Mercury Battery Replacement For Antique Instruments

by Craig Taylor

    I have recently been trying to restore a Radector radiation survey meter, vintage 1950s.  This instrument uses a 10 atmosphere argon filled ion chamber as the radiation detector and has an electrometer type vacuum tube sealed inside the chamber.  The design is simple and, for the time, quite elegant.  Unfortunately it uses three different mercury batteries.  One large 1.35V for the filament power, one smaller 1.35V for bias and one 5.4V for the plate supply of the electrometer tube.  Because of the design, the calibration depends on the voltage of the batteries staying constant over the useful battery life.  Mercury batteries maintain a very constant voltage over most of their life, a characteristic NOT shared by present day alkaline cells or any other easily obtained batteries. 

  So. What to do?  I decided that modifying the instrument itself was a bad idea;  I want it in original condition.  So I can’t modify the internal circuit or drill holes in it to install switches.  What I want is a drop in form, fit and function battery replacement.  That implies that when you switch off the instrument, the battery drain stops. Building a regulated battery powered supply is easy these days using integrated regulator-reference combination chips.  But, there is a problem.  All the existing packaged solutions have a bias current that is required to run the regulator IC.  If you simply disconnect the load ( turn OFF the existing switch on the instrument ), the bias current circuit still drains the battery.  This load might be only 100uA or so, but for a 100mA-hr battery that translates to a shelf life of 1000 hrs, which is 1.4 months.  Not good.  I don’t want to have to open it up and remove the batteries every time I finish playing with it, so this is a problem.  My solution was to design a supply that shuts itself off when the “battery” drain drops below the minimum current needed by the instrument.  That lower current drain implies the instrument has been turned OFF (the internal battery switch is open).  Ideally, the new supply should reduce battery drain such that the batteries are good for their shelf life, which is years for modern alkaline cells.

  The basic regulator design for the filament supply is shown in figure 1 .  It is designed for a load of 25mA maximum.

 

FIGURE 1

 

  Here is how it works:  Current through R1 tries to turn on Q3 such that the emitter voltage is Vbattery - 0.7V .  If the voltage at VR1-R is higher than 1.24V, VR1-C begins conducting more current and therefore diverting Q3 base drive.  At some point there isn’t enough current going into the base of Q3 to maintain the output voltage and it starts to fall.  When the divided output voltage starts to drop below 1.24V at VR1-R, VR1 conducts less current and leaves more for Q3 base.  A balance is reached with the output voltage just high enough to maintain 1.24V at VR1-R.  If the battery voltage drops enough so that the current through R1 is less than the 80uA (or less) needed by VR1 and enough left over for base current to support the load, regulation fails and the output voltage drops.  This will happen at a battery voltage of about 2.25 V.

   Note that, with a fresh battery, the bias current for VR1 and the voltage divider is about 1.4mA even with the load removed.  So this basic circuit doesn’t meet the design requirements.  Figure 2 shows the changes needed to make it work as desired.  

   schematic

 FIGURE 2

 

There are two more transistors in this design.  The actual ON-OFF switch is Q2.  When ON, it is saturated with a voltage of less than 20mV between Collector and Emitter.  When OFF, the collector-emitter leakage current at room temperature is less than 10nA, usually much less.  The only point of the regulator that is referenced to the current return point (Vreg Rtn.) is Q2 emitter.  If Q2 is OFF, there is no path for current to flow from battery plus to battery minus.  That’s about as “off” as I could make it.

  Q1 is the load sensor.  It can be almost any pnp silicon transistor you can find.  If less than about 0.5 mA is flowing through R4 (into the voltage divider and Load), there isn’t enough voltage drop (5E-4 x 6.8E2 = 0.34V ) to turn on Q1 (needs about .55V or so to start conducting).  Therefore no current is flowing out of the collector of Q1 and into the base of Q2.  Once the load current exceeds about 0.8mA, Q1 starts conducting and turns on Q2.  At loads over a few mA, both Q1 and Q2 are saturated and regulation becomes accurate.  For my application the load is about 22mA,  more than enough to saturate Q1 and Q2.

    If this gadget is going to be run at temperatures much greater than 40C, a 100K resistor should be connected between Q2base and Q2emitter to divert any leakage from Q1 and prevent Q2 from trying to turn on.

  Note that the 50uA needed by the voltage divider is not enough to “hold on” the regulator when the remaining external Load is removed.

  The saturation voltage of Q2 will vary by 7 to 12 mV as the base bias current changes with changes in battery voltage.  The result is a change in regulated output voltage of 7 to 12mV.  If that error is a problem in your application, you can replace Q2 with an N channel mosfet with a Vgs ON of less than 2.2 Volts and Rds ON of less than 5 Ohms.  It needs to maintain a relatively constant Vds over a gate voltage range of 2.2 to 3.0V to perform any better than the bipolar 2N3904.   The Vishay TN0200 looks like a good choice, but I haven’t actually tried.  If you decide to use a mosfet, connect a 470K resistor between the mosfet gate and source to sink leakage current.  Remember that the DC input impedance of a good mosfet is almost infinite and any leakage current could cause it to turn on.

5.4 VOLT, 120uA  MERCURY BATTERY REPLACEMENT

This supply replaces a battery of 4 mercury cells that produces 5.4 Volts.   It powers the electrometer tube plate circuit in the ion chamber amplifier of the Radector radiation survey meter.  The plate current varies as the ion chamber receives more or less radiation, but it stays within a range of  30 to 120uA for normal instrument calibration and operation.  This means that the choice of designs for the power supply is wider than for the previously designed 25mA filament supply.  For several reasons, mostly because I thought it would be more interesting, I decided to use a linear shunt regulator.

Shunt regulators have been around a long time.  TVs have used them to control the 15kV+ supply for the CRT electron gun since the 1950s, for example.  They are possibly the simplest of the linear regulator family.  The classic shunt regulator is a series current limiting resistor and a Zener diode to ground (see Figure 1 ).  The output voltage is limited to the Zener voltage with excess current “shunted” to ground.  If the supply voltage and load current are stable, it works great.

 

FIGURE 1

One  of the first things to notice is that the maximum current available to the load is [ Vbat - 5.4V ] / R1  .  If the load requires more than that, the output voltage drops and you no longer have a regulated voltage.  So you have to choose R1 to supply the maximum current ever needed by the load when the battery voltage is the lowest it will get.  Usually you choose a “dead battery” voltage such that about 80 to 90% of the energy has been extracted from the battery.  For an alkaline battery, that might be around 1.15V per cell ( 1.5V new ).  There is a small problem, though.  If the battery voltage is higher than the minimum or the load requirement is less than maximum, the extra current gets dumped in the Zener and turned into heat.  This type of regulator is not the most efficient type for widely varying line voltage combined with large load current changes.  You end up with too much power dumped in the Zener shunt element or the current limit resistor.  Also, 5.4V Zeners have a soft “knee”, which means their clamping voltage changes with current (and temperature!), especially in the sub 1mA range.

There is an integrated circuit available that acts something like a programmable Zener and can sink up to 20mA; the TLV431.  I used it in the 1.35V, 25mA design (see above) as the voltage reference and error amplifier for the series pass transistor.  It can also be used by itself as the shunt element in a shunt regulator.  I show it (as VR1) in a “simplest possible” 5.4V regulator design in Figure 2.

 

FIGURE 2

 

I put in the numbers for a 5.4V ,120uA  regulator with a battery voltage range as shown in the figure.

This won’t actually work in my application because it doesn’t disconnect the battery from the regulator when the load current goes to zero.  But it shows everything needed to understand the design variables in a shunt regulator.  Here’s how it works:  We assume the system is “in regulation”, which makes the load voltage 5.4V. VR1 requires a minimum of 80uA to run it’s internal circuitry.  The voltage divider requires about 15uA.  Added up, the worse case load current is 120+80+15=215uA.  I apply the engineering fudge factor and decide that 250uA is a good working number.  This 250uA has to come through the limiting resistor R1, so if the minimum battery voltage for operation is chosen as 7V, the resistor value has to be:

[7V-5.4V] / 250uA = 1.6/2.5E-4 = 6.4K  Ohms 

The next smallest 5% resistor value is 6.2K, so that’s our value for R1.  Now we can calculate the worst case current drain from a new battery:

[9.2 - 5.4] / 6.2E3 = 613uA

Oh my!  That’s about 5 times the maximum load requirement.  Seems a bit wasteful.  If we had a device that we could substitute for R1 that only passed 250uA no matter what the voltage difference across it, we could more than double the battery life.  There is a theoretical device called a Current Source that does just that.  Unfortunately, it doesn’t exist.  But we should be able to do better than a resistor, with a little thought. 

Figure 3 shows the regulator from figure 2 with an “imperfect” current source (R1,R2,D1,D2,Q2) in place of a resistor.

      

FIGURE 3

Here is how the current source works.  The collector of Q2 supplies current to the rest of the regulator, but that current depends on the base-emitter current in Q2, not the emitter-collector voltage (as in a resistor). The current in R2 splits to flow through R1+base-emitter diode or D1D2, more current going to the lowest resistance.  If it flows through R1 it is amplified by beta, the transistor gain, which is typically over 100.  So each uA flowing in the base causes 100uA to flow in the emitter-collector terminals and, of course, through R1.  This current causes the voltage across R1 to rise until all but a few uA of the R2 current is shifted to the two diodes.  The result is that the current in R1 is held constant and equal to the difference in the voltage drop between the base-emitter junction (0.63V) and the diodes (0.79V).  So the current is set to : [0.79 - 0.63 ] / 680, or about 240uA.

This will change some with battery voltage, which controls the current in R2.  R2 current flowing through D1D2 causes a change in the diode voltage.  Silicon diodes don’t make good voltage references, but the “sloppy current source” works much better than a resistor as the current limiting element.  Of course, it takes 4 additional parts.  Nothing’s free.

  

FIGURE 4 - final design

 Breadboard testing of the complete design (figure 4), including load sensor and disconnect switch, showed that the 431 installed in the circuit works quite well down to 36uA, so I set the VR1 minimum current to 40 instead of 80uA and ran the numbers again.  Final value of R1 was determined by breadboarding the circuit and testing it under maximum load with varying input voltage.

The experimental results for the current limiter output were 202uA for a Vbat of 7V and 240uA at 9V.  If we had used a resistor designed for 202uA at 7V, the current would have been 628uA at 9V .  That is enough savings in current to justify the added parts, I believe.

If you want to run the current regulator circuit at temperatures outside of the usual “room temperature” range, you need to test for temperature sensitivity.  Both the diodes and the Q2be junction will change voltage drop for a given current as temperature changes.  A quick and dirty test using my hand to warm up D1, D2 and Q2 caused the emitter current to drop by 11%.  If it gets hot, your design could stop regulating at lower battery voltage, so compensate by designing for higher current. 

An interesting experiment is to substitute an infrared LED for the D1-D2 pair.  The voltage drop should be almost the same, but the led should have a better “knee” than the diode pair.  There is an IR led in most optocouplers, if you can’t find a discrete part.  I measured  0.75V on a 4N25 from my parts box.  Have fun!

 

CRAIG TAYLOR  26 FEB 2004