re resistors, is it necessary to put negative feedback?
Click to expand...
Negative feedback from the output back to an inverting input is used on ALL half-decent and better audio amplifiers. Without negative feedback then the distortion will be horrible.
An audio amplifier is designed like an opamp to have VERY high voltage gain. The excellent LM3886 audio amplifier IC has a voltage gain of typically 115dB which is about 600,000 times! Negative feedback is added to reduce the gain to 21 times and then the distortion is reduced to only 0.03%.

Maybe as a previous stage I can put just a common emitter.
Click to expand...
The previous stage is called the driver stage. It might be emitter-followers as part of darlington output transistors. Bootstrapping (look at it in Google) of the load for a common-emitter driver stage is frequently added so that the output swing and the voltage gain are more than without them.
 
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