How do you measure VB in a Common Emitter Amplifier circuit with a bypassed RE2?

Question

I've got 300mV at 1KHz going in, but I only need to measure the DC values. These values are VB, VE, VC, and VCE. The main one I don't know how to measure with my meter is VB. I'm using a transistor with Beta of 100 and the specific transistor 2N3904.

I calculated a value of 3.8V for VB, but I just can't seem to get the right measurement.

Any help at all will be greatly appreciated. I will choose a best answer.

Answer 1

By passing RE2, is that a capacitor that is bypassing this resistor? If so, then for DC measurements, it is of little consequence. VB would be measured from the base to ground. The current flow through the bias network that connects to the supply voltage to ground sets the current through those 2 resistors. The voltage drop across the Base point junction of the resistors will set the bias voltage for the transistor. The thing about the 0.7 volts between the Base and the Emitter is a RELATIVE voltage. It may or may not be a voltage above the 0 volts base line. It could be a negative voltage, so long as the Base is more positive than the Emitter. A Base voltage of -0.5, and an Emitter voltage of -1.2 still produces a "positive" 0.7 volts since the Base voltage is more positive than the Emitter. My point, you don't need exactly 0.7 volts for the transistor to operate. What ever voltage you are getting from Base to ground is your VB voltage. I hope this answers your question.

Answer 2

Where's RE2?



Bypassing any of the resistors can definitely change your bias voltages. If you bypass the emitter resistor (which is what I think you must be talking about since you have an E after the R), then VB is going to be very close to 0.7V (you have to calculate the current going through the resistor to get an exact value and I can't remember the exact equation off the top of my head - 0.7 is usually a pretty good estimation though).



You might think the bias resistors can be used as a simple voltage divider, but you really have a parallel circuit with one branch running through the base to the emitter. The bias resistors will determine how much current is running through each of the bias resistors and from the base to the emitter, not necessarily the actual bias voltage.

Answer 3

Try this applet and plug in a few values.





7. Common Emitter Amplifier

This is basic circuit for common emitter npn BJT amplifier. This applets plots voltage gain, input resistance, frequency response, time response, and also calculate Q-point information.