4
\$\begingroup\$

Rod Elliot has a dual bench power supply project on his web site (project 223, see schematic below). I'm trying to understand his method for adding a low voltage negative rail without using an additional secondary winding. My questions are - What is the purpose of C2, C3 in series with the secondary winding and the second bridge rectifier creating the negative rail? Would this circuit work without them? Finally, does this circuit work because the negative voltage rail is only being used at small voltage and small current, or would a similar circuit work for a symmetric split rail supply?

Dual Power Supply - Rod Elliot Project 223 Dual Power Supply Elliot Sound Products

\$\endgroup\$
0

3 Answers 3

7
\$\begingroup\$

I haven't searched for Rod Elliot's article (and you didn't provide a link) but the design looks very like it has been taken directly from the LM317 datasheet.

The LM317 ajusts the Vout to 1.25 V above the voltage on ADJ. That means that with a single-rail supply the variable voltage output can only go down to 1.25 V. If you want to be able to adjust to 0 V then you need to somehow be able to pull ADJ down to -1.25 V.

  1. What is the purpose of R1, R2, C2, C3 in series with the secondary winding and the second bridge rectifier creating the negative rail?

The resistors, R1 and R2, give some current limiting and this may have been added to limit current into Zener D10. More on this later. C2 and C3 provide isolation between the positive and negative rail rectifiers.

  1. Would this circuit work without them?

No. Imagine they're left out. Now look what happens when the top of the secondary winding goes positive:

enter image description here

Figure 1. If R1, R2, C2 and C3 are omitted the secondary will be short-circuited.

  1. What is the purpose of D10?

D10 is a simple voltage regulator to limit the voltage into U2 to -15 V. Power dissipation in D10 is limited by R1 and R2.

  1. Finally, does this circuit work because the negative voltage rail is only being used at small voltage and small current, or would a similar circuit work for a symmetric split rail supply?

This circuit is specifically designed to overcome a limitation of the LM317. It could provide a larger (more negative) voltage but the power out will be limited by R1, R2, C2 and C3. Very often this is enough when the current required by the negative rail is modest in relation to the positive rail.

\$\endgroup\$
1
  • \$\begingroup\$ "I added the link". \$\endgroup\$
    – RobT
    Commented Dec 29, 2023 at 21:29
3
\$\begingroup\$

In this particular circuit, the LM317 used has a minimum output voltage of 1.2 V. A negative rail is used to bring the regulated voltage below that, so the output can get all the way down to 0 V.

The purpose of C2/3 is to support a DC shift to the lower rail's bridge rectifier, while still transmitting AC.

The positive rail is the 'power' rail. The negative rail is only used for biassing and control, so has been designed accordingly, for very low current. Regardless of C2/3, R1 and R2 are limiting the current to the negative rail, as part of the D10 shunt pre-regulator.

If you want a dual power rail supply, then omit R1/2 and D10, and increase C2/3 until they pass enough current. You can get 1000 uF, even 10000 uF caps. However, most people designing a dual rail output supply would use a centre-tapped secondary on their transformer. It barely changes the cost of a transformer, and big electrolytics can get expensive. The only reason you might go the big cap route for high current is if you already had the transformer, and didn't want to buy a new one.

You could also use the same technique to make a low current 'super-voltage' rail above the positive power rail, for instance for biassing NPN emitter follower power pass devices right up to the top rail.

\$\endgroup\$
2
\$\begingroup\$

R1, R2, C2, and C3 are used to decouple the two bridge circuits so that they are "almost" independent. Yes they need to be there or else the two bridge circuits are shorted. D10 is a 15 volt zener for regulating/clamping the negative rail to -15V WRT -Ve.

The purpose of this secondary bridge is to create a "below ground" (-Ve in this case) reference. The reason for this complication is so that the power supply can be adjusted all the way from the max output down to zero and everything in between. Without it, U1 can't regulate voltages between Max and +1.25V - The regulator tries to maintain +1.25 between "Vout" and "ADJ". Therefore, a bias circuit is required to extent the regulation range to cover voltages between 0V and +1.25V.

This circuit is designed for an adjustable single positive voltage output. It is much different than a symmetric split rail supply - The "negative" supply has much lower current capacity and only serves the purpose of a negative (below -Ve) voltage reference.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.