Dear Friends
Please help! I want to dim single 75W LED Module (24VDC and 3.2A)
I used TIP127 but it gets hot beyond operating temp and stalls.
Please help.
You are using PWM?
Odd - the transistor is rated at 5 amps.
Maybe try a MOSFET.
Can you post your circuit?
Yes, a BJT is going to get hot. A MOSFET with a fairly low Rds(on) value (<0.1 ohm) will be better. Circuits for doing this kind of thing are shown here. Scroll down to the part that talks about logic-level MOSFET's, replace 12V with 24V, the lamp with your LED module and you should be OK.
For an explanation of how to calculate power dissipation see:-
http://www.thebox.myzen.co.uk/Tutorial/Power.html
and
Hey! Thanks.. I had read them before in related posts. I need help on schematics too. I am not getting the transistor configured right. Is there text I could read on Transistor Calculations and Configurations.
The LEDs I am using are 24V and 3.5A max (totaling 75W). I need to Fade In and Out 4 rows of LEDs.
Have look at http://www.satcure-focus.com/tutor/page5.htm or just google Transistor circuits.
However there is not much to it, it's just ohms law.
Base resistor should be made so that when it goes up to 5v it supplies enough base current.
Base current = collector current / transistor gain. or slightly more for safety. Transistor power = collector current * turn on saturation voltage (from the data sheet)
To determine the temperature rise you need to calculate the junction
to ambient thermal resistance (Tja) ---
Tja = Tjc + Tcs + Tsa all the dimensions are DegC/W
Tjc thermal resistance between the IC junction and IC case. This is
from the IC datasheet.
Tcs thermal resistance between the IC case and your heatsink. This
should also be in the IC datasheet. Usually you place a thermal
compound between the IC case and the heat sink to decrease the
thermal resistance. The decrease occurs because the thermal
compound fills the air gaps between the two surfaces and conducts
heat better than the air.
Tsa thermal resistance between the heatsink and ambient air. This is
in the heatsink datasheet. Most heatsink datasheets specify Tsa
with and without airflow. You can determine if you need a fan by
reviewing this number.
For a single heatsource the junction temperature would be
Tj = Tja * P + Tamb (DegC)
where P is the power dissipated in the IC and Ta is the ambient
temperature. For a MOSFET P = rds(on) * Id (NB: use the rds(on)
value at the actually operating temperature). For BJT
P = Vce(sat) * Ic (NB: use the Vce(sat) value at the operating
temperature).
The maximum juction temperature is listed in the datasheet (usually
in the "Absolute Maximum" section). I would not run the device at
a temperature greater than 80% of the the absolute maximum rating.
Unless you have very short on times PWM is not going to help
much with power dissipation. To see how much look at the transient
thermal response graphs in the datasheet.
(* jcl *)
A PNP transistor is not ideal for your circuit. It would probably need an additional NPN transistor to drive its base, if used to switch 24V.
A darlington transistor dissipating ~6W (2V Vcesat * 3 A) will need a pretty substantial heatsink, even connected in ideal fashion. You didn't say whether you had one.
I had a heat sink on specs as below:
- Heat Sink
- Package/Case:TO-220
- Thermal Resistance:65°C/W
- Mounting Type:Screw
- Length:44.5mm
- Height:12.7mm
- Width:17.8mm
- Color:Black
- Sink Material:Aluminum
- Size:1.75L x 0.500H x 0.700W
I bought it from Farnell
http://in.farnell.com/wakefield-engineering/271-ab/heat-sink/dp/1699982
Still the Transistor gets so hot that it burns the PCB too.
I think I will start working on NPN driving the PNP.
Still the Transistor gets so hot that it burns the PCB too.
Yes it will do. As your heat sink has a thermal resistance of 65C/W that means at 3W there will be a 195C temperature drop between the transistor case and the heat sink. Therefore to get the case running at 100C the heatsink needs to be held at a temperature of -95C.
Put simply this transistor is burning too much power to use.
You need a FET not a transistor
It's not quite that bad. While the Farnell page says 65C/W, the datasheet says something more like "65C for 3W." I think. Still, this is meant (according to the datasheet) as a "booster" heatsink to be used in conjunction with another piece of metal, rather than on its own, so it's not really adequate for this application.
It's frequently depressing just how much heatsink is required for bipolar transistors!
I have make another schematics with a MJE13003 driving the TIP127.
http://www.scribd.com/share/upload/12034239/1tj9all7ps0ofnfx7w8z
Please see the attachment .pdf file with schematic and give suggestion.
To Update: The TIP127 does still get hot and a huge heat sink or fan is required to keep it going. Pls let me know if I have made any mistake in calculations.
No that won't work you will never be able to turn the Darlington fully off and there is nothing limiting the base current.
Darlington emitter to +24, load to collector to ground. Drive transistor 10K to 24v to collector and 1K in the base of the Darlington.