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I am trying to understand what is happening with this linear regulator circuit.

enter image description here

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

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  • \$\begingroup\$ Your image isn't working... \$\endgroup\$ – Ron Beyer Dec 21 '17 at 16:43
  • \$\begingroup\$ Explain why there would not be enough voltage dropped across R3 when the input voltage increases. Try to work out what changes when the input voltage increases. The voltage across R4 would increase as the voltage across Dz1 would remains almost constant, right ? So the voltage across R3 and/or Vce of Tr2 would increase. \$\endgroup\$ – Bimpelrekkie Dec 21 '17 at 16:49
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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.

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  • \$\begingroup\$ Thanks, this really helped. However, I'm struggling to fully understand what is meant by "sense chain". \$\endgroup\$ – D.MacLennan Dec 21 '17 at 18:18
  • \$\begingroup\$ @D.MacLennan the voltage is sensed by the chain R1, VR1, and R2. \$\endgroup\$ – Trevor_G Dec 21 '17 at 19:38
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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.

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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.

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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:

schematic

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:

schematic

simulate this circuit

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.

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