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Transistor Question

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fedail

New Member
we know the NPN transistor cut the wave from down when Vb<=VE+0.6

suppose the signal input is 5VPP
when i mesaure it by oscilscope by DC the range from 0 to 4.v
but by AC Measure is -1.2 to 3.2
 

crutschow

Well-Known Member
Most Helpful Member
When you measure with a DC connection you are measuring the actual waveform voltage values to common.

When you measure with an AC connection you are placing a capacitor in series with the measurement which blocks the DC in the waveform. This means that the average voltage must equal zero. The voltage will thus shift to make the average zero. The amount of shift depends upon the waveform shape and duty cycle.

The AC measurement mode is typically used when you want to look at a small AC signal that is riding on a large DC component.
 

fedail

New Member
i am still not understand for this osciliscope states

also i have question about transistor why it is noy cut the negative signal of we add dc signal
 

Sceadwian

Banned
Because the DC 'lifts' the AC signal so that while it still varys it may not vary down bellow the cutoff point of the transistor.
 

Sceadwian

Banned
AC measurements use a capacitor inline with the signal, assuming the waveform is symmetric it will appear to average 0 volts. When you measure it via DC you'll see any DC signals as well so the whole thing will be lifted up (or down)
 

ericgibbs

Well-Known Member
Most Helpful Member
assuming the waveform is symmetric it will appear to average 0 volts ?? why ?
hi,
Do you have a circuit diagram to post , showing where you are measuring the waveforms.?
 

Sceadwian

Banned
fedail I mean it will be centered around 0 volts, even if it has a DC component. This is what AC coupling means.
 
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