Seeking assistance with (seemingly) simple circuit analysis

be 4mA.

First step is to check if the schottky will be conducting with S1 closed..

With S1 closed the voltage at S1 will need to be...

= 4.0V + (4mA x 47) = 4.188V

so the diode won't be conducting with S1 closed and it can be ignored for the moment.

With S1 closed the voltage at the junction of Ra and Rb will need to be

= 4.188 + (4mA x 180) = 4.908V

Write two equations for the voltage at the RaRb junction Vrab...

1) With switch open...

Vrab (open) = 14.5V * ((Ra+Rb)//Rc)/(Rs + ((Ra+Rb)//Rc))

2) With switch closed the equation is the same except you replace Rb with Rb in parallel with 1227 Ohms

Vrab (closed) = 14.5V * ((Ra+(Rb//1227))//Rc)/(Rs + ((Ra+(Rb//1227))//Rc))

There is a third equation we can write..

3) Irs = 90mA = 14.5/(Rs + ((Ra+Rb)//Rc))

Vrab (open) < 5V + Vsch say < 5.3V

Now we have three variables Ra, Rb, Rc and three equations so it should be solvable (I think)!

Sorry that should be

3) Irs = 10mA = 14.5/(Rs + ((Ra+Rb)//Rc))

The requirement that is impossible to satisfy with just resistors is that the no-load voltage at the junction of Ra and Rb does not exceed 5V.

All specs can be met by going sideways though.....

Rs +14.5V

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CWatters, & Tony,

Thanks for your feedback. At the risk of being overly optimistic (or = perhaps overly lazy), it occurred to me that Rc in the originally posted = schematic was edundant, since the series path through Ra & Rb might = serve the same purpose. Also, since Rs is always 90 ohms, and is present = in all cases, the "equivalent" schematic can be further simplified by = combining Rs into Ra. So does the problem now simplify to two unknowns = in two equations? I suspect this requires simultaneous equations, which = I've used only infrequently, in one case using determinants, and in = another by forcing both equations to equal zero, and adding them = (substitution method? It's been a while). In both cases with the how-to = in front of me while I worked through the problem. At the risk of = evading a potentially intriguing task, It seemed as though the = switch-closed case simplified all the way down to a topology like a = potentiometer with an additional resistance "out of" the wiper, as = follows;

+14.5Vo

In article , Eng. Tech. wrote: [snip]

No, you also have to add in the current drawn by the Schottky clamp when S1 is open.

A slightly better resistive circuit is to make Rb= infinity and just work with Ra and Rc. The calculated values for Ra and Rc below assumed an exact 5V and Vdiode= 0.2V.

90 14.5V---/\/\---+----+ | | Rc \ \Ra -+-5V 2119.2/ /2164 _|_ \ \ /_\ SD, 0.2V Fwd drop. | | | | | 180 | S1 47 | +---/\/\-+-+/+--/\/\--+-->Vout | | | [500+500] | | -------------+--------------------------+----

With S1 closed there is 4mA of output current and about 10.4mA load current through the 90 ohm.

With S1 open there is 10mA of load current through the 90 ohm, and about 3.58mA Schottky current.

Whatever the simplification the sums end in a quadratic equation with one sensible root. There are *probably* values for Ra/b/c that gets the 10mA with S1 open or closed, and the 4mA with S1 closed, but also probably at a higher Schottky current.

The low output resistance of a 5V reference is the thing that removes the interaction between output current and load current. It also produces minimum Schottky current.

--
Tony Williams.

You mean the minimium value of Ra + Rb = 1.45K.

Unfortunately there is a problem which Tony spotted in his first post...

If Ra+Rb was as high as 1.45K then when S1 was opened and closed, the load provided by the 180+47+500+500 resistors would modulate the voltage at the junction of Ra+Rb too much. To satisfy all your requirements the junction of Ra+Rb must be 4.908V with S1 closed and less than 5.3V with S1 open (assuming the forward drop of the diode is 0.3V).

To prove it...lets do what you suggest and set Ra + Rb = 1.45K

Set the ratio of Ra:Rb to give 5.3V with S1 open so that the diode doesn't quite conduct...

The potential divider equation is...

14.5 x Ra/(Ra+Rb) = 5.3

Rearrange to give

Ra = 0.576 Rb

Now Ra + Rb = 1.45K

so substitite to give

Rb = 920 Ohms Ra = 530 Ohms

Now when S1 is closed you have 180+47+500+500 = 1227 in parallel with Ra

1227 // 530 = 370 Ohms

The potential divider equation is now

14.5 x 370/(370+920) = 4.16V which is too low to give Vtest = 4.0V

So either the Vtest voltage is too low or the diode conducts. It can't be done with just resistors as Tony points out.

Try making Ra a 5V zenner/band gap as Tony proposed.

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Colin & Tony,

You guys are amazing! I stopped by sci.electronics.misc after skipping = a day or two, to see that you'd predicted precisely what I'm seeing on = the bench. While first trying a 5.1V zener in place of Rb, and seeing = that it begins to conduct near 4.0V (which the plot in the datasheet = appears to indicate is normal for the available diode), I set this aside = for the moment in favor of the infinite-Rb approach, in the interest of = having some results to show for time spent (I'd started to feel a bit = self-conscious having spent so much time thus far). I approximated the = values of Ra & Rc (no 5V reference) for a first-pass estimate, and then = empirically iterated a couple times to zero-in on more precise values. = With some working values verified, I attempted a closer estimate by (as = Tony already pointed out) taking the schottky current into account. When = this appeared to yield very good first-pass results, I transferred my = steps into a spreadsheet and duplicated the approach for remaining test = voltage/current combinations. While this provides "target" hardware for = the test software developer, I can't help regarding it as cheating.

At the risk of parading my ignorance, what led you to conclude the = solution would be a quadratic? (I don't disagree, I'm just clueless.)

I'll paste in the spreadsheet sections once I've verified it for a = couple more test cases.

Thanks again for the constructive feedback,

post...

load

the

junction of

doesn't

with Ra

be

Here goes, although the low output resistance band gap reference is still the easier/better way to go.

Lucky I didn't throw the scribbles away. Please note that I am also known as Tony (Typo) Williams.

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1. S1 closed, current in the 1227 must be 4mA.

90 Rx 14.5V---/\/\---+----+ Vx--/\/\----+ | | | Rc \ \Ra \Ra / / / \ \ \ | | -----> | | | | | \ \ | /1227 /1227 | \ \ | | | -------------+----+ -------------+

The 90+Rc attentuator can be redrawn to an equivalent Vx and Rx.

Vx = 14.5*Rc/(90+Rc), and Rx = 90*Rc/(90+Rc).

The three resistors in series must draw 4mA from Vx.

So Vx/(Rx + Ra + 1227) = 0.004 amps.

Do some algebra on that and you should get.....

Rc = (90.Ra + 110430)/(2308 - Ra) ....... eqn.1

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1. S1 open, current through the 90 ohm must be 10mA.

90 14.5V---/\/\---+----+