Hi all I want to use a transistor has a switch but having a bit of a problem working out the resistance value.
I will be using a reed switch to do the triggering I’m working on 24 volts. I would like to use a 2n3055 or a Tip3055 now the load is at the max 4 amp. The in put out put voltage from the reed switch is 24v so I need a resistor to go between the reed switch and the bass of the transistor. Can any one help please.

I’m also looking for the same but on 12v same 4 amps
Any help will be appreciated.

I have looked at all the equations and illustrations on the net but it’s just got me even more confused.
So if it’s possible can anyone pleeeeeeease post some thing like, you will need a 680 ohms resistor?
Many thanks.

 

Hello Ron,
I took a look at ST Microelectronics' datasheet for the TIP2955/TIP3055.
Conveniently, for a 4A Ic with a Vce of 4V, hFE minimum is 20.
That means for a collector current of 4A, you need to put 1/20th of that current through the base, or 0.2A, or 200mA. That is really too much current to put through a reed switch, as it will very quickly burn up the contacts, causing them to either be permanently open, or welded together.

I suggest that what you need is a Darlington pair. It is better to buy them as a premanufactured unit, as they are more efficient that way.
Take a look at the TIP1xy series, where x=2, 3, or 4, and y=0, 1,or 2
"x" relates to their collector current capability; "y" is the max collector-emitter voltage where 0=60, 1=80,2=100.
These Darlingtons have (roughly) a gain of 2,500 when Ic is around 4A.
4A/2500=1.6mA, a nice low current to put through your reed switch for long life.

So for example, for the current limiter resistor for a TIP130, you're switching 24v, and we want 1.6mA current to the base of the Darlington.
Rlimit = E/I
Rlimit = 24v/1.6mA
Rlimit = 24/0.0016
Rlimit = 15,000 Ohms
Best check the power dissipation for the resistor.
P = EI
P = 24x0.0016
P = 38.4 milliwatts.
You could use a 1/10th Watt or larger resistor.

However, you may not be able to get TIP130's where you are. In that case, you may need to build your own Darlington.
You might use a 2N2222 or BC546/BC547/BC548 for the driver transistor, and a TIP3055 or the like for the power transistor.
But the TIP1xy option is better. See if you can get something like that. If not, we'll figure out what you might do with what you can get.

 

However, you may not be able to get TIP130's where you are.

Pssst... Hey Sarge... He's in the UK.

 

Pssst... Hey Sarge... He's in the UK.

I know that! What I don't know is if TIP130s (or the rest of that series) can be obtained there. I have never tried to order anything from a UK parts distributor, and it seems that Euro semiconductor markings are quite different from the US.

 

OK, just checked Maplin. They do carry a few of that series, even though they've typoed in their catalog (page 908); the capacities should be in Amperes rather than mA's.

A TIP120 or TIP121 would work, but they would be rather marginal.
A TIP142 (their stock# UJ30H) is considerably more expensive @ £2.26 including VAT, but would likely be a better choice.
Too bad they don't carry a TIP130 through TIP132.

Oh, the TIP1x5, 6, 7 are PNP Darlingtons. Since you originally specified NPN transistors, you probably wouldn't want a PNP Darlington.

[eta]Just checked on the BDX33C (stock# N80AH) they carry for £0.69 including VAT, this is a UK equivalent for a TIP132! This should work absolutely fine with the 15k current limiting resistor, and save you £1.57 in the bargain.

You can order it right from this page:
http://www.maplin.co.uk/Module.aspx?ModuleNo=33874&TabID=1&QV=Y&C=EBook&U=Ecat

Now for the 12V current limiting resistor, just use 1/2 the value for the 24v current limiting resistor. They don't have to be exact; for example you could use a 14.7k for the 24v resistor. In this case, it's better to go down a bit in resistance to make sure the Darlington is saturated ON.

For the 12v resistor, use 6.8K to 7.5k.

 

hFE is not used for a switching transistor. The hFE of a 2N3055 transistor is a minimum of 20 at 4A but only when its collector to emitter voltage is 4V or higher (a linear amplifier). Its max saturation voltage is 1.1V at 4A when its base current is 1/10th which is 400mA when it is used as a switch.

The same rules apply to any transistor including darlington transistors.

 

Thanks for your response.
So a bit like this one I’ve just setout.

Sorry needed to edit the pic it was to big for the forum.

 

hFE is not used for a switching transistor. The hFE of a 2N3055 transistor is a minimum of 20 at 4A but only when its collector to emitter voltage is 4V or higher (a linear amplifier). Its max saturation voltage is 1.1V at 4A when its base current is 1/10th which is 400mA when it is used as a switch.

The same rules apply to any transistor including darlington transistors.

No no no no no. Just because a transistor has the hfe (gain) specified, it doesn't mean that it has to be used in its active region. The 2N3055 transistor can be driven in its saturation region as well, as a switch. The guideline here is that the collector current is about (at least) 10 times greater than the base current. Guaranteed that condition, the transistor can be used in that region.

Thanks for your response.
So a bit like this one I’ve just setout.

Sorry needed to edit the pic it was to big for the forum.

Since R1 and R2 are equal, Q1 will be over saturated. Using coarse figures, R1 should be 10 times greater than R2. On real calculations you have to take into acount Vbe and Vce for the biasing currents. Start by calculating the current biasing the load. Then Ib (Q2) should be 10 times smaller than that [Il = Ic (Q2)], and Ib (Q1) should be 10 times smaller than Ib (Q2) = Ic (Q1).

 

Actually, more like the attached. R2 and R3 are there to ensure that it's cut off when S1 opens.

You have the TIP3055 shown as a PNP; it's an NPN transistor.

Your base resistors are so high that you wouldn't get much current from them.

 

Now it's just got even more confusing for me. I want to build a test switch for experimenting that will be able to switch on a 24 volt positive rail by the means of a reed switch, but having the current handling capabilities from milliamps to 4 amps, is this possible. Or am I just going crackers again.
Now I bet your going to ask what the hell would you want to do it like that for? Well in a nutshell HHO environment. So no sparks during experimenting. But at this stage I don’t know the currant I will be using all I know is I will not be going over 3 amps. I have had 2 pop off’s, this has happened when I was switching the sell off.
It’s not nasty but I want to eliminate the possibility. And a reed switch is sealed and all the switchers I have used
hasn’t bean.
Thanks for all your help on this.

 

Yes you are correct again I just spotted that one, just as I hit the post button for the post above this one.

 

OK. You use NPN for switching the ground (current sink) on and off.

You generally use PNP transistors to switch source (current source) on and off.

This does change things a bit.

I give up for now; my circuit emulator has now blown up on me twice in a row.

 

No no no no no. Just because a transistor has the hfe (gain) specified, it doesn't mean that it has to be used in its active region. The 2N3055 transistor can be driven in its saturation region as well, as a switch. The guideline here is that the collector current is about (at least) 10 times greater than the base current. Guaranteed that condition, the transistor can be used in that region.


Since R1 and R2 are equal, Q1 will be over saturated. Using coarse figures, R1 should be 10 times greater than R2. On real calculations you have to take into acount Vbe and Vce for the biasing currents. Start by calculating the current biasing the load. Then Ib (Q2) should be 10 times smaller than that [Il = Ic (Q2)], and Ib (Q1) should be 10 times smaller than Ib (Q2) = Ic (Q1).

Audioguru's point was that the hfe at Vce=4V is almost irrelevant to the base current required for saturation. The curve below is the one which is relevant in this case.

 

Actually, more like the attached. R2 and R3 are there to ensure that it's cut off when S1 opens.
...

The transistor will cut off by itself (it is a bipolar transistor). Remember that the switch is connected to the positive and not to ground, so current biases R1 when the switch is closed. R1 prevents Q1 from over-saturating.

Audioguru's point was that the hfe at Vce=4V is almost irrelevant to the base current required for saturation. The curve below is the one which is relevant in this case.

Yes, indeed. He is right then. Although, Vce will be close to 0.4V in that situation. After all it was a typo. I just assumed that he was thinking that the transistor would forcibly be in its active region. Sorry!

 

OK, take this schematic.
Change the number on the TIP3055 to TIP2955. You really should be using a PNP instead of an NPN. You have depicted a PNP transistor, which is what's needed.
Change R1 to 15k Ohms. This limits the current through the reed switch to about 1.5mA.
Change R2 to 240 Ohms. This limits the maximum current through the BC548 to 100mA.

 

Thank you all for your help.
I have amended the mistake in my first diagram. Again thanks to all.

 

OK, same circuit for 12v, except:
1) R1 should be 6.8K to 7.5k
2) R2 should be 120 Ohms.

 

Thank you all for your help.
I have amended the mistake in my first diagram. Again thanks to all.

That's more like it.

 

That circuit leaves Ron with a base drive of less than 100mA for his 2955, when he wants to handle up to 4A collector current. That requires a saturated beta of 40, which the 2955 is not guaranteed to have. I think R2 needs to be changed to 56 ohms, and R1 to 1k. Keep power dissipation in mind for both resistors. BC548 will not handle 400mA, so it should be changed to something like BC639.

 

Hi Ron H is that for 12 volts or 24 volts. Thanks to everyone.