Transistor Mystery

Thread Starter

hensle

Joined Dec 31, 2011
46
Hi again,

I have another simple but strange problem with transistors. I have attached a schematic. I used pspice to create it, but it is actually built and I am discussing real results (not a simulation). It is an emitter follower and as it is shown it works exactly as it should. The 20V signal goes into the transistor and comes out (at probe V) shifted so that the bottom of the wave just sits near zero, thanks to the biasing resistors.

However, when I remove the biasing resistors, the full wave still comes out but centered around zero. I would think that the negative side of the wave would be clipped and I would only see the positive part of the wave on output.

Does anyone know why this is happening? Is it supposed to?
 

Attachments

hobbyist

Joined Aug 10, 2008
892
Haven't did any experimenting, but this is my guess,

The transistor is not acting as a diode clipper through the base, Probably, if you disconnect the collector lead than the signal is going to be clipped through the base emitter diode.

With the collector connected the output signal is not a true representation odf the input signal, because the output is being established through the collector circuit.

Inother words the output is coming through the collector supply, the signal to the base is not the output signal.

But if the collector was not connected, than the output would be the signal through the base,

Just as a diode would work in that condition.

I just throwing this out there as a guess.
 

steveb

Joined Jul 3, 2008
2,436
Does anyone know why this is happening? Is it supposed to?
Question? Are you loading the output of the amplifier with a resistance (in other words is that 10K emitter resistor actually connected)? If not, perhaps you are just measuring a floating voltage on the negative cycle. Since no current is flowing through the base circuit, even that high impedance might not be enough to prevent you from tracking the floating voltage with a scope.
Try loading the output with a resistor. It doesn't need to be too small, but just enough to prevent the floating voltage by providing a path to ground.
 
Last edited:

DickCappels

Joined Aug 21, 2008
10,152
When the base voltage swings negative far enough (perhaps 4 volts), the base-emitter junction acts like a zener diode and passes negative current from the base to the emitter circuit.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
Haven't did any experimenting, but this is my guess,

The transistor is not acting as a diode clipper through the base, Probably, if you disconnect the collector lead than the signal is going to be clipped through the base emitter diode.

With the collector connected the output signal is not a true representation odf the input signal, because the output is being established through the collector circuit.

Inother words the output is coming through the collector supply, the signal to the base is not the output signal.

But if the collector was not connected, than the output would be the signal through the base,

Just as a diode would work in that condition.

I just throwing this out there as a guess.

Thanks for the reply. I don't know what:
"The transistor is not acting as a diode clipper through the base"
means though.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
Question? Are you loading the output of the amplifier with a resistance (in other words is that 10K emitter resistor actually connected)? If not, perhaps you are just measuring a floating voltage on the negative cycle. Since no current is flowing through the base circuit, even that high impedance might not be enough to prevent you from tracking the floating voltage with a scope.
Try loading the output with a resistor. It doesn't need to be too small, but just enough to prevent the floating voltage by providing a path to ground.
Thanks....The resistor is connected. I did wonder about this as it would be the current that is clipped. So I tried smaller resistance and no change.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
When the base voltage swings negative far enough (perhaps 4 volts), the base-emitter junction acts like a zener diode and passes negative current from the base to the emitter circuit.
Thanks. I failed to mention that the zero'ed output wave was undistorted, an exact match of the input wave. I would think a zener effect would clip until the wave reached 4 volts and that would be observed in the output.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
Did you also short out the capacitor, or leave it in the circuit?

Thanks. I left it in the circuit. It is needed to unbias the signal (The actual sine wave generator is an oscillator which provides a wave biased to the positive).
 

Audioguru

Joined Dec 20, 2007
11,248
I simulated it in LTspiceIV and the transistor acts like a rectifier, passing only the positive part of the input sinewave minus a base-emitter voltage drop.
I did not attach it because it wrongly does not show avalanche breakdown of the reverse-biased emitter-base junction when the input goes more negative than -6V.
 

Audioguru

Joined Dec 20, 2007
11,248
If the base-emitter of the transistor is shorted then the signal generator will drive the load through the coupling capacitor and the output will have the same waveform as the input.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
I simulated it in LTspiceIV and the transistor acts like a rectifier, passing only the positive part of the input sinewave minus a base-emitter voltage drop.
I did not attach it because it wrongly does not show avalanche breakdown of the reverse-biased emitter-base junction when the input goes more negative than -6V.
Thanks for the response.

This is funny, because you actually bring up a point which I was not going to ask about, because it would be another issue and I did not want to lose focus.

I also ran it in pspice based on the attached diagram in the OP. And it did rectify the signal, but in the real world it did not. There was even more strangeness as the simulation with the bias resistors failed to shift the the wave even though in real life they worked fine. The simulation worked only when the upper bias transistor was made much less than the lower, but again in real life the resistors shown in the diagram worked correctly.

Now, since I did not have a model of the actual transistor I was using, I wasn't sure what to think....but the biasing should have worked in the simulation in any case.

The bottom line: I could not get a simulation to show this problem, but in real life, it is happening.

I did replace the transistor (in real life) with another type just at random and still it was not rectifying, but in that case there was distortion especially with the negative part of the wave...so it looked like it was making a feeble attempt to rectify the wave.

The transition frequency of the random transistor was less than the NTE transistor I intended to use, so perhaps that was an issue. This is a high frequency circuit so I selected the NTE transistor with a 30MHz transition frequency. I also needed higher current...about a 1.6 amp peak. I was starting to design a B-Class amplifier, but if I can't even rectify the wave I am sort of stuck.
 

Thread Starter

hensle

Joined Dec 31, 2011
46
The base-emitter junction is shorted?
Thanks.

This is an Idea since the wave forms are similar (as Audioguru mentions below). But I had just used this transitor in a voltage amplifier and it worked as expected. I will try some other (identical) transistors, and at lower voltages to test this.
 

Audioguru

Joined Dec 20, 2007
11,248
I don't think that NTE make transistors. I think they buy some real ones, remove the part numers and stamp on their own part numbers. Now they are called "replacements" so they raise the price a lot.
The transistor is an NPN emitter-follower. When the base is about +0.65V then the transistor turns on and its emitter voltage follows the base voltage but is less 0.65V.
When the input signal voltage goes negative then the emitter base is reverse-biased and nothing happens. When the base reaches about -7V then the emitter-base junction has avalanche breakdown and conducts. If the generator has a low output impedance then it drives the base and emitter more negative.
 
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