Author Topic: Feedback follower circuit  (Read 710 times)

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Offline Burning CircuitsTopic starter

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Feedback follower circuit
« on: May 08, 2020, 06:43:38 pm »
Hi,
I have a question regarding the attached circuit (a power supply).
The LT8612 is the switching pre-regulator of the two linear LT3081 regulators.

I have understood the purpose of Q3: it's a feedback follower, that keeps the output of the pre-regulator 1,7 V higher than the output of the LT3081 (VOUT).
The voltage at the FB pin of the LT8612 is 0,97V

I don't understand the mathematical relation between the resistors value and the 1,7 V value.

Someone can help me?
 

Offline unitedatoms

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Re: Feedback follower circuit
« Reply #1 on: May 08, 2020, 07:42:55 pm »
With reference voltage of feedback known from datasheet 0.97V and resistor 4.99k the collector current is 0.19 mA.
The high beta transistor makes base current negligible. So then dividing pair of 100k resistors is loaded to very low current of base. Divider is then equal: Makes transistor to open up to necessary 190 uA at doubled opening voltage.

So
1.7v=2*(voltage pn + o.ooo19*1000 ohm). Where voltage pn is 0.65v for slightly opened silicon transistor.
Interested in all design related projects no matter how simple, or complicated, slow going or fast, failures or successes
 

Online mawyatt

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Re: Feedback follower circuit
« Reply #2 on: May 08, 2020, 08:13:21 pm »
A quick circuit analysis shows that

Vx = Vo + 2(Vfb*R1/R2 + Vbe)

Where Vx is the voltage in question, Vo is the output, Vfb is the feedback voltage of Lt8612 (0.97V), R1 =1K, R2 =4.99K, and Vbe is base emitter voltage of Q3 PNP.

Hope this helps.

Best,

Mike
« Last Edit: May 08, 2020, 08:16:10 pm by mawyatt »
Curiosity killed the cat, also depleted my wallet!
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