Super Diode (Active Rectification) Question

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I think this would work for you.

use op amp comparator circuit.

An opto coupler would give you the required reverse polarity "protection" and put the diode out of the series motor current path. You seem to be pretty hung up with the forward voltage drop over a semiconductor's junction. I can't see how you can avoid this with a solid state solution. The only way to get clean contacts is to use a mechanical device (forget the contact resistance). These miniature relays can have polarised coils and are available in 1.5V variants on request. http://www.techno-star.biz/relays/v23026.pdf

Actually I am not at all married to doing this via a semiconductor and if a mini-relay does the trick I would use it. So, is it possible that by reversing polarity to the relay I can enable/disable my motor?

The R_DS(on) can be low but at a gate voltage not available in your ckt.

This is an FET with an integrated free-wheeling diode, not an optocoupler.

Seems that a relay will do. There are some with biased magnetic circuits so they close with the right polarity only. Or else you can use that diode you tried before and hope the relay will close with such a low voltage.

brgds

Snel

Hi Randall,

power supply is 1½ V.

Hi Molecool--You need to be careful with devices with low "PIV" junctions as the back EMF off the motor can destroy them when power is disconnected and you may not see it without a storage scope.

If you are able to use 2 smaller 1.5V batteries just place a diode on both motor leads

reversing polarity-Thus drop 1.2V and you still have whatever current is provided by the state of the batteries. Make them at lest 600-PIV diodes-Good Luck----

Regards.

There are two points of view in solving any problem.

1. Simple

or

2. Problematic or advance-technology usage.

If your problem is for a simple device or experiment then you have to bear its limitations.

Like a series diode to block reverse connection - will drop 0.7V Si and 0,2V Ge diode

Or polarize the connections to the cell; i.e. cell is not coinnected electrically though inserted without any problem.

This is a most common way used.

Or cell cannot inserted in reverse. It needs special design cells and cell base.

You have a lot of options.

The simplest solution to your problem is the following circuit.

If you apply three volts to B then the transistor will turn on and give you about 2.8 to 2.9 V at the motor. The diode placed in parallel with the motor is to get rid of any back EMF when the power to the motor is removed. If you apply a negative voltage at B the transistor will not turn on and the motor will not run.

However, this is a very simple circuit and doesn't have any protection for the driving transistor or motor. If you have a problem with the motor for example the resistance drops too low then the transistor will blow up. If the transistor is overloaded then the chances are that you will not be able to turn the motor off without disconnecting the battery.

Anyway, this circuit is intended as a starting point so you can add what other safeguards you need later, just make sure that the NPN transistor that you purchase is capable of taking the start-up current the motor draws otherwise it will blow up the first time you turn it on and you won't be able to turn it off.

You could also build a similar circuit using a MOSFET transistor instead of a bipolar transistor again make sure that the transistor can handle the start‑up current and has a low enough on resistance from the source to the drain that it won't cause too much of a voltage drop.

Regards.

A good answer, but I suggest not to lift the emitter and use an emitter-follower but to connect the load at collector.

Now control signal polarity will change but at lower level.

If you move the load to the collector then you have to add a resistor to the base the value of which needs to be calculated according to the gain of the specific transistor and that can vary dramatically even with identical transistors.

By having the load collected at the emitter you effectively have a base resistor that has a value of the load multiplied by the forward bias current gain of the transistor. As a result you don't have to worry about the gain of the transistor all you need to be sure of is that it is capable of handling the load.

Very good solution.What if you use a LM117HV ? A National Semiconductor (now TI) 3 pin regulator. Now you have current and thermal protection... the input can be any voltage and output will be where you set it, even down to 1.2 Vdc.

That might be a good solution, however, I haven't read the data sheet on the LM117HV and from past experience three terminal regulators don't take kindly to the input voltage being reversed so if the drive was set to run the motor in reverse i.e. a negative voltage on the unregulated input leg, I don't know what would happen to the regulator chip. It may have internal protection for it but I'm not sure. I haven't got time at the moment to read through the data sheet for the LM117HV but if somebody has the time maybe they would check if it has anything to say about reverse polarity input protection.

Of course you have to work for stabillity, gain etc. etc.

emitter follower has unity gain and lowest Zo

The KISS principle has gone right out the window now......

"The KISS principle has gone right out the window now......"

More like out the window, down the street, off to the nearest airport and up up and away into the wild blue yonder.

You really mean : http://www.irf.com/product-info/datasheets/data/irlml6401.pdf

But the Mosfet requires a higher Gate-Source voltage to properly turn it ON.

You need to indicate what current the small motor draws at 1.5 volts to see what can be done and probably you can use a PNP transistor if the current is low below 100 milliamps like the 2N4003 which is a PNP device that can be driven by a just a resistor to ground ( negative ) When the voltage is the reverse the Gain of the PNP is around 0.03 so you may have some leakage.

To warrant a single polarity is necessary to Have two small MOSFETs high current and low Rdson, like 0,002 ohms -- N type and 12 volts break-down is OK, connect both sources together and both gates together but again you need a voltage higher than the 1.5 volts ( like 5 volts if you use low level Mosfets ).

There are other solutions though we need to know what you are doing exactly to know what to suggest instead.

A simple solution.

1. Get a PNP silicon transistor, preferably a small power type designed for low voltage with high gain. Actually I first thought of germanium transistor which have base-emitter voltage ~0.2V better than silicon ~0.7V. But difficult to get germanium cheaply nowadays - American Microsemiconductor (AMS) make some - search GlobalSpec for "germanium transistor"
2. Connect battery + to transistor emitter.
3. Connect transistor base to battery negative via 1 kohm resistor.
4. connect motor between transistor collector and battery negative.

I do not know what current your motor takes, but it should be possible to get an "on" voltage drop of 0.1 volt with correct transistor and adjustment to "1 k" resistor value. An inverse diode 1N4001 across the motor might be needed to avoid voltage spikes on break.

The emitter reverse breakdown voltage of transistors is usually 5 V or more, so there is no problem with "Off leakage" if battery is reversed.

You can use an NPN transistor to get ON for negative battery.

67model

I tried a PNP silicon BC461 (TO39 metal can) transistor with 1 kohm base resistor, as per post #23, with 82.5 ohm load at 1.39 V supply.

0.072 V drop, base-emitter about 0.7V, ~16 mA load.

Modern pnp 12 volt bipolar transistors can give 0.01 V saturation at 0.1 amp with 5 mA base current e.g.

http://www.diodes.com/datasheets/ds31618.pdf - and Vbe <0.7V typical @ 100 mA collector.

Also tried PNP germanium transistors, with base shorted to collector, as "super-diodes".

Same supply & load ~14 mA.

1. ancient unmarked, might be near 50 year old, 0.25 V drop.
2. ACY22, general purpose. 0.20 V drop.
3. ACY187, output pair type for battery radio - 0.175 V drop. AMS supply these, but not cheap.