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Can I stack these on top of each other http://www.ti.com/product/TPS7A47/datasheet without degrading AC performance? Will the output impedance of the lower ones affect the output impedance of the higher ones since they are serving as the new ground connection or does it not matter because it is an active circuit? Like this enter image description here

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  • \$\begingroup\$ AC performance? Those are DC linear regulators. \$\endgroup\$ – C_Elegans Mar 4 '18 at 1:00
  • \$\begingroup\$ The output of that device is limited to one volt below the input, so stacking them would produce a lower voltage. \$\endgroup\$ – Russell Borogove Mar 4 '18 at 1:06
  • \$\begingroup\$ Updated my question for clarity. \$\endgroup\$ – coinmaster Mar 4 '18 at 1:34
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    \$\begingroup\$ Often it is better to use a zener diode on the ground pin. Connect a 1 K (per volt) resistor from ground pin to output so zener is always biased, and output has a minimal load. \$\endgroup\$ – Sparky256 Mar 4 '18 at 1:38
  • \$\begingroup\$ But how does this affect the output impedance vs frequency of the regulator? In a passive filter the impedance to ground is the impedance of the regulation. Does this apply in any way here? Doing this will be pointless if I lose performance. \$\endgroup\$ – coinmaster Mar 4 '18 at 1:41
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Cascoded (stacked) circuits are not something unheard of. Here is an example from Texas Instruments,

enter image description here

However, to have a stable (elevated) ground for the upper regulator U2, the bottom regulator U1 should be able to source AND sink the current, which is not the case with TPS7A47 nor LM1084.

Good thing is that the ground current for TPS7A47 is no more than 1 mA, so you might want to load the U1 with more than 1 mA extra load, so you will have some regulated output both ways. Obviously you would need proper load capacitance on U1 and U2.

Obviously the output noise will likely go up 40%, because the noise of U1 will be added. Also you might have overall stability issues, due to modifications to loop transfer function due to extra impedance in ground point. Here is one example when this kind of circuit was not successful.

The concept of "output impedance" is not used in LDO regulator technology, so you would need to look for quality of regulation (ripple rejection, load transient response, etc.) I don't think they will be affected much if the overall circuit will be stable.

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Can I stack these on top of each other without degrading AC performance?

Consider the regulator at the top of the stack. It will naturally pass a small current through its GND connection. This can be tens of uA to a few mA (like the LM78xx type). For your regulator the GND current is 0.5 mA to 6 mA typically so it is significant.

This type of regulator also cannot prevent its output voltage rising if there is a load connected to a higher voltage and this is the case with the regulator at the bottom of the stack. The upper regulator is pushing 0.5 mA to 6 mA into the lower regulator's output and this will cause problems unless the lower regulator has a load resistor fitted to ground that can balance this current out. So you need a balancing resistor: -

enter image description here

Will the output impedance of the lower ones affect the output impedance of the higher ones since they are serving as the new ground connection or does it not matter because it is an active circuit?

Yes, the output voltage at the top of the stack is affected by the stability of output voltages at the bottom of the stack. In my words above I gave the solution for the DC scenario but, for AC scenarios you have to live with this problem and ensure output capacitors (and balancing resistor) are all present.

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I'm trying to regulate the power supply of an audio circuit that is above the voltage rating of the regulator IC I want to use.

Be careful about voltages during turn-on and turn-off. For example, this is a 7805 sitting on top of a zener, which turns it into a 7815 which can take 10 more volts at the input.

schematic

simulate this circuit – Schematic created using CircuitLab

However, if the output caps are large enough, at power-on, the input voltage will rise faster than the output voltage, and the max rating of the regulator will be exceeded. Also in case of output short circuit, the max voltage rating will be exceeded, so the regulator's internal protections no longer work.

This schematic is better:

schematic

simulate this circuit

Here a very simple BJT/Zener pre-regulator brings the voltage down.

Regulator input/output caps depend on reg's datasheet. R1 should have proper value to give Q1 enough base current (you can also use a darlington or a MOSFET).

Filtering the zener reference with a cap isn't necessary, but it'll give you a bit of extra HF PSRR. Global PSRR will be improved by 20-30dB relative to the original regulator, as Q1 will filter the input voltage.

Output impedance, noise and other characteristics like load response will be the same as your regulator.

Also this splits the dissipation in two. You should try to set the zener voltage for equal dissipation in transistor and regulator. The same amount of power dissipated in two components is easier to get rid of than in one component (you get 2x the area of silpad).

The regulator's internal protections still work and will protect Q1.

But if voltage allows, you can use a LM317HV, it's the simplest option.

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  • \$\begingroup\$ 2N3904? Sounds a bit whimpy. \$\endgroup\$ – Oskar Skog Mar 4 '18 at 11:56
  • \$\begingroup\$ I put D44H11 in the part number! argh, I'll edit it. \$\endgroup\$ – peufeu Mar 4 '18 at 12:21
  • \$\begingroup\$ I think you put it in the component designator field. :) \$\endgroup\$ – Oskar Skog Mar 4 '18 at 12:23

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