Relationship between transistor base current and emitter resistor voltage  - Page 1
Forum section: electronics / beginners

author: Chris.M
author_rank: Contributor
timestamp: May 05, 2013, 01:10:30 pm
content: Hello All,I would like to understand the relationship between the base current and voltage drops across the base resistor and emitter resistor in the second circuit attached.I think I have a grasp on whats happening in the first circuit attached.The supply voltage must be dropped across both Rb and Vbe, so since we know Vbe to be 0.7 volts, then the remaining supply voltage must be dropped across Rb.VRe= Vcc - Vbe = 5.6 - 0.7 = 4.9VUsing ohms law we can determine the current through Rb and therefore through the base of the transistor.Ib= VRb/RRb= 4.9/1000 = 4.9 mANow we can determine Ic to be 100*Ib = 490mAHowever for the second circuit, I don't understand why there is a large voltage drop across the emitter resistor and only a small voltage drop across the base resistor.Would someone be able to explain what are the relationships here, and what's going on practically in the circuit and how I could use ohms law to calculate the currents and voltages.Regards,Chris

author: Rufus
author_rank: Super Contributor
timestamp: May 05, 2013, 01:28:11 pm
content: In the second circuit the emitter current in Re will be HFE + 1 times the base current in Rb.The circuit will stabilize at a base current Ib where (Ib * Rb) + Vbe + ((HFE + 1) * Ib * Re) = VCC.

author: MikeK
author_rank: Super Contributor
timestamp: May 05, 2013, 01:42:05 pm
content: Well, since current flows through Re, it has to have a voltage across it, right? It certainly can't have the 490mA through it that you calculated for the first circuit, because that would be a voltage drop of 490V. Certainly not possible with a 5.6V supply.Take the extreme hypothetical example where Re drops all of the supply voltage. It can't, but let's say it does. Ohm's Law says that the current through Re then is 5.6mA. Well, if the transistor has a beta of 100 then the most current through Rb would be 0.56mA, right?That's just to give you an understanding of the why. Rufus gave you the real calculation.And another "why" thing to realize is that Re reduces the collector current. If this current is reduced then the base current must also drop, because of the relationship between Ib and Ic.

author: Chris.M
author_rank: Contributor
timestamp: May 06, 2013, 10:18:53 am
content: Quote from: Rufus on May 05, 2013, 01:28:11 pmThis is the connection I was looking for and mathematically makes sense. Looking at the equation above, it is clear that most of the voltage is dropped across Re because of ((HFE + 1) * Ib * Re).I still have some questions that I'm hoping to clear up.Is the +1 in the (Hfe + 1) because the emitter current is Hfe * Ib and the +1 is for the Ib current flowing out of the base and into the emitter adding to Ic?So the transistor has two independent series paths where the supply voltage must be expended?Path A:From the supply through to Rb, then into the transistor base, through the base emitter diode, then through Re and back to the supply. So in this path, because of "transistor action", the voltage across Re is determined by ((HFE + 1) * Ib * Re).Path B:From the supply through the transistor collector and out of the transistor emitter, then through Re and back to the supply.If the above is correct, that there are two independent paths then is it safe to say that you came up with (Ib * Rb) + Vbe + ((HFE + 1) * Ib * Re) by following all of the voltage drops in the first path? Or is that equation from something else?I'm currently reading Tab Electronics Guide To Understanding Electricity And Electronics, and in the transistor chapter, if I understood it correctly, it states that the voltage on the emitter will rise to the same voltage as the supply minus a small drop across the transistor. Is this because the base current has put the transistor into saturation or because from the equation we can see that most of the supply voltage will be dropped by ((HFE + 1) * Ib * Re) and Vbe, so therefore Vrb is ignored?Rufus, what do you mean by the circuit will stabilize at a base current Ib ...I always thought that we define what the base current will be, which then determines what the collector current will be. By stabilize, do you mean there is something else going on behind the scenes? Can you elaborate a little on this?Quote from: MikeK on May 05, 2013, 01:42:05 pmAgree.Quote from: MikeK on May 05, 2013, 01:42:05 pmIndeed this is true. That's a good way of thinking about it.Quote from: MikeK on May 05, 2013, 01:42:05 pmYes, this also makes sense.Thanks Rufus and MikeK. Both of your posts have helped to understand what's going on.

author: Rufus
author_rank: Super Contributor
timestamp: May 06, 2013, 01:22:21 pm
content: Quote from: Chris.M on May 06, 2013, 10:18:53 amThe collector current is HFE * Ib, that is the definition of HFE. The collector and base current both have to flow through the emitter.Quote from: Chris.M on May 06, 2013, 10:18:53 amIn this circuit the base current is determined by the voltage across Rb but that voltage depends on the voltage across Re. There is only one value of base current where the voltages add up to the supply voltage. That can be called the 'operating point' of the circuit, the emitter resistor can also be said to provide negative feedback.

author: Chris.M
author_rank: Contributor
timestamp: May 06, 2013, 10:33:25 pm
content: I could see that Ib was influenced by Re, and with ((HFE + 1) * Ib * Re) I can now see how to calculate that influence.Got it, so the operating point or quiescent point is the Icq and Vceq you get once the collector current and voltage rise to Icq and Vceq and settles down (what you said as stabilize), for the given base current.Thanks, this makes sense now.