# Regulated power supplies current limiting

Discussion in 'Homework Help' started by Saviour Muscat, Jul 9, 2015.

1. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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Hello
I have a problem regarding this circuit , I don't know clearly how it works especially bolded part the quoted part of the book when says this "When the load current is greater than 600mA, the voltage across R4 is more than 0.6V, which turns on Q3. The collector current of Q3 flows through R3 and decreases the base voltage to Q2"
Thank you for your continuous help
SM

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2. ### WBahn Moderator

Mar 31, 2012
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First, make sure you understand how the circuit works without Q3 and R4.

Do you understand how Q2 is controlled by the base current that comes through R3?

Once you have this, then you can see that we can control the output current by bleeding off current coming out of R3 and sending it elsewhere. The voltage regulation action takes place by shunting this current through Q1 and down through the zener diode. The current regulation (or limiting) action takes place by shunting it through Q3. In either case we starve Q2 of base current.

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3. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
82
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" The current regulation (or limiting) action takes place by shunting it through Q3. In either case we starve Q2 of base current"
I came to a conclusion please correct me if am wrong, R3 and Q3 are like as a potential divider, Q2 base connected between them,
When Q3 is at cutoff(Vce resistance is maximum) is there more voltage across base of Q2, on the other hand when Q3 starts conducting(Vce resistance drops) there is less voltage across the base of Q2

4. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
82
0
I think my statement is wrong, but I can't get how the voltage on the base of Q2 is decreased, I understood the remaining text said on the book!

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
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In this case it is better to think in terms of a Q2 base current. When Q3 starts conducting the Q3 collector current also starts to flow. So Q3 steals the current that normally will go to Q2 base.
Or we can think in terms of a Q2 Vbe. Simply Q3 short the Q2 base-emitter junction with Q3 emitter and this reduces Q2 Vbe voltage .

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6. ### Saviour Muscat Thread Starter Member

Sep 19, 2014
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ok I understood many thanks WBahn and Jony130