# Op amp volt reg transfer function

Discussion in 'Homework Help' started by laguna92651, Jan 23, 2010.

1. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
Is this transfer function for the attached voltage regulator correct? I was expecting to have to include the transistor, but it doesn't seem to have any affect on the transfer function.

The regulator is to output 6.1 volts for loads ranging from 0 to 200 mA. What current should I use to calculate Rsc? The TIP29 has a min gain of 40 at 200mA and Vce=4v.

Thanks

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2. ### thyristor Active Member

Dec 27, 2009
94
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Your transfer function is indeed correct. The reason that the feedback is taken from the output of the transistor (its emitter in this example) is specifically to be able to exclude the transistor from the transfer function which, as you can see, clearly happens in your illustration.

So any Vbe loss is taken into account by the fact that the output, after being fed back by R1 & R2, will adjust itself so that the two inputs to the op-amp are both at 5.1v as set by the zener.

3. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
Any suggestions on calculating Rsc, I assume this is just to limit the current if the output is short circuited?

4. ### thyristor Active Member

Dec 27, 2009
94
0
Calculate the voltage drop across Rsc. Clearly, when RL is at maximum, with 200mA through it, what is the maximum value of Rsc such that Rsc's volts drop + the RL volts drop = 10v (ie: the supply voltage).

If the total comes to more than 10v, what will happen to Vo?

5. ### laguna92651 Thread Starter Active Member

Mar 29, 2008
101
0
Rsc=(10-6.2)/.2 = 19Ω, are we assuming Vce=0v?