NPN transistor usage
Javalin
Posts: 892
Chaps,
Im using a TIP31A·NPN transistor·(·http://www.st.com/stonline/products/literature/ds/4137.pdf·) to switch on/off components on a power board upto 3A.
The NPN Collector to Emittor is between the +12V and the load.· The load is wired to ground.· This is opposite the usual diagramed way of using the C-E to switch to ground.
It works, but for example the 9V/1A regulator and the 3A capable NPN get quite hot with only 1/4 amp load.
Is this normal or am I missing something?
Also - what voltage should be applied to the Base to switch the transistor?· 12V?
Thanks
James
Im using a TIP31A·NPN transistor·(·http://www.st.com/stonline/products/literature/ds/4137.pdf·) to switch on/off components on a power board upto 3A.
The NPN Collector to Emittor is between the +12V and the load.· The load is wired to ground.· This is opposite the usual diagramed way of using the C-E to switch to ground.
It works, but for example the 9V/1A regulator and the 3A capable NPN get quite hot with only 1/4 amp load.
Is this normal or am I missing something?
Also - what voltage should be applied to the Base to switch the transistor?· 12V?
Thanks
James
Comments
Preferably 5V on the base is normal, but I think you can go from 1 to 12V with a protection (not sure about the value, I use 1K) resistor FOR ALL VOLTAGES.
Rafael
What voltage is supplying the·base lead?
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Ken
You have answered the heat question yourself, but perhaps don't realize it.
You are using the NPN transistor as a "high side" switch (abnormal configuration) rather than as a "low side" switch, which as you say is normal, and I suspect therein lies the answer.
In order to saturate (fully switch-on) the NPN transistor in this high side configuration, the base needs current supplied trhough a limiting resistor from a supply more than 0.6 volts higher than the collector voltage; 12+.6=12.6 volts. Likely you are just switching it to only 12 volts, hence you are not saturating the transistor, and therefore develop heat.
Finding a source for switching the base to higher than the collector voltage may be more trouble than its worth. Can you not rewire the load to be connected from the 12 volt source to the collector, and ground the emitter?
If not, then you should use a PNP transistor with its emitter connected to 12 volts, and the load between the collector and ground, and switch the base voltage (current limited by a resistor of course) from above 11.4 volts (12-0.6) for OFF, to below 11.4 volts for ON.
Cheers,
Peter (pjv)
Thanks for the reply - thats made it a lot clearer.
A further question if I may;· I will change the NPN's to PNP's as you suggest.· If then I want an SX (or stamp) to be able to switch the PNP on/off in order to get the 12V i'd use a darlington or another smaller transistor.· The thought then occurrs that the 0.5 to 1.5V voltage drop across the controlling transistor would leave me back in the original position?!
I've attached a diagram to clarify.
Cheers
james
If your load is not hard wired to ground I suggest the following simple circuit (IRL520). No heat sink, no extra transistor, done.
If you must source 12 volts to your load, and you want to control it with the SX or stamp (a 5 volt signal) an additional transistor is needed.
The 0 volts stamp output will certainly be able to provide a ground to turn your load on with a properly selected PNP transistor, but the 5 volts will not be able to turn the load off.
A typical configuration is also attached.
Regarding heat, with a typical ~1.2volts CE of a darlington and a conservative 2 amps for your application, the transistor will need to dissipate over 2 watts(1.2 * 2 = 2.4 watts), a heat sink would be needed.
The mosfet solution would not need a heatsink unless you were switching it on and off rapidly for more than 5 minutes or so.
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Ken
Post Edited (KenM) : 9/5/2005 8:01:09 PM GMT
The drawing in your schematic is incorrect. You have the emitter of the PNP transistor at a lower voltage (through the grounded load) than its collector. You have the emitter and collector reversed. In my post I said that the PNP emitter should be connected to 12 volts, and the load between the collector and ground.
As further stated, this does require you to pull the PNP base to over 11.4 volts for shutting the transistor OFF, and below that to turn it ON. This is best accomplished with a second small NPN transistor with a grounded emitter configuration.
If you follow Ken's second drawing, you will get the desired results. Depending on load current, make R1 somewhere between 22K to 47K, and R2 from 1K to 2.2K (neither are critical).
Also, it is best NOT to use a darlington transistor for the PNP power transistor as by their internal configuration they cannot saturate (switch on hard) to better than about 0.7 volts. Non darlingtons can saturate to better than 0.1 volts. This means that at 1 Amp, a darlington would need to dissipate 1 Amp x 0.7 volts = 0.7 watts, whereas the non-darlington would dissipate 1 Amp x 0.1 volts = 0.1 watts; the former requiring a heat sink at that current, and the latter probably not.
Hope this helps.
Cheers,
Peter (pjv)
Thanks for the info!! Everytime I get the feeling "yeah I know what i'm doing.." I always get slapped back......
What do I need to do to calculate the values of the resistors R1 and R2? I guess R1 is the pull up for the PNP and R2 is to limit the flow to ground. What specs on the transistors would I need to look at to arrive at these values?
Cheers
James
No one is wanting to "slap you back", we're just trying to answer legitimate questions.
The purpose of R2 in Ken's drawing is to provide base current to switch the PNP transistor. It's value is determined by choosing the base current which is dependent on the required collector current. For a 1 Amp collector current, and assuming a saturated current gain (gain varies with current) of 50, that suggests a base current of 20 mA. So calculate the resistor value as 11.4 (that is 12 VCC - 0.6 VBE) divided by 0.02 amp = 5,700 ohms, and then pick a lower value such as 1K or 2.2K to ensure sufficient current in case of lower gains or lower temperatures (gain drops significantly with lower temperatures). As said before, for switching purposes, the values are not critical. For analog amplifiers, that statement is not true.
The purpose of R1 is two fold; it bleeds off any leakage current in the PNP transistor's collector-base junction, as well as charge stored in the base junction capacitance. Any such current in the PNP base region would be multiplied by the transistor's gain (perhaps as high as 100 or more at elevated temperatures), and that would slow down switching speeds and be undesirable. Generally the lower the resistance value the better, but not to excess; the NPN transistor and resistor R2 need also to pass this current, so using 5% or 10% of the PNP base current required for the switching function is in the reasonable range. Sometimes with good low leakage transistors you can get away without the R1 resistor, but it's not recommended especially if fast switching is desired.
Hope this reasoning makes some sense.
Cheers,
Peter (pjv)
Post Edited (pjv) : 9/5/2005 8:25:04 PM GMT
""No one is wanting to "slap you back", we're just trying to answer legitimate questions.""
No, No, No - not what I meant!! More a personel feeling of "yes I know it all" then you realise there's still a lot you dont know.
Just reading the rest of the post......
James
If you did not know, the base current of 20 mA given by Peter was found by dividing the 1 Amp collector current by the transistor gain of 50.
In your original post, it indicates that you will need to switch up to 3 amps, is that correct?
If indeed 3 amps is your worst case collector current, the base drive current must increase,·and using the same transistor gain of 50, the value of R2 becomes much smaller, 190 ohms. (11.4 / [noparse][[/noparse]3/50]) = 190 ohms.
It gets worse....That 190 ohm resistor now has to dissipate over 1/2 watt of power.
And.....by the time you are·done the transistor will likely·need a heat sink at 3 amps of collector current unless you can drive it "deep" into saturation.....Vce =·0.1v or less.
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Ken
Yeah the purpose of all this is to have one 12V battery (7AH) to power a robotics application, with a power distribution board able to switch components on/off and provide different voltages for different bits. I.e. It could be programmed to switch off most things overnight and switch back on in the morning...
So 3A is what i've spec'd for each load, but in reality more like < 1A.
This is all getting a bit above me, i've ordered the electronics book Jon Williams keeps recommending and will read. I'm reluctant to make yet another board which does'nt work as expected - its very frustrating....
I may well scale the individual loads down to 1.5A or 2A following this....
Many thanks for the help - I will be refering back to this post later in the week when I have read up a bit.....
Cheers
James
You will still need to provide some sort of higher voltage (~+/-12V) to activate.
Nate
Yeah I looked at them, but was a bit put off by what happens when the fail - hence (so I read) their nickname "Flame Effect Transistors".
As this app will be (eventually) running 24*7 and unattended, ........
Cheers
James
·· Poor design can result in failure of almost any semiconductor component, some in quite interesting ways.· That shouldn't stop you from using the most appropriate part for the job.· If FETs were that bad they wouldn't be used in so many devices.
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Chris Savage
Parallax Tech Support
csavage@parallax.com