Linear power supply series transitor

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Hi,

In linear power supplies there is a transistor (usually darlington) that is current-controlled depending on the output voltage.
The problem is I cant seem to find the Vout - current-to-darlington-base equation.

Do I operate it in the linear region?
When Vout < Vout_desired do I saturate it?
What is the range of current in its base? (i.e. 100uA to 500uA)?

Thanks
 

A Darlington acts basically like a single transistor except it has a higher base-emitter voltage and much higher current gains (typically in the neighborhood of 1000 or more) so requires much less base current (equal to the output current divided by the current gain).

It would operate in the linear region for a linear regulator.

Don't understand your question about saturating the transistor. (?) You only saturate it if you want the output voltage to be near the input voltage, otherwise you just adjust the base voltage to give the output voltage you want (for an emitter follower output circuit). The base current will then be whatever it needs to be based upon the load current.
 
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Thank you for your answer.

What I need to know is, how do I calculate the base current depending on the Vout.

i.e. when Vout=15V and I want Vout = 16V, I have an error of 1V. How does this translates to additional or less current to the base? Ib= K*(Vdes-Vout). How I calculate K?
End when Vout=Vdesired does that mean that Ibase = 0?

Best
G
 

You don't need to calculate the base current depending on Vout. You need the base current for I out.
Say you want 16V out and the Vbe of the transistor is 1.2V then you need to set the base to 17.2V to get 16V out.
If your load at the emitter draws 1A at 16V , then your base current (assume hfe of 1000) will be about 1mA.
Whatever the base is connected to must be able to deliver 1mA.If there is no load current (emitter open cct)
then there will be no base current.
Neddie
 
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One more question:

How can a adjustable linear supply be controlled down to 0 volts?
I see many commercial supplies claiming that.
 

GeoAVR said:
One more question:

How can a adjustable linear supply be controlled down to 0 volts?
I see many commercial supplies claiming that.
Click to expand...
In your design with a Darlington emitter-follower output you just adjust the base voltage down to 0V.
You seem to be overly concerned about base current but it's the base voltage that determines the output voltage.
The base current will take whatever it needs to supply the output current (in an emitter-follower configuration).
You just need to make sure that whatever is generating the base voltage can provide more than enough base current needed for the maximum output load current. There's no need to be exact. The base will take only what it needs.
 
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I am trying to figure out this simple configuration.



The idea is, controlling the V2 (vref)voltage to control the Vout.
I am thinking of using a microcontroller and a DAC to create the V2.

How do I select the resistors values, is every resistor needed?
Is the opamp + - inputs ok, or I need to invert the signals?
When I want to go down to Vout=0V the circuit produces very large Vout. I think its because the opamp saturates.

Thank you.
 

That circuit does not allow the output voltage to go below the reference voltage, V2 ( achieved when R7 = 0Ω).

How are you going to power the op amp?

The op amp polarity is correct as it provides negative feedback for the loop (due to the signal inversion of Q2).

The loop will likely be unstable and oscillate due to the added loop gain from Q2, so some compensation will be required, such as a capacitor from collector to base of Q2.

- - - Updated - - -

Here's a way to adjust to 0V out:
Connect a pot across the V2 reference and the connect the wiper to the (-) input of the op amp (removing the connection shown of course). That will allow adjustment of the output voltage between 0V and 20V for the values of V2, R7 and R8 shown. The voltage divider action of R7 and R8 determine the maximum voltage.
(Note that to go to zero volts the op amp must be a rail-rail type).
 
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Thank you for the answer.

The R7, R8 will remain constant. I will not use a pot to change the voltage but I will control the V2 from 0 to 5V with a DAC and microcontroller.

Basically, because of the 1/10 voltage divider whatever i put at V2 gets multiplied by 10. ie. Vout = 10*V2
Q2 is used to draw base current from Q1 depending on the "error" output of the opamp.


It works great from V2 = 0.5 to 5V but bellow 0.5V the opamp saturates to 15V(supply rail).
If I decrease the R5 I think it works better and it makes sense because it allows more current to be drawn from Q1 base. Also the opamp doesn't saturate.

Still, I don't have any equations of how to calculate all this and find the limitations.

ps.
I think I will use a rail to rail opamp from 0 to 5 or 10V from a secondary power supply. With a 14bit DAC I will have a theoretically resolution of 300uV at Vout.
 

R4 provides the base current for Q1 so its maximum value is (V1-Vout-0.7v) / (Iomax/Q1beta).

R5 is not needed unless you need a compensation cap between Q2's collector and base. If so it needs to be small enough so that the base voltage at Q1 due to R4 can be pulled to below about 0.7V if you want a 0V output.

R6 is just to limit the output current of the op amp and also help with the compensation for the above cap. It should be small enough to provide sufficient base current to Q2 and large enough to limit the op amp current below its rated value.

R7 and R8, along with V2 determine the output voltage as you know. The values of R7 and R8 are chosen so that the op amp input bias current,as given in its spec sheet, doesn't significantly affect the regulated output voltage value.
 
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R4 is much too low, at minimum output (Vc/2) the base of the series transistor will be at 30.8V, as Vc is 60V this leaves 29.2V to be dropped across R4, P = 29 ^2/100 ~ 10W, also Q2 has to take 60/200 ~ .3A when there is no load. I would replace Q1 with a Darlington pair and make R4 10K, you will also get a higher gain from Q2. This will also reduce the minimum output voltage to 100/10,100 X 60 ~.6V?
Frank
 
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