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A DC to DC converter ?

TaylorTronics

Hello smart people, engineers, mathematicians, scientists, physicists, women and men. image

 

I have an interesting electronic design, and I need help. image

 

It involves pulsing an LCR circuit with a constant current source, that can supply any voltage and current to a DC motor, from a battery source who's voltage is well below the motors' rated voltage (i.e. a 12V brushed motor running off a high current capacity 1.2V rechargeable battery).  image

 

I am fresh out of uni, at 48, and found that I truly didn't understand a constant current source, until I began designing a 'Reflective Light Proximity Detector". image

 

I found that in theory, I could pulse an LCR circuit at resonance with a constant current source, and get any current and voltage I desired. Free amplification and sensitivity, without complicated amplifiers. image

 

I then posted a thread on this forum, asking "Where can I get a constant current device ?". image

 

I was duly informed by some extremely clever people, who had understood the concept of a constant current source in uni, what a constant current source really was. image

 

Whilst a constant current source can induce thousands or millions of volts, and thousands or millions of amps, it must be capable of supplying those millions of volts, in order to get those millions of amps. Not possible with a 1.2V battery. I would need a million volt battery for this device to work, which defeats the purpose.

 

So, how do I induce a high voltage and current at resonance, using a bipolar BJT (diode) as a constant current source, using a low voltage source with high current ? image

 

I was told that if you put a spanner across the terminals' of a submarine battery, it would vaporize. They are low voltage cells, 1.2V, I think. image

 

So, how do I get that current into a 12V motor, with a 1.2V battery, using the motors field and armature wingdings as part of the LCR resonant circuit ? image

 

Keep it pumping image

 

Thanks in advance image

 

Peter Taylor image

  • Hi,

     

     

    The Low Voltage battery does not have the voltage (pressure) to force enough current through the motor winding resistance.

     

    To get enough current flowing through your motor you need to have 12Volts of pressure.

     

    So, we can boost the low voltage from the battery up to a high voltage using a DC to DC converter:

     

    This uses an inductor (coil of wire), which works a bit like a fly-wheel.

    We connect the coil to the low voltage battery, and current starts to flow, as the coil 'charges up' the current gets higher and higher until it is limited only by the resistance of the coil.

    So if you have a low-resistance coil, you will get a massive current flowing through it, even if you are only connecting your low-voltage (high current) battery.

     

    So, now the coil has a massive current flowing through it, and a big magnetic field around it. (It is like a spinning fly wheel).

    When we disconnect the battery,  The coil will keep trying to push current..The magnetic field will collapse 'Squeezing' out all its' current. But this current has nowhere to go (because the battery is disconnected) So the voltage increases between the coil terminals..This voltage will increase to whatever it takes to keep the current flowing..so if the coil is completely disconnected, the voltage will be infinity (almost). Or, if the coil is across your 12V motor, it may be a reasonable voltage = Motor resistance * Current.

     

    The current is set by the Coil (fly-wheel), it will try and push this current no matter what.

     

    So, you need  a "DC to DC boost converter"

    You can convert your low-voltage high-current into a high-voltage low-current.

    Power in = Power out.

    (Voltage * Current ) IN  = (Voltage * Current) Out

     

    Hope this helps.