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I am trying to understand what is happening with this linear regulator circuit.
Is it right to say: It would be very hard for this circuit to manage large changes in line voltages because there wouldn't be enough voltage dropped across R3 if the line voltage increases.
Or am I thinking about this in the wrong way?
What solutions should I be looking at to improve the line regulation of a power supply with this configuration?
Thanks
The circuit is very dependent on the applied voltage.
In particular the zener voltage will vary considerably with bias current which is of course directly proportional to the supply voltage and also changed by whatever current TR2 is driving through R3.
Similarily with higher input TR2 has to pull more current for a specific output voltage, but since the base current for TR2 is through the sense chain, that will also affect the output voltage.
Furthermore, since the current through the zener changes markedly with applied voltage, the power dissipated and temperature of the diode will also vary greatly. This again affects it's voltage.
"What solutions should I be looking at to improve the line regulation of a power supply with this configuration?"
I'm tempted to say "Throw it away and use a proper adjustable regulator."
However as a minimum you need a proper voltage reference not a plain zener, and you need to change TR2 to something that does not draw so much base current from the sense chain, perhaps a darlington.
I assume this question is rather academic than practical. So it worth talking about improvements of the circuit.
First step is to remove direct dependence of zener current (defined by R4 and Vin) from Vin. Connect R4 to the stabilized Vout:
simulate this circuit – Schematic created using CircuitLab
But even after that, variation of zener current remains due to variation of R3 current (due to input voltage change). So, second change is to remove R3 current from zener, using fully differential stage:
Another advantage of this diff circuit is that Vbe of Q2 and Q3 are substracted, so if Q2 and Q3 are of same type and of same temperature, Vbe is excluded from equation for output voltage.
One could optimize this design by increasing the base and zener current. But it is hardly worth it considering chances you will design something better than a good LDO are slim to none.
The other way around. It would be the lowest voltage input that causes the current regulator in the base of the series regulator to stop conducting. It is called the dropout voltage. This circuit is typically what is inside of your standard 3 terminal IC voltage regulators (7805,7812ct,7815...etc) btw.
