FL5150 (LED dimmer) circuit explanation

Question

I have been working on an IC from Fairchild semiconductor named FL5150, whose electrical working I have tested by controlling the brightness of a lamp. Until now I have been using the circuit given in the datasheet with a manual potentiometer.

But one thing which I am not able to understand is the circuit in this portion. .

When the current passes through the MOSFET, the mid-point of both the MOSFETS is in direct line of AC supply. But that part is also connected to the Ground of the IC. How is this thing working? Is the voltage on the mid-point of both the MOSFETs at a constant voltage?

Answer 1

When the current passes through the MOSFET, the mid-point of both the MOSFETS is in direct line of AC supply.

No it isn't. It is a node with a GND symbol on it and this doesn't automatically mean that it connects to true ground of the incoming AC supply. If this happened, the MOSFETs would become ashes in milliseconds.

The ground symbol is just a way of hinting that the GND pin on the chip is connected to the centre-point of the MOSFETs. It connects nowhere else and it certainly should not be assumed that it connects to earth.

How is this thing working? Is the voltage on the mid-point of both the MOSFETs at a constant voltage?

Yes it is relative to the VS pin on the chip. Regard the two MOSFETs' internal diodes as being the lower half of a bridge rectifier. D1 and D2 form the upper half. R1 and C2 smooth the voltage for the chip.

Answer 2

It is a pretty weird circuit to get your head around, but think of a bird sitting on a power line doesn't get fried because there is no voltage or more accurately difference in potential between the 2 points, it's feet. Like wise the there is no different between the two grounds in the schematic above, because they are literally connected together by the one wire, so they remain the same. There is potential difference(voltage) between VS and Gnd of 17v this comes from what the bridge rectifier created by Diodes D1 D2 and the 2 internal IGBT Diodes as Andy Aka pointed out, between load and hot actives there is no potential difference unless the IGBTs are off, when they are off you will measure the potential of main voltage from hot through the load to neutral. Disconnect the load neutral and you will have no potential(no voltage at all) with in the circuit in fact it will appear safe to touch, but if you are standing on ground and you touch hot active there will likely be a path to earth which will return to neutral via a ground link in the electrical installation. The main point being voltage is simply the difference in potential between two point, voltage is not simply at one point.

Answer 3

You said in a comment

Circuit ground must be at a constant voltage so that all of the rest circuitry works wrt to that voltage. Voltage wrt to which all of the rest circuit works

Why must circuit ground be constant? It definitely does not need to be constant. Constant to 'what'? To EARTH/0V?

Let's look at the circuit. Assume the AC neutral is tied to EARTH/0V. Measure the voltage at Vs to Neutral/Earth. Measure the voltage between the FL5150 'GND' (labelled 'gnd_a' in my circuit) and Neutral/Earth. Neither of these pins are 'constant'.

Now let's look at voltage between Vs and gnd_a.

It is ~17V (created by the internal Zener). So the FL5150 is happy. But I would NOT go connecting a wire to the FL5150's GND pin and grabbing a hold of it. Between the FL5150's GND and ACTUAL EARTH is ~240VAC.

Answer 4

Update

The FETs share a common Source so only 1 FET is active at one time and they alternate conduction to the load.

In the trailing edge mode a Zero Crossing (ZC) pulse is created with a time delay to stop conduction in the middle of the phase.

______|——-|.

In both cases volatile alternates about the earth ground=0V.

Rather than leading edge which starts in the middle and stops at the zero crossing of voltage.
|————|_______

ZC Monitor input is a standard 5 V logic threshold input such as 2.5V so line or neutral voltage thru > 1M R ought to detect the near 0V crossing signal for dimmer control.

It doesn't show how but it could be an XOR gate input with a delay on one side of xx pF to generate a ZCS pulse in both polarities.

How is this thing working? Is the voltage on the mid-point of both the MOSFETs at a constant voltage?

NO It is connected to midpoint of Line Hot and load when enabled. Since Load is relatively high impedance compared to FET, you can say the 0V floating ref. called the gnd symbol is connected line hot. and the PE earth symbol allows for a ZCS input.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 5

The two ground symbols are completely different! The triangular ground here is simply the IC supply and gate drive common point- NOT AC Neutral, or PE ground. The zero cross input DOES connect to PE (or Neutral, in a 3-wire configuration).

There are two FETs in a phase dimmer because of the intrinsic body diode present in MOSFETs (and IGBTs). A single FET would result in phase control during one half of the AC period, and body diode conduction during the entire other half so the minimum ON time would be 50%, rather than zero. The two FETs are in series with each other and with the load and connected so the body diodes are "back to back" (AC is blocked when they are gated OFF).

Answer 6

There is no "mid-point". Both mosfets are connected to GND line with their Source pins if you look closely. This is not a push-pull configuration. The mosfets are just two switches, each working during each half-wave of AC sine. Whilst one mosfet is working, the other one is shorted with the internal diode. As AC polarity changes, they switch places. That means the point that you call "mid-point" is actually Ground. So, this point is tied to chip's ground pin so that it can control the mosfets. To switch a mosfet on one must apply positive voltage to its Gate pin relative to Source pin.