# Regulator Output

Discussion in 'Homework Help' started by gisdude, Feb 24, 2015.

1. ### gisdude Thread Starter Member

Oct 30, 2008
16
0
Hi all,
I saw this circuit on a website and was wondering if you could guide me on this. The question is "what is the expected output voltage at RL?" The answer is 5.5V. How?

I just attachedppicture I. drew up.

Thanks,

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2. ### MrChips Moderator

Oct 2, 2009
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3,282
The base-emitter voltage is approx. 0.7V.
Hence the voltage at the emitter is 6.2 - 0.7 = 5.5V

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3. ### gisdude Thread Starter Member

Oct 30, 2008
16
0
Is that .7V for ANY transistor?

Thank you, BTW

Last edited: Feb 24, 2015
4. ### MrChips Moderator

Oct 2, 2009
12,228
3,282
This is true for any PN junction made from silicon. But it also depends on the forward bias current.

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5. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
For silicon bipolar junction transistors (BJT), this is a commonly used approximation. Small signal transistors generally operate at higher base-emitter voltages than power transistors owing to the much larger area usually used in power transistors for heat management. The voltage drop typically increases by about 60mV for every order of magnitude (10x) increase in base-emitter current (and, in the active region, the corresponding collector current). So if a given transistor had a collector current of 100 mA at Vbe=0.7V, it would have less than 100 uA at 0.5V and nearly 10 A at 0.9V. As a result, for most purposes, it is reasonable to treat the base-emitter drop as a constant as long as the transistor is in the active region and values of 0.6V, 0.65V, or 0.7V are the most commonly used values and it almost never matters which value you use (generally, but not always, if it does matter, then the circuit was poorly designed).

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6. ### Papabravo Expert

Feb 24, 2006
10,021
1,757
Provided the resistor, with no value specified, allows sufficient zener current to fix the base at 6.2 volts. If the resistor is too big and the zener diode does not have enough current then it will not draw current from the base of the transistor, and the transistor will be cutoff, dropping the output to a low value. You want the sum of the current through the zener and into the base of the transistor to be equal to the current through the resistor. Without sufficient reverse bias there will only be a small leakage current drawn from the base of the transistor.

Last edited: Feb 25, 2015