# Question Regarding Current Boosters (Electronic Principles)

Discussion in 'Homework Help' started by SkiBum326, Jun 18, 2014.

1. ### SkiBum326 Thread Starter Member

May 16, 2014
33
0
Hi Everyone,

I'm working through Electronic Principles 7th ed by Albert Malvino. I have a quick question about material presented on page 973 regarding current boosters. To offer better context, I took a picture of the diagram.

The book states that as the load current increases, the current increases through the LM78xx, producing more voltage across the current-sensing resistor. In this way, the outboard transistor produces the bulk of any increase in load current above 1A, with only a small increase in the current through the 78xx.

Is this statement correct? I thought that once the base-emitter voltage reaches ~.7v, it remains constant. I thought what would happen is that as the load current increases and as the current through the LM78xx increases, more current is drawn from the base of the transistor. Consequently, due to current gain, the majority of the current is contributed by the transistor. Nevertheless, the current through the resistor and consequently the voltage across it remain the same.

Any thoughts?

2. ### SkiBum326 Thread Starter Member

May 16, 2014
33
0
I was unable to create the thread with the full title. Could an admin change it to "Question Regarding Current Boosters (Electronic Principles)" please, thanks.

3. ### SkiBum326 Thread Starter Member

May 16, 2014
33
0
Here's the diagram.

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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115

Both of you are right. You assume that BJT is "ideal" one, and work as a current controlled current source.

But the author simply assume the real-world bipolar transistor was use in this circuit. And if so, the Vbe voltage is not constant but will very with the Ic current. An increase in base-emitter voltage Vbe by about 60 mV will increase the collector current Ic by about a factor of 10.

Last edited: Jun 18, 2014
5. ### #12 Expert

Nov 30, 2010
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Click on the red triangle (upper right corner of the post) when you want a moderator to intervene.

6. ### #12 Expert

Nov 30, 2010
16,665
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Summarizing what Jony130 said, your basic mistake is the idea that there is a turn-on voltage for a bipolar transistor that stays the same when the current changes. It doesn't. The Volts, base to emitter, changes by about .06 volts for a collector current change of about 10 to one. This seems a very small change, but it is dependable.

7. ### crutschow Expert

Mar 14, 2008
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3,362
More generally, the base-emitter junction looks like a forward-biased diode with its typical logarithmic relationship between voltage and current. This current then appears to be amplified by the beta or Hfe of the transistor at the collector-emitter terminals. This non-linear relationship gives an apparent cutoff of the transistor at a Vbe of about 0.6V-0.7v depending upon what low level of collector current is considered "cutoff".

8. ### daviddeakin Active Member

Aug 6, 2009
207
27
Yes it is essentially correct.
Yes, more or less the same. The transistor may turn on at around 0.6V and be fully saturated around 0.7V, so the voltage across the resistor will always be somewhere within this window of operation. (Except when the current is so small that the transistor cannot be turned on, of course).

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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This may require further qualification. Simply having a |VBE| of >0.7V doesn't necessitate saturation condition. I'm not even sure of the circumstances in which the transistor could go into saturation with the topology shown.

The attached image shows the variation in VBE vs Ic for a medium power PNP transistor series BD136,138,140. Shot comes from the Fairchild data sheet.

• ###### BD136-140 Vbe shot.jpg
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Last edited: Jun 21, 2014