Maker Pro
Maker Pro

220v LED DIMMER??

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
Wouldn't a triac type dimmer work okay if you took out the smoothing capacitor after the bridge rectifier?

Bob
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
I guess a triac type dimmer would be worth a try. You might not even need to remove the 4.7 uF cap in the light bulb circuit. It doesn't provide much hold-up time. There might be a problem with insufficient current to keep the triac turned on, and because the load current and voltage are out of phase with each other (because the load is largely capacitive), but I'd say it's worth a try - would be an easy solution if it works!
 
Last edited:

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
I guess a triac type dimmer would be worth a try. You might not even need to remove the 4.7 uF cap in the light bulb circuit. It doesn't provide much hold-up time. There might be a problem with insufficient current to keep the triac turned on, and because the load current and voltage are out of phase with each other (because the load is largely capacitive), but I'd say it's worth a try - would be an easy solution if it works!

Unfortunately it doesn't work I’ve tried it before posting the first post. What it does is that it makes a hum after some point (i believe it's when i try to dimm it to less than 50%) it begins to flicker and behave strangely. The effective dimming is just 2% - 5%.
 
Last edited:

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Well, I've had a go at designing something. The circuit simulates OK but it's pretty large. Also I don't know whether it would comply with any electrical requirements - in your area, or anywhere else.

I've drawn it up in LTSpice. I don't have models for the high-voltage transistors I specified; I used generic LTSpice transistor models called 2N5401 (for the PNPs) and 2N5550 (for the NPNs). I used NPNs for both current sources because at high voltage and significant power dissipation, NPNs have much better performance.

I'm posting it here just for interest's sake. I DO NOT make ANY claims about this design.

attachment.php


Here's a circuit description.

The LED light bulb is shown at the right, starting from Cin. The LED light bulb responds to current of either polarity. The higher the current, the brighter the LED will light. Also, the higher the current, the more rapidly the input capacitor (Cin, 1.6 uF) in the bulb will charge or discharge, and the more rapidly the voltage across the bulb will change. The circuit generates alternating positive and negative currents into the bulb. These currents are supplied by "Darlington-and-a-half" transistors QP1/2/3 and QN1/2/3

Starting from the left of the schematic, the AC mains voltage (230V RMS nominal) is rectified by D1~4 and smoothed by C1, and becomes the main positive rail (across the top of the diagram). The voltage on this rail, measured relative to the 0V rail (along the bottom of the diagram) is nominally about 324V and the design maximum is 400V.

C1's value should be 2 uF multiplied by the LED current (in mA) at the maximum brightness setting, in order to keep the ripple on the main positive rail to less than about 5V peak to peak. Ripple doesn't affect the circuit much; you could use 1 uF per mA for C1 to save space or cost.

The circuit comprised of DD1, DD2, DS1, DS2, RD and DS3 sets up several voltages, relative to the positive and negative rails. DD1/DD2 and DS1 set the voltages at both ends of the "brightness" trimpot, which is shown here as two resistors, POTA and POTB. These voltages are about 1.4V and 7.6V negative relative to the positive rail.

DS2 adds another 15V to create a voltage reference about 23V below the positive rail, which is used by QL1, described later. RD drops the bulk of the voltage and sets the current in that path to about 1 mA. DS3 sets another reference voltage at about 15V above the 0V rail, which is used by QL2, also described later.

The "Darlington-and-a-half" transistors formed by triplets QP1/2/3 and QN1/2/3 have a relatively high current gain (I have assumed at least 2000 total beta for each compound transistor) and can withstand 400V DC. Almost all of the power is dissipated in the final transistor of the triplet, which is specified as a BUX85G and dissipates on average about 60 mW for each mA of LED bulb current. For example, operating at an LED current of 50 mA, QP3 and QN3 will dissipate about 3W each.

The current limiters are identical and use a simple, standard design, which is documented at http://en.wikipedia.org/wiki/Constant_current_source
The current is controlled by current from the collectors of QPS and QNS. This current develops a proportional voltage across RPB or RNB, and the emitter follower action of the QP1/2/3 triplet ensures that the voltage across RPE or RNE follows. DP1~3 or DN1~3 provide compensation for the three base-emitter drops in the transistors. The result is that the collector-emitter current is limited to a value which is roughly equal to the current from QPS (for the top current source) multiplied by RPB divided by RPE.

QPS and QNS are also adjustable current sources. They provide the current that controls QP1~3 and QN1~3 respectively. Only one is enabled at a time; this is ensured by DPS and DNS, which are driven by QFF1 and QFF2, only one of which is ON at a time. When the appropriate QFFx transistor is ON, the appropriate QxS transistor is enabled as a current source with its current controlled by the voltage at the bases of QPS and QNS, which is set by the Brightness potentiometer.

In this diagram, the Brightness potentiometer is represented by resistors POTA and POTB, which are calculated according to the pot_position parameter in the simulation, which represents the position of the Brightness potentiometer in percent, in the range 0~100. This produces values for POTA and POTB which add to 100 kilohms, the resistance of the Brightness potentiometer in a real (not simulated) circuit.

The feedback loop is completed by the flip-flop made from QFF1 and QFF2. The outputs of the flip-flop, on the collectors of those transistors, control the QPS and QNS current sources, which in turn control the two large current sources that supply the output.

The flip-flop is set and reset when the voltage at the output point gets closer to the positive or negative supply rail than the "limit" reference voltages set by DS1~3. When the output voltage goes higher than the voltage at DS2 anode, which is roughly equal to the positive supply rail minus 22V, QL1 conducts and forces QFF1 ON, forcing QFF2 OFF. When the output voltage goes lower than the voltage at DS3 cathode, which is roughly 15V higher than the 0V rail, QL2 conducts, causing QL3 to conduct and turning on QFF2, forcing QFF1 OFF.

The circuit alternates between these two states, with the output connection slewing alternately positive and negative at a rate determined by the current provided by the active current source (positive and negative currents are roughly equal) and the load capacitance, which mainly consists of Cin in the LED light bulb. The current that flows through Cin also flows through the LED(s) in the light bulb.

The actual LED bulb current that corresponds to the maximum brightness setting is determined by the values of RPE and RNE. The value of these resistors is chosen according to the formula R = 5.7 / ILED(max).

The voltage applied across the LED lamp is about 270V peak to peak and unipolar. In the real world running from AC mains the LED would see about 650V peak to peak, bipolar.

attachment.php


The waveforms show the current into the LED bulb at five different settings of the Brightness potentiometer:
0% (pink)
25% (cyan)
50% (red)
75% (blue)
100% (green)

Comments anyone?
 

Attachments

  • schematic.gif
    schematic.gif
    24.7 KB · Views: 3,593
  • output current waveforms.gif
    output current waveforms.gif
    13.1 KB · Views: 1,249
Last edited:

CDRIVE

Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Joined
May 8, 2012
Messages
4,960
Kris, if in the future, discrete circuit design achieves the collectivity of the art world's minimalist and impressionist periods this will have to fit in your 'Mega-Maximumist' period. :p

GottaI love it for it's "WTF!" value! :D

Chris
 

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
:eek: WOOW how much time did it take you to sketch this HUGE schematic and to write that description?

I don't have the components that are required in you circuit so i went ahead and tried mine :)

It works fine (although not linearly) the poor IRF740 gets the most dissipation when i'ts dropping from 240v to 200v. After running for half a minute i can't touch it for more than half a second so it should be aprox 50C?. I also tried a fet that is rated at 50amps and has the 1/10 RdsON rating (0,055ohm) and it is THE SAME.

is there a way of making it less hot? many fets in parallel perhaps?
Thanks.
 

Attachments

  • CYCLE 1.jpg
    CYCLE 1.jpg
    78.7 KB · Views: 1,588
  • CYCLE 2.jpg
    CYCLE 2.jpg
    79.8 KB · Views: 559
Last edited:

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
You are using the MOSFET in the linear region, so Rdson has nothing to do with it, that comes into play only when the MOSFET is fully on.

Bob
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
You went ahead and tried your circuit? What circuit is that? I ask because as far as I can tell, you do actually need a current generator with alternating polarity, like the circuit I designed. A simple linear regulator won't work because of the capacitor in the LED light bulb.

As BobK says, the heating in your MOSFET is because it's operating in its linear region.

Power dissipation in any component is equal to the voltage across the component multiplied by the current through the component. When an active component like a transistor or MOSFET is operated as a switch, it spends almost all of its time either fully OFF (with no current flowing through it) or fully ON (with no voltage across it), so it dissipates very little power. When you operate the active component in its linear region, there is some current through it, and some voltage across it, so the power dissipation in the active component is relatively high.

You can mitigate this heating by using several MOSFETs in parallel and/or by using a heatsink.

But please show me the circuit you're using!
 

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
As the (now) updated circuit shows (post 26) the led bulb (in the box on the right) works very good the circuit on the left does not restrict AC movement as shown by the red and blue arrows. It even works with electronic lamps :cool: (CFL). It has the same effect as a VARIAC.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Ah, I see what you mean. That's a good way to simplify the design - use the mains itself as the frequency source. If repeatability or control aren't good, you can change your poorly defined variable resistance element to be a current source. Just adding a source resistor will help. And of course, heatsinking will be needed. There's no way to avoid power dissipation using a linear regulator like that.
 

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
If repeatability or control aren't good, you can change your poorly defined variable resistance element to be a current source. Just adding a source resistor will help. And of course, heatsinking will be needed. There's no way to avoid power dissipation using a linear regulator like that.

Sorry i didn't got what you mean here?

Any way of making a DIY switching regulator with a few components like FETs/IGBTs without large coils and transformers? I still can't find a switching regulator that has over 300v rating.

I noticed that the VDS of the FET was not what the lamp received as it dropped only few volts. So i may use commercial liner regulators for less dissipation and better linearity.
 

CDRIVE

Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Joined
May 8, 2012
Messages
4,960
So i may use commercial liner regulators for less dissipation and better linearity.
As Kris has explained repeatedly you can't avoid power dissipation losses when using a linear current limiter, voltage regulator etc. You mentioned a variac earlier. Though large, it would be an efficient method with very low power losses. A multitap transformer would also be efficient.

Chris
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Yeah, what CDRIVE said.

A commercial linear regulator (needs to be a current source/sink) will have better linearity but it won't have lower dissipation. A certain amount of power needs to be dissipated, i.e. converted into heat. Any linear regulator will have to dissipate this heat.
 

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
Yeah, what CDRIVE said.

A commercial linear regulator (needs to be a current source/sink) will have better linearity but it won't have lower dissipation. A certain amount of power needs to be dissipated, i.e. converted into heat. Any linear regulator will have to dissipate this heat.

Ok so how about a DIY switching regulator?
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
A switched-mode current source? There may be such a thing but I haven't heard of it.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Ok so how about a DIY switching regulator?

Yeah, sure. many (most) LED drivers these days are switchmode.

The simplest ones use a current mode regulator and set the maximum current through the inductor. This provides a rough (but adequate) control over output current.

A "better" approach senses the voltage across a current sense resistor in the output, but practically it doesn't make a lot of difference.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Steve, the problem is that he wants to drive LED light bulbs, which contain a series capacitor and bridge rectifier as well as the LED string (and a small smoothing capacitor). You can't drive them with DC.

He wants to use a circuit in post #26 on this thread, which puts the current source across the output of a bridge rectifier which is connected between the mains supply and the LED light bulb.

So it's not clear how it would receive power. Also it would need to be configured as a current sink with a wide output voltage compliance range.
 

BlinkingLeds

Feb 23, 2013
180
Joined
Feb 23, 2013
Messages
180
Steve, the problem is that he wants to drive LED light bulbs, which contain a series capacitor and bridge rectifier as well as the LED string (and a small smoothing capacitor). You can't drive them with DC.

He wants to use a circuit in post #26 on this thread, which puts the current source across the output of a bridge rectifier which is connected between the mains supply and the LED light bulb.

So it's not clear how it would receive power. Also it would need to be configured as a current sink with a wide output voltage compliance range.

I just need an 300v switching voltage regulator or power supply (adjustable down to 100v). it makes no difference if it's DC or AC i can easily strip away the rectifier bridge and capacitor circuit.
 
Last edited:

CDRIVE

Hauling 10' pipe on a Trek Shift3
May 8, 2012
4,960
Joined
May 8, 2012
Messages
4,960
He wants to use a circuit in post #26 on this thread, ....

Yeah, but I sure wish BL didn't overlay all those current squiggles all over those schematics. They obscure and confuse what I want to see. Makes me dizzy too! :eek:

I just need an 300v switching voltage regulator or power supply (adjustable down to 100v). it makes no difference if it's DC or AC i can easily strip away the rectifier bridge and capacitor circuit.

BL, is this dimmer going to replace a wall (240VAC) switch? In other words it's not plugged into a wall receptacle but rather becomes part of the house wiring that's installed in a standard wall box? Do you know where I'm going with this?

Chris
 
Top