Using a Current Transformer to charge a NiCD

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I'm trying to design a circuit that utilises a Current Transformer to
create a low current charging circuit on the secondary, for a 2.4V
NiCD with the addition of a few components .

primary 4 - 6A load
1 : 100 = Isec approx 40 to 60mA into a short circuit

Isec ________
-->----O | O--->--| |-----||+
ILoad O | O |Low I | ||NiCD
O | O |Charger |-----||-
--<-----O | O------|__cct___|

Any advice appreciated!



Jim

 

[happyhobit]

Hi Jim,

Bridge rectifier to convert AC to DC and an LM317 hooked up as current
limiter (see datasheet) with 25 ohm resister

http://www.national.com/ds.cgi/LM/LM117.pdf

Jay

 

[mike]

I'm trying to design a circuit that utilises a Current Transformer to
create a low current charging circuit on the secondary, for a 2.4V
NiCD with the addition of a few components .

primary 4 - 6A load
1 : 100 = Isec approx 40 to 60mA into a short circuit

Isec ________
-->----O | O--->--| |-----||+
ILoad O | O |Low I | ||NiCD
O | O |Charger |-----||-
--<-----O | O------|__cct___|

Any advice appreciated!



Jim

Use a diode and a light bulb to limit the current.
But there are issues. Do you really need the current
transformer thing?
mike

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Jay

Thanks for this.

I've tried to simulate a similar circuit with little success. I think
it is due to rectifier diodes' forward voltage combined with CT
(remember only one turn on the primary) saturation?

 

[[email protected]]

Thanks mike

Yes I do need to use a CT since all I have is the live feed to the
load on the primary side. Interesting idea, but since this has only
one diode will the secondary effectively be open cct every half cycle
with potentially disastrous results? What are the issues you have
thought about?

 

[Fred Bloggs]

Yes I do need to use a CT since all I have is the live feed to the
load on the primary side. Interesting idea, but since this has only
one diode will the secondary effectively be open cct every half cycle
with potentially disastrous results? What are the issues you have
thought about?

If by "live feed" you mean the AC line voltage then a CT is the last
thing to use. Did you ask yourself about the 4A in and 0.04A out? This
means something like 4x 120VAC=480 Watts in to charge 0.04A x 2.4=0.096
Watts out. Does this strike you as ridiculous?

 

[happyhobit]

Hi Jim,

Have you measured the output of the CT into a load (100-ohms maybe?).

Jay

 

[mike]

Thanks mike

Yes I do need to use a CT since all I have is the live feed to the
load on the primary side. Interesting idea, but since this has only
one diode will the secondary effectively be open cct every half cycle
with potentially disastrous results? What are the issues you have
thought about?

Yep, thems is the issues.
Use a center tap and two diodes, or a voltage doubler with 2 diodes.
You're gonna need about a hundred to 1 turns ratio. Put back to back
zeners across the secondary. If the secondary load
opens, you've got a big pile of smoking electronics.

The whole idea seems generally....ill conceived.
mike



--
Bunch of stuff For Sale and Wanted at the link below.
laptops and parts Test Equipment
4in/400Wout ham linear amp.
Honda CB-125S
400cc Dirt Bike 2003 miles $550
Police Scanner, Color LCD overhead projector
Tek 2465 $800, ham radio, 30pS pulser
Tektronix Concept Books, spot welding head...
http://www.geocities.com/SiliconValley/Monitor/4710/

 

[Chris1]

If by "live feed" you mean the AC line voltage then a CT is the last
thing to use. Did you ask yourself about the 4A in and 0.04A out? This
means something like 4x 120VAC=480 Watts in to charge 0.04A x 2.4=0.096
Watts out. Does this strike you as ridiculous?

I assumed that 4A would go on to do some work in some other load. His
charger would only drop the voltage going to the other load by some tiny
amount. I sure don't know why he needs to do it this way, though.

Chris

 

[John Fields]

If by "live feed" you mean the AC line voltage then a CT is the last
thing to use. Did you ask yourself about the 4A in and 0.04A out? This
means something like 4x 120VAC=480 Watts in to charge 0.04A x 2.4=0.096
Watts out. Does this strike you as ridiculous?

---
Actually, I think it's pretty clever. He's got an ostensibly constant 4
to 6 amp load on the mains, (I don't think it's a 500 watt resistor he's
hooked up just to be able to charge his NiCd cell!^) and by using the
CT he can get make an isolated 40 to 60mA supply to charge his NiCd
cell. As long as he full-wave rectifies the output of the CT and puts a
Zener across the DC output of the bridge to limit the voltage rise when
he disconnects the battery, what's wrong with it?

 

[Fred Bloggs]

---
Actually, I think it's pretty clever. He's got an ostensibly constant 4
to 6 amp load on the mains, (I don't think it's a 500 watt resistor he's
hooked up just to be able to charge his NiCd cell!^) and by using the
CT he can get make an isolated 40 to 60mA supply to charge his NiCd
cell. As long as he full-wave rectifies the output of the CT and puts a
Zener across the DC output of the bridge to limit the voltage rise when
he disconnects the battery, what's wrong with it?

Sounds like a PIC project shoved into a switch box.

 

[Tony Williams]

I've tried to simulate a similar circuit with little success. I
think it is due to rectifier diodes' forward voltage combined
with CT (remember only one turn on the primary) saturation?

It does sound like saturation.

You don't say what core you are using, what type of material and
the dimensions. It is possible for a 1/100 turn CT to work, but
only when Vdc is designed to be a minimum, with a magnetic
material that has a high saturation flux density, and a certain
minimum (magnetic) cross-sectional area (csa).

In this app, the sec is connected to a bridge rectifier and then
clamped or smoothed to produce a Vdc. Therefore the sec waveform
is going to look like a heavily truncated sine, for which the
nearest approximation is a square wave. The transformer sum to use
here then is the "volts-seconds sustaining" version.

-8
V.t = 2.N.A.Bmax.10

V is the peak secondary volts, (Vdc + bridge drop).
t is the half-cycle time of the mains, = 1/2F.
N is the number of turns.
A is the csa of the core, in square-cm.
Bmax is the peak flux density, in gauss.

Let's see what A we need for a fixed Nsec= 100 turns.

Suppose you are going to go straight into the 2.4v battery,
unregulated. Secondary peak volts, V= (2.4 + 1.6) = 4v.
To get the high Bsat it has to be silicon steel material,
and assume we are going to work it up to 10000 gauss.
I don't know which country you are in, but assume 50Hz.

8
A = 10 *4/(2*50*2*100*12000) = 2 sq-cms.

That is a *big* core for this job. It strongly suggests that
instead of 1/100 turns it might be better to go for something
like 5/500 (or even 10/1000) turns on a much smaller core.

 

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John

Spot on

cheers

 

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Thank you to all for your replies

Jim