Pocket Oscillator

[mediatechnology]

I think I can do the power control and timer with a single or pair of enhancement mode MOSFETS.
No counters or 1 Hz clock...

A BS170/BS250 pair and a long RC network at one of the MOSFET gates provides both battery disconnect and timer.
It doesn't have to be accurate: Just long.

Right now I'm trying a 10M 470 uF to a single BS250 in series with the battery lead.

As it turns out there's no need for a split supply - looks like just using the 18V without a center tap and using a resistive rail splitter does the trick.
This reduces the On/Off to a single MOSFET.

[flyboy71]

Something like this maybe? https://www.edn.com/a-discrete-auto-pow ... teristics/

[mediatechnology]

Something like this maybe? https://www.edn.com/a-discrete-auto-pow ... teristics/

It's similar to that. Instead of the capacitor being discharged to turn it off I'm doing the opposite and having it charged from the unswitched supply to provide turn off.
The reason for that is to keep the electrolytic full-formed and leakage current low.
My concern was, with a large value resistor and cap, is that dielectric absorption might cause the cap to re-charge and turn itself back on.

Right now I'm using a single BS250 on the positive rail.

I'm thinking a three position toggle switch.
On
Off
Timer

I just saw it turn off: With a 10M/220 uF it was about an hour and eight minutes.
I'll post a schematic once I get it tuned...

[Tubetec]

Id been wondering about running op amps off a dual battery supply in this way , did the centre tap need to be grounded .
So what your saying is no , it doesnt ,but it does need a resistive divider to create a reference point . Obviously you dont want that divider across the battery itself permanently , so its placed after the fet switch ?
Will it result in a DC offset at the output if the battery voltages drift appart ?

[mediatechnology]

Id been wondering about running op amps off a dual battery supply in this way , did the centre tap need to be grounded .
So what your saying is no , it doesnt ,but it does need a resistive divider to create a reference point . Obviously you dont want that divider across the battery itself permanently , so its placed after the fet switch ?
Will it result in a DC offset at the output if the battery voltages drift appart ?

The divider is after the MOSFET switch.
Since the divider is referenced to 0V and Vbat it will always be half of the total battery voltage.

[ruffrecords]

I find a lot of people end up confused when running an op amp from a single supply. Personalty I find split supplies confusing but that's just because I am very old school. Back in the day, before semiconductor op amps were used in audio, everything was single supply. However, so your output could swing as far negative as positive, you had to BIAS the output stage to about half the supply volts. And of course you had an output coupling capacitor because your ground reference was some way below the dc output of the amplifier.

Op amps are no different. You can operate them from a single supply and bias the input to about half the supply volts with a simple decoupled pot divider.. Ground is the 0V of your supply but you will need to couple the outputs with a capacitor. An op amp's ground reference does not have to be at half the total supply voltage.

The one exception is when you want the mid rail bias point to be able to sink or source significant current. In that instance you simply buffer the bias voltage with a unity gain op amp stage and use its output as the bias.

The original pocket oscillator schematic illustrates this.

Cheers

Ian

[mediatechnology]

Some may have noticed the absence of output coupling capacitors in the schematics I've posted.
I haven't drawn the "single supply" version yet so bear with me...

With a "center-tapped" 18V battery stack, +/-9V, it's obvious that the caps are not needed.
But with an 18V "single supply" battery stack they're still not needed.

Only three points in the oscillator connect to the rail-splitter mid supply.
Two are the non-inverting op amp inputs. One is directly connected to mid supply, the other gets it's mid supply reference from the bottom of the AGC attenuator.
The third is the center tap of the output pad where the 49R9 resistors meet.

The mid supply reference sets the Q-point of the op amps to 1/2 the supply nominally +9V.
If you measure from the circuit 0V DC to the output you'll see +9V.
But +9V relative to what else?

Since it's battery-powered it floats.
Oscillator AC ground, audio 0V reference, XLR pin 1, assumes whatever potential it is connected to.
In one sense it's analogous to a center-tapped output transformer with an almost infinite CM impedance to system ground. (System ground being whatever it's connected to.)
Oscillator DC 0V, measured relative to AC ground is thus -9V.

The rail splitter is not buffered for a reason and I may also not bypass it conventionally.

Two of the rail splitter's loads are low current. (The op amp non-inverting inputs.)
The third "load" isn't really a load at all until it see's signal return current.
For balanced loads the center-point of the pad is driven to 0V AC ground.
For unbalanced loads, with one leg shorted to pin 1, the current imbalance is relatively small owing to the output pad.
(316+49)/316 or about 15% neglecting load impedance.

The finite impedance of the rail splitter is driven by the center-point imbalance but it appears at both output legs in common mode because the non-inverting inputs are driven.
It would appear that common mode voltages at the output, from the load the oscillator is connected to, will also drive the rail splitter which may or may not be advantageous.
At this point it would appear to be.

In this design bypassing the rail splitter by placing a C from the mid point to DC 0V may not be the best idea.
Instead, bypassing the rail splitter at its' +18V and 0V connections seems to be optimal as it really is a single-supply.
Plus, on power up a large thump is avoided as that C charges.
Bypass will be from +Vcc to -Vee on the one op amp.

EDIT: I just re-checked the ProtoBoard and the rail splitter in fact has bypass caps from +18V to +9V and another from +9V to 0V. These 10uF normally live on the Protoboard and are all but invisible to me. Thus, the divider center point is bypassed into the batteries.

I may rethink some of this but running it on the bench, floating, seems to work just as well as using a split +/-9V supply.

On another note the standby current for the auto-off appears to be about 10 nA.
I'm going to add a couple of MOSFETs so it may increase but I'll take anything under a uA.

[mediatechnology]

This is the battery-saver timer for the Simple Oscillator.



The basic timer and power switch is Q3, a BS250 enhancement mode MOSFET.
When C2 is discharged, and is more negative the Gate Source threshold voltage, Q3 conducts and the oscillator runs.
Vgs(th) is -1V to -3V. With the BS250 Source referenced to +18V, the turn-off voltage is +15V to +17V relative to 0V. Below this voltage Q3 conducts.
When C2 is discharged it takes about an hour for C2 to fully-charge through R6 and R7. This turns off Q3.

A pushbutton switch could be used to discharge C2 to turn the timer on, but the peak current demands would rapidly damage the switch.
Instead, Q1 is used to discharge C2 and turn the oscillator "On" in either a continuous or timed mode depending on the positions of S1 and S2.

In the power Off position Q2 is used to force Q3 off and charge C2. This also resets the timer.
R3 causes Q2 to conduct.
When S1 is switched to "On" or "Timer" Q2 is turned off.
D1 prevents the oscillator from running continuously when S1 is switched to "Timer."

When S1 is "On" Q1 conducts continuously causing Q3 to conduct and C2 to discharge.
S1 in the "Timer" Position places S2 in the Gate control line of Q1.
Pressing S2 starts the timing cycle; pressing S2 during the timing cycle extends it.

The timing length will not be particularly accurate owing to the leakage current of C2 and the MOSFET threshold voltage and leakage currents.
Having C2 charged when batteries are connected and while switched Off allows its electrolyte to remain fully-formed.
The current to maintain charge is quite low; after 24 hours it measured around 10 nA.

The BS170 and BS250 have a maximum +/- 20V Vgs.
R7 limits Q3 gate current in the event the supply voltage were to exceed 20V (e.g. operating on an external supply) and it limits current between Q1 and Q2 in the event of a device or wiring fault.
R7 also allows Q2 to rapidly turn off Q3 without having to fully-charge C2.

R1 and R2 reduce Vgs for the BS170 and, along with C1, provide a delayed turn-on of Q2 with some added noise immunity.

When switched to "Off" static current drain is << 1uA.
When the timer has completed its' cycle and is Off the current drain is <<2 uA.
"On" draws about 15 mA most of which is 5532 and signal current.

The added current burden of the LED is worth it IMHO.
The approximate 1 mA burden with the values shown might be reduced.
I may place a TL431 in series with the LED and resistor to provide a low battery indication.

I'm thinking I'll add an AC adapter input and sub-regulator to allow a 12-24 VDC input.
Since the current is low I may use a TL431 shunt regulator for the external supply.

[mediatechnology]

Added circuitry for an external DC supply.

Note that the component identifiers have changed since the circuit description was written.

[ruffrecords]

I am seriously tempted to add your timer circuit to my existing pocket oscillator. Would it need much modification to work from a single 9V battery?

Cheers

Ian