Rail Splitting Chips...

Ricky · Mar 12, 2023

On Sunday, March 12, 2023 at 4:34:44â¯PM UTC-4, Lasse Langwadt Christensen wrote:

sÃ¸ndag den 12. marts 2023 kl. 18.29.26 UTC+1 skrev Ricky:
On Sunday, March 12, 2023 at 9:57:00â¯AM UTC-4, bitrex wrote:
On 3/12/2023 3:07 AM, Ricky wrote:
On Sunday, March 12, 2023 at 1:54:46â¯AM UTC-5, bitrex wrote:
On 3/11/2023 4:41 PM, Ricky wrote:
On Saturday, March 11, 2023 at 3:51:23â¯PM UTC-5, whit3rd wrote:
On Saturday, March 11, 2023 at 12:25:59â¯PM UTC-8, Ricky wrote:
On Saturday, March 11, 2023 at 12:10:21â¯PM UTC-5, bitrex wrote:
On 3/9/2023 7:32 PM, Ricky wrote:
On Thursday, March 9, 2023 at 6:18:06â¯PM UTC-5, Lasse Langwadt Christensen wrote:

npn+pnp follower, https://imgur.com/NhlLGGc

Does your drawing need a couple of diodes? Otherwise the output would have a dead band 1.something volts wide, no?

The small-signal AC impedance looking into the virtual ground is always
fairly low, at most whatever the small-signal AC impedance of the two
caps in parallel is.

The transistors just keep the DC operating point from getting too out of
whack. So there\'s no \"dead band\" per se, both the small and large signal
impedance looking in from the VG terminal always going to be something
relatively low...
Why is there no deadband? The transistors BE junction barely turn on until there is already movement in the set point. This movement is exactly what needs to be minimized.
If one is using one rail as a voltage reference, that\'s significant. If, however, one is just
using the rails for power, and all signals are ground-referenced, then it just works.
Common-mode tolerance is built into op amp designs, and at high frequency the
capacitors handle it, while at low frequency the high differential gain
dominates the tiny common-mode effect.

It is a matter of designing the circuitry to tolerate a bit of variance in the raw supply
voltage absolute values, while removing the irritations that come from trying
to use a negative power rail as signal ground. In the worst case, one can add
positive and negative regulators after those filter capacitors.

Not sure what distinction you are trying to make by saying, \"If, however, one is just using the rails for power, and all signals are ground-referenced, then it just works\". The problem is it doesn\'t work. See my simulation in the previous post. It not just doesn\'t work, it doesn\'t work *horribly* with 1.4Vpp variation in the reference output. It\'s pointless.
For the price of two transistors, two caps, and two resistor it can sink
or source appreciable DC current from either rail while drawing zero
quiescent current.

Your circuit
\"My circuit\"?
I guess it\'s Lasse\'s circuit. Either way, it doesn\'t work. It\'s pointless.
maybe in your application

if you need a reference that can sink and source, not a virtual ground then only thing that will work
is two resistors and an opamp

Hmmm... was there any issue of clarity in my statements of the application???

From my initial post.

> I read some of the discussion on \"rail splitting\" and it occurred to me that there should be a market for these devices. I am working on a design that will run op amps from a single rail and I need a 6V level, that can both source and sink current.

In any event, the transistor circuit without the diodes has a 1.4V deadband in which neither of the transistors are turned on to any significant degree. I can\'t think of a use for such a circuit, unless you actually want the level to shift in that manner for some reason. That would be a very specialized circuit, indeed.

--

Rick C.

+++ Get 1,000 miles of free Supercharging
+++ Tesla referral code - https://ts.la/richard11209

Ricky · Mar 12, 2023

On Sunday, March 12, 2023 at 4:40:47â¯PM UTC-4, whit3rd wrote:

On Sunday, March 12, 2023 at 10:27:40â¯AM UTC-7, Ricky wrote:
On Sunday, March 12, 2023 at 5:19:55â¯AM UTC-4, Lasse Langwadt Christensen wrote:
sÃ¸ndag den 12. marts 2023 kl. 09.40.48 UTC+1 skrev Ricky:
On Sunday, March 12, 2023 at 4:13:25â¯AM UTC-4, whit3rd wrote:
On Saturday, March 11, 2023 at 1:41:44â¯PM UTC-8, Ricky wrote:
On Saturday, March 11, 2023 at 3:51:23â¯PM UTC-5, whit3rd wrote:

It is a matter of designing the circuitry to tolerate a bit of variance in the raw supply
voltage absolute values, while removing the irritations that come from trying
to use a negative power rail as signal ground. In the worst case, one can add
positive and negative regulators after those filter capacitors.

Not sure what distinction you are trying to make by saying, \"If, however, one is just using the rails for power, and all signals are ground-referenced, then it just works\". The problem is it doesn\'t work. See my simulation in the previous post. It not just doesn\'t work, it doesn\'t work *horribly* with 1.4Vpp variation in the reference output. It\'s pointless.
depends on what you consider to be ground
I don\'t get to choose the ground. The 12V power supply negative rail is ground, along with the 5V negative rail and the 3.3V negative rail and the -12V positive rail. This is a daughter card on a main board. Take a look at your power rails, the 12V positive and negative (you didn\'t even name the negative rail). They are bouncing 1.4Vpp. Is this the sort of amplifiers that you design???
What does the variance of power rails do to a transimpedance amplifier? Nothing, really. There\'s
an input current, an operational amplifier, and a resistor in feedback... as long as you can take out
a signal wire and make a ground connection, the power solution with two transistors DOES work.
There\'s no significant problem with power rails there.

If power rails \'bounce\', so what? In most of my work, a bunch of boxes all route signals
for various operations, and sometimes I want a variant of one of those boxes... so a single-voltage
wall wart and a few parts go into the new unit.

Sorry, I have no idea how this turned into your design. Ok, whatever. I wasn\'t discussing some unrelated design. I was discussing an audio op amp design on a 12V power supply that required a virtual ground.

I have no idea what you are thinking.

The transistor circuit without the diodes does nothing useful. It absolutely does not provide a stable output voltage.
It certainly does something useful; that TIA won\'t have a lack of power. What OTHER requirements you
might want to satisfy, we have no clue to.

The transistor circuit does nothing useful. The 6V virtual ground doesn\'t source power. It\'s there to provide the reference voltage for the rest of the circuit. The two transistor circuit without diodes is not useful for this, as it wobbles all over if you draw any current. You would probably be better off with just resistors and no transistors.

I don\'t know what you are trying to say. I have no idea why you are talking about a completely different design than what is being discussed here. I have no interest in continuing this extraneous discussion. Thanks for your comments.

--

Rick C.

---- Get 1,000 miles of free Supercharging
---- Tesla referral code - https://ts.la/richard11209

whit3rd · Mar 12, 2023

On Sunday, March 12, 2023 at 3:19:39â¯PM UTC-7, Ricky wrote:

On Sunday, March 12, 2023 at 4:40:47â¯PM UTC-4, whit3rd wrote:
On Sunday, March 12, 2023 at 10:27:40â¯AM UTC-7, Ricky wrote:
On Sunday, March 12, 2023 at 5:19:55â¯AM UTC-4, Lasse Langwadt Christensen wrote:
sÃ¸ndag den 12. marts 2023 kl. 09.40.48 UTC+1 skrev Ricky:
On Sunday, March 12, 2023 at 4:13:25â¯AM UTC-4, whit3rd wrote:
On Saturday, March 11, 2023 at 1:41:44â¯PM UTC-8, Ricky wrote:
On Saturday, March 11, 2023 at 3:51:23â¯PM UTC-5, whit3rd wrote:

It is a matter of designing the circuitry to tolerate a bit of variance in the raw supply
voltage absolute values, while removing the irritations that come from trying
to use a negative power rail as signal ground. In the worst case, one can add
positive and negative regulators after those filter capacitors.

Not sure what distinction you are trying to make by saying, \"If, however, one is just using the rails for power, and all signals are ground-referenced, then it just works\". The problem is it doesn\'t work. See my simulation in the previous post. It not just doesn\'t work, it doesn\'t work *horribly* with 1.4Vpp variation in the reference output. It\'s pointless.
depends on what you consider to be ground
I don\'t get to choose the ground. The 12V power supply negative rail is ground, along with the 5V negative rail and the 3.3V negative rail and the -12V positive rail. This is a daughter card on a main board. Take a look at your power rails, the 12V positive and negative (you didn\'t even name the negative rail). They are bouncing 1.4Vpp. Is this the sort of amplifiers that you design???
What does the variance of power rails do to a transimpedance amplifier? Nothing, really. There\'s
an input current, an operational amplifier, and a resistor in feedback.... as long as you can take out
a signal wire and make a ground connection, the power solution with two transistors DOES work.
There\'s no significant problem with power rails there.

If power rails \'bounce\', so what? In most of my work, a bunch of boxes all route signals
for various operations, and sometimes I want a variant of one of those boxes... so a single-voltage
wall wart and a few parts go into the new unit.

Sorry, I have no idea how this turned into your design.

But it\'s clear from the discussion that you think the solution \"doesn\'t work\", when it fact it
is a useful gizmo. Your \'audio op amp\' comment doesn\'t specify why you think there\'s some
significant problem, nor does my example of a TIA have a problem with a volt or three
wandering.

I can\'t run your simulation on this machine, and wasn\'t referring to it.

John Larkin · Mar 12, 2023

On Sun, 12 Mar 2023 13:34:39 -0700 (PDT), Lasse Langwadt Christensen
<langwadt@fonz.dk> wrote:

søndag den 12. marts 2023 kl. 18.29.26 UTC+1 skrev Ricky:
On Sunday, March 12, 2023 at 9:57:00?AM UTC-4, bitrex wrote:
On 3/12/2023 3:07 AM, Ricky wrote:
On Sunday, March 12, 2023 at 1:54:46?AM UTC-5, bitrex wrote:
On 3/11/2023 4:41 PM, Ricky wrote:
On Saturday, March 11, 2023 at 3:51:23?PM UTC-5, whit3rd wrote:
On Saturday, March 11, 2023 at 12:25:59?PM UTC-8, Ricky wrote:
On Saturday, March 11, 2023 at 12:10:21?PM UTC-5, bitrex wrote:
On 3/9/2023 7:32 PM, Ricky wrote:
On Thursday, March 9, 2023 at 6:18:06?PM UTC-5, Lasse Langwadt Christensen wrote:

npn+pnp follower, https://imgur.com/NhlLGGc

Does your drawing need a couple of diodes? Otherwise the output would have a dead band 1.something volts wide, no?

The small-signal AC impedance looking into the virtual ground is always
fairly low, at most whatever the small-signal AC impedance of the two
caps in parallel is.

The transistors just keep the DC operating point from getting too out of
whack. So there\'s no \"dead band\" per se, both the small and large signal
impedance looking in from the VG terminal always going to be something
relatively low...
Why is there no deadband? The transistors BE junction barely turn on until there is already movement in the set point. This movement is exactly what needs to be minimized.
If one is using one rail as a voltage reference, that\'s significant. If, however, one is just
using the rails for power, and all signals are ground-referenced, then it just works.
Common-mode tolerance is built into op amp designs, and at high frequency the
capacitors handle it, while at low frequency the high differential gain
dominates the tiny common-mode effect.

It is a matter of designing the circuitry to tolerate a bit of variance in the raw supply
voltage absolute values, while removing the irritations that come from trying
to use a negative power rail as signal ground. In the worst case, one can add
positive and negative regulators after those filter capacitors.

Not sure what distinction you are trying to make by saying, \"If, however, one is just using the rails for power, and all signals are ground-referenced, then it just works\". The problem is it doesn\'t work. See my simulation in the previous post. It not just doesn\'t work, it doesn\'t work *horribly* with 1.4Vpp variation in the reference output. It\'s pointless.
For the price of two transistors, two caps, and two resistor it can sink
or source appreciable DC current from either rail while drawing zero
quiescent current.

Your circuit
\"My circuit\"?
I guess it\'s Lasse\'s circuit. Either way, it doesn\'t work. It\'s pointless.

maybe in your application

if you need a reference that can sink and source, not a virtual ground then only thing that will work
is two resistors and an opamp

I assume that the downstream rails will be bypassed, so one needs a
c-load stable opamp, or more parts.

Ricky · Mar 12, 2023

On Sunday, March 12, 2023 at 6:30:44â¯PM UTC-4, whit3rd wrote:

On Sunday, March 12, 2023 at 3:19:39â¯PM UTC-7, Ricky wrote:
On Sunday, March 12, 2023 at 4:40:47â¯PM UTC-4, whit3rd wrote:
On Sunday, March 12, 2023 at 10:27:40â¯AM UTC-7, Ricky wrote:
On Sunday, March 12, 2023 at 5:19:55â¯AM UTC-4, Lasse Langwadt Christensen wrote:
sÃ¸ndag den 12. marts 2023 kl. 09.40.48 UTC+1 skrev Ricky:
On Sunday, March 12, 2023 at 4:13:25â¯AM UTC-4, whit3rd wrote:
On Saturday, March 11, 2023 at 1:41:44â¯PM UTC-8, Ricky wrote:
On Saturday, March 11, 2023 at 3:51:23â¯PM UTC-5, whit3rd wrote:

It is a matter of designing the circuitry to tolerate a bit of variance in the raw supply
voltage absolute values, while removing the irritations that come from trying
to use a negative power rail as signal ground. In the worst case, one can add
positive and negative regulators after those filter capacitors.

Not sure what distinction you are trying to make by saying, \"If, however, one is just using the rails for power, and all signals are ground-referenced, then it just works\". The problem is it doesn\'t work. See my simulation in the previous post. It not just doesn\'t work, it doesn\'t work *horribly* with 1.4Vpp variation in the reference output. It\'s pointless..
depends on what you consider to be ground
I don\'t get to choose the ground. The 12V power supply negative rail is ground, along with the 5V negative rail and the 3.3V negative rail and the -12V positive rail. This is a daughter card on a main board. Take a look at your power rails, the 12V positive and negative (you didn\'t even name the negative rail). They are bouncing 1.4Vpp. Is this the sort of amplifiers that you design???
What does the variance of power rails do to a transimpedance amplifier? Nothing, really. There\'s
an input current, an operational amplifier, and a resistor in feedback... as long as you can take out
a signal wire and make a ground connection, the power solution with two transistors DOES work.
There\'s no significant problem with power rails there.

If power rails \'bounce\', so what? In most of my work, a bunch of boxes all route signals
for various operations, and sometimes I want a variant of one of those boxes... so a single-voltage
wall wart and a few parts go into the new unit.

Sorry, I have no idea how this turned into your design.
But it\'s clear from the discussion that you think the solution \"doesn\'t work\", when it fact it
is a useful gizmo. Your \'audio op amp\' comment doesn\'t specify why you think there\'s some
significant problem, nor does my example of a TIA have a problem with a volt or three
wandering.

If you don\'t care about the wandering, why would you need any circuit at all? Two resistors would be very adequate from what you\'ve said.

> I can\'t run your simulation on this machine, and wasn\'t referring to it.

Ok, so you were conversing with yourself. Sorry to intrude.

--

Rick C.

---+ Get 1,000 miles of free Supercharging
---+ Tesla referral code - https://ts.la/richard11209

Don · Mar 13, 2023

Don wrote:

<snip>

Does your simulation serendipitously stay balanced due to its zero
voltage offset sine \"load?\"
Will an unbalanced load, with, for instance, 100 mA pulled from the
positive rail and nothing drawn from the negative rail, sink virtual
ground downward? Because, for the unbalanced case, no current flows
through Q2 to balance things out?

With an unbalanced load, 56KR on the positive rail and 560KR on the
negative, you see this: <https://crcomp.net/misc/vgnd.png> The black
DVM shows Vin while the burnt-orange DVM shows the voltage between
the positive rail and virtual ground. It\'s difficult to see, but
there\'s a decimal point between the zero and the nine.

Danke,

--
Don, KB7RPU, https://www.qsl.net/kb7rpu
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.

Don · Mar 13, 2023

Please ignore previous followup. 56R needs to be connected between the
positive rail and virtual ground, and not 56 KR.

Danke,

--
Don, KB7RPU, https://www.qsl.net/kb7rpu
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.

Ricky · Mar 13, 2023

On Monday, March 13, 2023 at 1:50:54â¯AM UTC-4, Don wrote:

Please ignore previous followup. 56R needs to be connected between the
positive rail and virtual ground, and not 56 KR.

You might fix the link while you are at it.

--

Rick C.

--+- Get 1,000 miles of free Supercharging
--+- Tesla referral code - https://ts.la/richard11209

bitrex · Mar 13, 2023

On 3/12/2023 6:48 PM, Ricky wrote:

> If you don\'t care about the wandering, why would you need any circuit at all? Two resistors would be very adequate from what you\'ve said.

Here are two example situations where two resistors doesn\'t work, and
works, respectively:

<https://imgur.com/a/vaYuw3Z>

Adding the two transistors turns the =( face circuit into at least a =|
face, which is a non-negligible improvement for almost negligible cost

Ricky · Mar 14, 2023

On Monday, March 13, 2023 at 6:06:16â¯PM UTC-4, bitrex wrote:

On 3/12/2023 6:48 PM, Ricky wrote:

If you don\'t care about the wandering, why would you need any circuit at all? Two resistors would be very adequate from what you\'ve said.
Here are two example situations where two resistors doesn\'t work, and
works, respectively:

https://imgur.com/a/vaYuw3Z

Adding the two transistors turns the =( face circuit into at least a =|
face, which is a non-negligible improvement for almost negligible cost

You say that, but what voltage do you get when you add the two transistors? It\'s no longer 6V, that\'s for sure. It would be 5.3V, I believe. So it\'s still a frowny face. That\'s the point!!! The transistor circuit is not much better than the two resistors.

I have no idea why you are pursuing this. It\'s obvious that a 6V regulator that has 1.4V of deadband is not of much use. Somehow, you seem to want to distract from the reality of this with silly circuits.

In the simulation of the two transistor circuit without the diodes, the emitters junction follows the load voltage, until about 0.7 volts is reached. Only then does it start to pass any real current. Run the simulation. You will see what I mean. This is a crazy circuit.

--

Rick C.

--++ Get 1,000 miles of free Supercharging
--++ Tesla referral code - https://ts.la/richard11209

boB · Mar 15, 2023

On Mon, 13 Mar 2023 05:50:47 -0000 (UTC), \"Don\" <g@crcomp.net> wrote:

Please ignore previous followup. 56R needs to be connected between the
positive rail and virtual ground, and not 56 KR.

Danke,

Don, time to renew your ham license !

boB

Rail Splitting Chips...

Ricky

Guest

Ricky

Guest

whit3rd

Guest

John Larkin

Guest

Ricky

Guest

Don

Guest

Don

Guest

Ricky

Guest

bitrex

Guest

Ricky

Guest

boB

Guest

Welcome to EDABoard.com

Sponsor

Online statistics

Forum statistics

Rail Splitting Chips...

Ricky

Guest

Ricky

Guest

whit3rd

Guest

John Larkin

Guest

Ricky

Guest

Don

Guest

Don

Guest

Ricky

Guest

bitrex

Guest

Ricky

Guest

boB

Guest

Log in

Welcome to EDABoard.com

Sponsor