2
\$\begingroup\$

For powering a RS485 circuit, I am requiring a Negative Power Source.

I was thinking to split a standard power source with 0V, +6V and +12V, using the +6V as ground with LM7805, and using some LM7805|LM7905 (or similar) devices for regulate into +5V|-5V.

In this case, the +6V could become a sinking terminal if the negative voltage circuit is unloaded, which I think should not be correct and could break the source(?).

The arbitrary change of GNDs confuse me, because the GND should always be a sinking terminal, not a sourcing.

So, how should I generate +5V -5V sources in a proper way?

Also, it is a choice to generate a low power -5V from a +5V source?

enter image description here

Edit.

  • Agreed regarding RS485 not actually requiring -5V source. Please keep the example valid for other application.
  • Consider 0V, +7.5V and +15V for the voltage dropout.
  • Consider only resistive loads, DC output signals.

In the best case, I am afraid this question depends on the nature of the power source (?). For example,

  • Case 1: if we have two sources, one from GND to +7.5V and other from GND to +15V, then loading only the 7805 will negatively polarize (charge) the 0 +7.5V source (?), hence this design is wrong in this case?

  • Case 2: The proper design should have two sources from GND to +7.5V, and putting them in series (?).

enter image description here

\$\endgroup\$
  • \$\begingroup\$ What makes you think that "the GND should always be a sinking terminal, not a sourcing."? \$\endgroup\$ – The Photon Jun 29 '19 at 22:53
2
\$\begingroup\$

This is a great question. the LDO’s have a 1.5 to 2V dropout so it won’t work with 6Vin and 5V out. But you do not need V- for a differential RS 485, only bipolar RS232.

You only need 5V and 0V to create RS485 signals.

Other info for future RS232 considerations. we normally use a charge pump for +/-V.

——x-x-xxx— ———

When we have floating DC sources, we can call anything Gnd=0V but use it for two different nodes on a “non-isolated” LDO.

  • so use a different Gnd symbol like Gnd A and Gnd B, keeping mind the DC difference could be anywhere from 5 to 7 V difference.p and not always 6V! why?

Furthermore, must be aware of the load currents with respect to 6V, 12V and 0V. Op Amps have a constant current between Vcc and Vee or V+ /V- and only use then middle-level voltage and impedance of that voltage for things like input bias currents, which are small. However, the output load will have some RLC values with respect to each rail so we can estimate these currents and their effect on each node voltage.

For background info, consider each LDO an Emitter Follower with internal feedback. This means the V+ out is NPN and V- regulator uses PNP outputs ( usually are Darlington’s consequently the output can be driven above the regulator’s |Vout| with very little current such as in a flyback Voltage).

So each regulator only pulls the load voltage towards the input. (+5 = delta V towards input +12V & -5 = delta V towards input 0V).

Now if your Load is connected to “common V” shared between both LDO’s what is the actual source impedance of this node before a load is applied? This depends on the LDO’s internal bias currents as an R divider so if not equal then what? Each LDO will be fighting to get + or - 5V out and if there is any mismatch in current to each regulator they will be noisy one may possibly pull itself towards its input rail thus saturating then error amplifier inside and getting zero feedback gain in the process so then current drops and voltage swings back towards some equilibrium near the LDO input cut-off voltage.

So will it work if the load is connected to input 0V?

No, because as the background info I gave says, the PNP emitter output becomes reverse biased in the LM 7906.

But putting the load like an Audio speaker Amp to mid-point Voltage could work but may have Noise if the other load currents are not balanced.

So how you fix this? Or does it need fixing? Well, that depends on your exact RLC load to each of 3 rails. this includes RS232 Cable, target SMPS leakage noise current, connections to Earth ground at both ends etc, etc.

Is floating actually infinite impedance. No it depends on the supply. 50,60 Hz transformers are different than 50kHz SMPS transformers and 1nF of leakage capacitance can cause issues.

Finally, if the input is 6V and output is 5V what is the regulator dropout Voltage at desired load current and how will 0V be connected to the target RS232 Earth gnd, assuming yours is isolated but with 1nF of leakage capacitance and Possible SMPS noise current coupling thru.

Summary of wrong assumptions

  • you can specify different gnd symbols to mean different 0V relative potentials as this is not connected to earth gnd (yet)
  • These BJT LDO’s need 2V headroom, not 1V, while FET-based LDO's use low RdsOn to get <<1 V yet not needed for 12 to 5V.
  • you don’t need V- for RS485 just 5V, 0V
  • bipolar LDO’s need a low impedance midpoint reference ( e.g. active V/2 driver) otherwise they do not balance well with unknown loads
\$\endgroup\$
  • \$\begingroup\$ Thanks for the answer. I would safely assume DC | resistive signals | loads. My doubt is if I load only the 7805 output, I will be inputting current into the +6V terminal, and thus the power source at that terminal will be reversely biased?. I am at least requiring to know which is the nature of the source?; if the source can handle negative DC currents? \$\endgroup\$ – Brethlosze Jun 30 '19 at 0:10
  • 1
    \$\begingroup\$ currents to and from LM7805 NPN out are always positive, even if Load comes from a negative voltage, but not from a voltage greater than output. E.g. 100 Ohms to -5V is OK if you had that would be OK. But again RS485 won’t need that. A negative current to LM7805 out implies V> 5V and more current than load to gnd. But since that current will be low with Vbe reverse biased a low impedance load say to 12 V would damage Ic and Voltage rises to 12V If that condition should ever occur and reverse diode from out to IN protects the IC. To prevent driving output higher than regulated or <-5 for other \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 30 '19 at 0:33
  • 2
    \$\begingroup\$ "These BJT LDO’s need 2V" - they are not LDO's, just standard dropout voltage regulators. \$\endgroup\$ – Bruce Abbott Jun 30 '19 at 4:47
  • 1
    \$\begingroup\$ OK @BruceAbbott LDO came from the 70's when LDO meant 2V dropout. Now in the 21st century FET LDO's are really ULDO's but the names haven't changed Note I said BJT LDO's which is accurate and you are implying the names have changed on the datasheets \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 30 '19 at 5:00
  • 1
    \$\begingroup\$ Fascinating TI site on an article written by an TI R&D engineer who was promoted to Apps Mgr wrote an article explaining essentially what you just said ( 2 kinds of regulators Linear and FET LDO) and made a major faux pas in design explaining how FET LDO's work using an Nch (instead of Pch) for a +ve regulator. from Michael Day exactly 10 yrs ago It is archived here web.archive.org/web/20190209124352/http://www.ti.com/lit/ml/… but removed off TI website. " you seem to have found some resistance---/\/\/\/--" FWIW @BruceAbbott \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 30 '19 at 5:18
3
\$\begingroup\$

First of all, RS-485 does not normally require a negative power supply. It uses a pair of balanced signals that individually swing between 0V and +5V. Only a +5V power supply is required.

But to answer your actual questions:

the GND should always be a sinking terminal, not a sourcing.

Not at all. "Ground" just implies a voltage reference point. It doesn't imply anything about the direction of current flow.

The negative terminal of a power supply "sinks current", but whether that's tied to ground is entirely up to you.

I'm not sure what the nature of the +6V source shown in your diagram is, but it's likely that it can't sink significant current.

So, how should I generate +5V -5V sources in a proper way?

Also, it is a choice to generate a low power -5V from a +5V source?

Sure. There are plenty of DC-DC converters out there that can do exactly that. Some of them use switched-capacitor techniques for very low power circuits; others use inductor or transformer based SMPS techniques.

\$\endgroup\$
  • \$\begingroup\$ RIght about the RS485 about swinging the 0V 5V signals, instead of creating a -5V signal. Considering only DC signals and loads, my intuition says I should not have a sinking current into a positive sourcing terminal i.e. I should not "charge" a power source by design. I don't have a confirmation of course, rather than batteries, and some simpler cases. Should be consider a good design having an SMPS with 0, 7.5V and 15V terminals, and putting a resistive load between the 7.5V and 15V? \$\endgroup\$ – Brethlosze Jun 30 '19 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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