can anyone tell me what this means? cd4047b

[JAMES IRVINE]

I've been reading this description of the cd4047b an IC which Uncle scrooge gave me https://www.electro-tech-online.com/custompdfs/2008/04/cd4047b-1.pdf the thing is I can't understand what the maximum input voltage is. I think it's 0.5 volts so if I am correct then does anyone know of a voltage regulator I could use for this circuit. Also am I right in thinking that I would need a couple of transistors to boost up the output voltage to drive a couple of mosfets. Can I have this exact circuit **broken link removed** excluding everything to the left hand side of the sn7400.

 

[Torben]

Hi James,

From the datasheet:

Maximum Ratings, Absolute-Maximum values:
. . .
Input voltage range, all inputs: -0.5V to Vdd + 0.5V

. . .which means that the inputs can handle voltages between -0.5V and the supply voltage plus 0.5V. So if you're running the circuit on +5VDC, the safe input voltage range is -0.5V to +5.5V.

Someone who's actually used a MOSFET will have to help you with the second question, sorry.


Torben

 

[ronsimpson]

1) The cd4047b could be run at 12 volts not 5 volts. That way it will drive MOSFETs better.
2) The FETs could be 'logic level' type that will work at 5 volts.
I would do option 1.

 

[jpanhalt]

A couple of comments about the CD4047. The CD family is nice in that it takes up to 18 V supply (the CD4047B shows a 20-V rating). The downside is that its fan-out for logic devices (number of devices it can feed) is low. The datasheet says just 2, and the current is also low, about 1 to 2 mA for the logic devices. For the CD4047B from TI, the max output is listed at 6.8 mA at 15V. The HEF family (e.g., HEF 4047) can also take supplies of up to 18V and provides at least 10 mA per output. It is not an expensive chip either. Digikey sells the chip for $0.50 USD, but has a minimum.

You can power a mosfet from the HEF family directly. You might be able to use the CD4047 directly too, but I have not tried that. If you are using a 5V supply, then you need a logic level mosfet. If you use a 12V supply, you can power a regular mosfet. I would be a little cautious powering more than one mosfet with either, particularly the CD4047. Finally, if you are going to be working at 5V, then there are other logic families available that will give you up to 20 mA per pin output. Neither chip compares favorably to dedicated mosfet drives for mosfets with large gate capacitances.

In summary, the CD4047B might work for driving a mosfet directly, but I would prefer to use the HEF 4047 at 12 V and another logic family, if at 5 V. John

 

[crutschow]

A couple of comments about the CD4047. The CD family is nice in that it takes up to 18 V supply (the CD4047B shows a 20-V rating). The downside is that its fan-out for logic devices (number of devices it can feed) is low. The datasheet says just 2, and the current is also low, about 1 to 2 mA for the logic devices.

The low fanout is for operating at the maximum device frequency where operation is limited by the capacitance drive capability of the output current. At low frequencys or dc the fanout is high, limited by the gate leakage of the MOSFETs being driven, which is typically microamps. So the fanout can be scaled according to the operational frequency, from 2 logic devices at maximum frequency to, limited by leakage of the devices being driven, at dc.

 

[audioguru]

All 4047 ICs are the same. The HEF4047 has exactly the same specs as the CD4047.

I think a simple square-wave inverter using a CD4047 driving Mosfets should be like this:

 

[jpanhalt]

AG, You are right and I offer my apologies. Thank you for correcting that misinformation I had stored away. No good excuse, except my confusion resulted from misreading the split datasheets Philips puts out vs. the single publications from TI. John

 

[JAMES IRVINE]

It's an amazing looking circuit Uncle scrooge. So is the nener diode present to ensure that voltage transients don't blow the CD4047.
So I am back to where I was with a previous circuit in not fully comprehending what is happening at the supply, am I correct in thinking that the battery can supply two sources with enough voltage because they are in paralell to each other i.e. the CD4047 and the two n-channel mosfets and transformer. I suspect that a much higher proportion of current will be drawn by the mosfets and the tranformer so this is the explanation for me that fits, is there another one.
I'm also guessing that the capacitor and resistor to the right hand side of the mosfets must be present to absorb transients again which would possibly come directly from the battery or be caused due to the inductive nature of the transformer.
Thanks Uncle scrooge I was feeling like a dog chasing his own tail.

 

[audioguru]

The CD4047 operates with a very low supply current.
The Mosfets do the hard work of switching many amps from the battery into each side of the transformer winding, alternating.

This is a very simple square-wave inverter. Many electronic products don't work from a square-wave. They expect the much higher peak voltage of a sine-wave. But this square-wave has the same average voltage as a sine-wave to drive incandescent lights the same.

This simple inverter does not have voltage regulation so with a fully charged battery and with not much load, the voltage is too high. When the battery is almost dead and with a heavy load, the voltage is too low.

 

[JAMES IRVINE]

Thanks you must be reading my mind cos this is exactly what I need for one of my alternative solutions, just out of curiosity do you think it could supply a 60watt load such as a television set. In the UK at 50Hz 230-240 volts of course.

 

[blueroomelectronics]

Thanks you must be reading my mind cos this is exactly what I need for one of my alternative solutions, just out of curiosity do you think it could supply a 60watt load such as a television set. In the UK at 50Hz 230-240 volts of course.

Why not trash the old TV and get a LCD TV, some LCDs will work on DC and they all use much less power than a tube TV.

 

[Someone Electro]

I recomend you use some kind of boost to drive the gates of the mosfets.Tipicaly 1A peak is used to drive them nice and fast.Thing is you need to get the mosfet in to saturation and back to fully closed in the shortest time.This is becuase the half open state creates a lot of heat.So if you drive the gate faster the mosfets will heat a lot less.That means better efficency and a smaller inverter.

Common solution is a 2 trasistor buffer.Works realy well and certanly worth the few cents for them.

As for mosfets i would recomend IRF3205. Its in a common TO-220 and has only 0,008 Ohm resistance when saturated.Should result in almost no heating.

 

[Hero999]

But if the MOSFETs are only being driven at 60Hz the switching losses will be pretty low, even if the switching speed is slow so it doesn't matter.

Using a 12V transformer is also pretty pointless because the output voltage will be much less than the rated input voltage at full load. Even if you powered your 12V transformer from a 12VAC sinewave you'd only get 230V on the secondary off load, when a load is connected it could drop as low as 180V which is totally useless.

Build a modified sinewave inverter and use a 9V transformer rather than a 12V transformer as it'll give you a high enough peak voltage, even at full load.

 

[audioguru]

A problem with a simple square-wave inverter circuit using good Mosfets is when both Mosfets switch at the same time and conduct a momentary high current at the same time. A modified sine-wave circuit turns on one Mosfet at a time with a pause between them.

 

[Someone Electro]

But when switched fast enugh this time is next to nothing.There is a simple solution to it tho. A series resistor on the gate and then a parallel diode on that resistor that lets current out of it. This was it thurns off in a instant but has a tiny delay for a thurn on.

Anyway i agree this cirucit can be inproved a lot.But this one is ment to be as simple as it gets, and i dont see why wouldnt it work, just not work the best