High Value Bias Resistors/Emitter Current of transistors (Help?!?!) - Page 1
Forum section: electronics / beginners

author: shanecy
author_rank: Contributor
timestamp: July 11, 2013, 08:16:56 pm
content: I'm reading my textbook and it tells me the following information. However I do not understand what they are trying to say and why it would avoid this problem. Why is this? What does this have to do with dissipation and efficiency?"Using high value bias resistors results in the emitter circuit loading down R2. Using low bias resistors would avoid this issue."Here is a picture of the circuit I constructed along with voltage and current readouts.

author: c4757p
author_rank: Super Contributor
timestamp: July 11, 2013, 08:24:23 pm
content: Quote from: shanecy on July 11, 2013, 08:16:56 pmYou have a poorly written textbook.With a resistor on the emitter, you get a voltage follower effect, causing the voltage on the base to appear at the emitter, minus the VBEdrop. Ohm's law then says the emitter resistor will draw current determined by that voltage. When a transistor operates in the linear region, the base current will always be approximately the collector current divided by the DC current gain. This means that the emitter current (which also flows through the collector) will increase the base current proportionally, and eventually you'll reach a point where the current drawn through the base resistors causes a substantial voltage drop. If they are smaller, they'll lose less voltage for the same current.If you account for the base current when choosing the base resistors, you won't have to worry about this.

author: shanecy
author_rank: Contributor
timestamp: July 11, 2013, 08:42:32 pm
content: So if i understand this right.....Having high value base resistors will cause less heat dissipation on R2 but will put the load on the emitter resistor.Having low value base resistors will cause more heat dissipation on R2 but will not put the load on the emitter resistor.I see a good and a bad side to both scenarios. I would think we would want the load to be on R2 since we don't want the emitter current controlling the collector current. But if we do this we have a increase in power "heat" dissipated from R2 which now we have to resolve. How do we find a medium? Do we just find a way to optimize both without creating this unbalance.

author: c4757p
author_rank: Super Contributor
timestamp: July 11, 2013, 08:59:01 pm
content: It's not about heat dissipation, it's the fact that if you assume the base resistors are unloaded (like they seem to), but then you load them significantly, the transistor won't be biased the way you planned for and might not work right.Here's a usual way to bias a common-emitter amp which takes base current properly into consideration:0. The power supply voltage is VCC.1. Choose the collector voltage VC(typically VCC/2) and current IC(your choice).2. The collector resistor is therefore (VCC- VC)/IC.3. If you will place a capacitor on the emitter, choose the emitter resistor to drop 5% to 10% of VCC, at the same current as the collector resistor. Otherwise, select a gain A for your amplifier, and A = RC/RE(so RE= RC/A).4. Calculate the base current IB, which is IC/hFE. hFEis the DC current gain of the transistor, found in the datasheet.5. Calculate the base voltage VB, which is VBE(about 0.65V) + VE.6. The lower base resistor has VBacross it and should have about 5 IBthrough it, so RBL= VB/(5IB).7. The upper base resistor has (VCC-VB) across it and should have about 6 IBthrough it (one more because it supplies the actual base current), so RBH= (VCC-VB)/(6IB)There is no reason for a signal amplifier like this to have any substantial heat dissipation. Keep all the currents low and you don't have to care about heat.

author: hlavac
author_rank: Frequent Contributor
timestamp: July 14, 2013, 09:26:40 pm
content: It may help you to understand if you realize the base-emittor junction is a diode in forward direction, leading thru emitor resistor to ground.It effectively puts the emitter resistor (multiplied by beta due to amplification of the transistor) and the base-emittor diode in parallel with the lower resistor of the biasing voltage divider, making the base biasing voltage lower than it would be with the biasing divider unloaded...

author: c4757p
author_rank: Super Contributor
timestamp: July 14, 2013, 09:32:30 pm
content: You know.... it never occurs to me that people don't always realize that. Thanks