Hi

I need 100mA on the collector for my loads
I need to use a 1.3V for my power supply.
I don't need to use a LED. becuase they drain current.
and i have a low supply voltage.
Are my calculations correct!
I set this circuit up.But do not know if it working.
How can find out if this transistor is on and working as a switch.
I can't use a LED as a indicator do to my low Power supply.

How can i test this with a multimeter do see if the transistor
turns on and turns off.

Forgot to mention in the datasheet I use a gain of 215 0r 220 for my calculations. in the chart at 100mA Is this correct.

 

Hello,

The transistor will not pull down the load completely.
There will ALWAYS be a voltage across the transistor called Vce(sat).
At 150 mA this voltage is 0.4 Volts according to the attached datasheet (see page 2).

To reach the saturation, the base current must be about 1/10 th of the collector current.

Bertus

 

To see if it is acting as a switch (and how good a switch, at least at DC), simply measure the collector voltage relative to ground and see if it is close to 1.3V when off and close to Vce(sat) when on.

Does the 13Ω resistor represent your load, or is it an additional resistor that you think you need?

Since the collector will only get down to about 0.4V, you will have about 900mV to 1V across the load. That means that the total effective resistance in the collector path has to be below 10Ω.

 

It looks like the 13 Ohm resistor is the load. So, you are assuming 0v saturation voltage.

How accurate does the collector current have to be?

You cannot drive an LED with 1.3v -- so no 5v supply?

You should add a resistor from base to emitter to ensure the transistor can shut off completely. 10K or 50K should do it.

 

Hi mcasale

I was using the 13ohm for the load to try and test to see if i can get 100mA out of the collector.The two devices i have one put out 10mA and the other 68-70mA so i will need to compensate for these loads.
How can i get 100ma out of the collector.And yes i will use another resistor on the base to ground for shutoff.I think its 10X the base resistor.Is that correct!

Bertus
How do i get the base at 1/10 of the collector.to get me close to 100mA.

 

How do i get the base at 1/10 of the collector.to get me close to 100mA.

130Ω base resistor

 

I simulated the circuit on a simulator. Looks like the best way is to use a darlington transistor TIP122.

Fig 1 is the original circuit giving 27.9mA
Fig 2 would give 35.1mA if the load resistance is reduced to 1Ω
Fig 3 is using TIP122 to replace 2N4401 with a load resistance of 11Ω would give an Ic of 100mA.
Fig 4 uses two 2N4401 connected as darlington pair with 3V battery and RL of 6Ω would give an Ic of 92.6mA. The Vce on the 2nd transistor is 2.43V.

The internal resistance of the battery is set to 0.1 Ohm. See the attached for yourself.

Allen

 

Hi mcasale

I was using the 13ohm for the load to try and test to see if i can get 100mA out of the collector.The two devices i have one put out 10mA and the other 68-70mA so i will need to compensate for these loads.
How can i get 100ma out of the collector.And yes i will use another resistor on the base to ground for shutoff.I think its 10X the base resistor.Is that correct!

Bertus
How do i get the base at 1/10 of the collector.to get me close to 100mA.

By "two devices" are you referring to the transistors? Are these two the same part number, or different?

Be sure you are using a power resistor for your load. At 100mA a 10Ω resistor is going to be dissipating 1W, which is a surprising amount of heat.

To get 100mA from a 1.3V source when the collector saturation voltage is 0.4V means that the load resistance has to be below 9Ω. With a 13Ω load you might well not be able to get it above 0.9V/13Ω=69mA.

To get 10mA into the base, just recognize that the base voltage is going to be at about 0.7V to 0.9V. Call it 0.8V. You are applying 1.3V on the other side of a resistor connected to the base. What size does the resistor need to be in order to get 10mA? Be sure to consider how much that could vary if the base voltage is only 0.7V or if it is 0.9V.

 

I simulated the circuit on a simulator. Looks like the best way is to use a darlington transistor TIP122.

Fig 1 is the original circuit giving 27.9mA
Fig 2 would give 35.1mA if the load resistance is reduced to 1Ω
Fig 3 is using TIP122 to replace 2N4401 with a load resistance of 11Ω would give an Ic of 100mA.
Fig 4 uses two 2N4401 connected as darlington pair with 3V battery and RL of 6Ω would give an Ic of 92.6mA. The Vce on the 2nd transistor is 2.43V.

The internal resistance of the battery is set to 0.1 Ohm. See the attached for yourself.

Allen

Hi Allen
I was thinking to use a darlington in the first place.But? didn't know how to calculate for the resistors.Because i never used one before.That's way i when with the 2n4401.And even if i used two 2n4401 it will get near the 100mAs but? still.I would need 2.43Vs or more for my power supply.Which i can't use.Since you have show me,that a darlington will get me 100mA for my two loads.This would get me 100mA at the 1.3volts i can use for my power supply.This is a the way to go.Seeing those four simulations that you have shown me.

The enclosed diagram showing what i want to do.I want to use a SW to turn on my two loads at a specific timer.This timer I'm using can only work on 1.3V-1-5Vmax. This will make better sence to what i'm trying to do.

Any and all help is appreicated.And to all who posted in the thread. I Thank you all for you help and advice.

absf
Which simulation program did you use to get those results.Please post the program you used. Thanks. And any further Help on my circiut is appreciate. Thank very much.

 

This is noticeably different than what you originally posted. In that diagram, you had your loads being powered by 1.3V, not 3.7V. That makes all of the discussions about this 13Ω load resistor pretty much meaningless.

How much current can your timer source? More to the point, can it source at least 10mA?

If so, then do what I suggested in my previous post - put a resistor between the 1.3V timer output and the transistor base that will result in 10mA of base current. Hook the collector directly to the load (don't put additional resistance there).


Also, note the mA and V are the correct units for millamps and volts. Don't use Vs or mAs because 's' is the unit for time (seconds) and mAs is a unit of charge (namely 1 mC or millicoulomb) while Vs is a unit associated with the product of inductance and current change (somewhat analogous to impulse in mechanical systems).

 

I totally agreed with WBahn, If we knew that there was a 3.7V supply, your problems would be solved long ago without going through all these discussions.

What is the purpose of the LDO? Is it really needed? Is your timer running on 1.3V? If you can switch 3.7V directly through the timer, it would be much easier to get 100mA into your devices without using the Darlington. The TIP122 is an overkill as that is the only darlington in my simulator.

Which simulation program did you use to get those results.Please post the program you used.

The simulator I used is proteus. You can use multisim or LTSpice which would work the same for you.

Allen

 

I totally agreed with WBahn, If we knew that there is a 3.7V supply, your problems would be solved long ago without going through all these discussions.

What is the purpose of the LDO? Is it really needed? Is your timer running on 1.3V? If you can switch 3.7V directly through the timer, it would be much easier to get 100mA into your devices without using the Darlington. The TIP122 is an overkill as that is the only darlington in my simulator.



The simulator I used is proteus. You can use multisim or LTSpice which would work the same for you.

Allen

What can i use to switch on my two loads with. Please help.
Allen.Sorry. I already throught to do that. But? the timer cannot run on any voltage above 1.5V. If it did.There would've been no reason for this Thread.

The timer can not go above 1.5Volts max. I tested the timer.If it gos above 1.5V the timer dose not work correctly.To much voltage running
through the timer above 1.5V.This is why i used a LDO to drop the votlage to 1.3V. The timer is need and must be used in the circuit. So i can program the output to shut off/on any time i need.
If the tip122 is overkill.
Is there a way to get 100mA on the collector with any transistor or darlington.And keep the 1.3or1.5 Volts on the output of the Timer.
And with out using a heatsink. I should of posted the diagram when
i posted the thread. Sorry.

Your's and everybodies help on this is appreciated.

 

What can i use to switch on my two loads with. Please help.

I have already told you TWICE what to do!

Take your 2N4401 transistor and put a resistor between the base and the timer output that limits the base current to about 10mA. This will result in the transistor being placed firm in saturation for collector currents below 100mA. The collector voltage will be about 0.4V and the base voltage will be about 0.8V.

Is there a way to get 100mA on the collector with any transistor or darlington.

You keep shifting requirements. Is the goal to have a transsitor that acts like a switch that can handle at least a total of 100mA from the combination of two loads, or is it to control the collector current so that it IS 100mA?

When you turn on a light switch in your house it may be rated to handle 15A but the actual current flowing in it is dictated by the load. A 60W bulb will pull about 0.5A. So it doesn't make sense to ask what switch you should use to get 15A. You want a switch that can handle at least 15A.

Is that what you want here? A transistor switch that can handle at least 100mA? If yes, then that is what everyone else here has been talking about.

 

This is noticeably different than what you originally posted. In that diagram, you had your loads being powered by 1.3V, not 3.7V. That makes all of the discussions about this 13Ω load resistor pretty much meaningless.

How much current can your timer source? More to the point, can it source at least 10mA?

If so, then do what I suggested in my previous post - put a resistor between the 1.3V timer output and the transistor base that will result in 10mA of base current. Hook the collector directly to the load (don't put additional resistance there).


Also, note the mA and V are the correct units for millamps and volts. Don't use Vs or mAs because 's' is the unit for time (seconds) and mAs is a unit of charge (namely 1 mC or millicoulomb) while Vs is a unit associated with the product of inductance and current change (somewhat analogous to impulse in mechanical systems).

Sorry. Should of posted the full diagram when i posted this thread.
I do not know how much the timer can source. I do know it can't
take more then 1.5Volts max through it.If it dose the timer dose not
work correctly. Which transistor should i use to get 100mA.
Should i use the TIP122 as absf suggested. or the two 2n4401.
the TIP122 is overkill. please reccommend a transistor.

What value resistor should i put between the timer the transistor base. 9ohms, 11ohms, or 6ohms.

 

If the timer is putting out 1.3V and the base of the transistor is 0.8V, then there is 0.5V across it. In order for this to happen when 10mA is flowing through it, the resistance has to be, by Ohm's Law, R = V/I = 0.5V/0.01A = 50Ω. Since 51Ω is the nearest standard value, use that. This resistor will dissipate about P=(0.5V)(10mA)=50mW, so an 1/8 watt resistor should be fine.

The transistor itself will dissipate about (0.8V)(10mA)=8mW in the base circuit and (0.4V)(100mW)=40mW in the collector circuit, so about 50mW, which is way below the 650mW that a 2n4401 is rated for (elbeit when mounted on an FR4 circuit board).

So, as asked before, the real issue is whether you can tolerate the fact that the collector will not get all the way down to 0V like an ideal switch should but will, instead, be up around 0.4V. This will have basically the same effect as reducing the 3.7V to 3.3V and connected the low side of the loads directly to ground.

 

In my datasheet the 2N4401 is only a 325mW device. Better choices are MPSA42/43, 2N3019 or BC337.

fig 1 Using 2n4401 with base resistor 50Ω, VCE(sat) is 3.12V to get Ic=100mA.
fig 2 Uses MPSA43 with base resistor 330 ohm to get Ic 101mA, the Vce is only 0.84V.
A 1Ω resistor is put in series with R4 27Ω. So you can adjust the load current to 100mA by putting a mV meter across the 1Ω resistor. When 100mA is flowing in the load the meter should read 100mV.
fig 3 is using BC337 with Rb 220Ω but the Vce was also also quite high at 2.09V.

So make your choice wisely.

Allen

 

I don't know that I buy that Vce(sat) for the 2n4401 is 3V+ at 100mA.

Here's the datasheet I'm referencing:

2n4401 Data Sheet

The limiting power dissipation is given as 630mW (I had closed the document and so misremembered it a bit). But that is the limiting power dissipation. The thermal resistance from junction-to-ambient (mounted on FR4) is 200K/W. With a 150C max junction temp and assuming a 25C ambient, that would be 125K/200K/W or 625mW.

The Characteristics give a max Vce(sat) of 400mV with Ic=150mA and Ib=15mA, so it should be less than that at 100mA and 10mA respectively.

Now, that's based on the Philips data sheet. Is it at odds with others?

 

You're absolutely right! Now that I read the datasheet, I found out that it is a 625mW 600mA device. But I cant understand why my simulator listed it as 350mW 150°C device....

 

Your simulator does not know anything about arithmatic or electronics.

 

WBahn

Thanks! You guy's rock.

You are correct! In post #13 Both of your replies. I want at lease 100mA on the collector. The two loads will only draw what they can handle.
As you stated i will need 51ohms on the base to get 100mA.
Using the 2N4401.And no resistor on the collector.
I will also put another resistor from base to ground for shutoff.

2N4401
Onsemi Data 625mW
Philips Data 630mW
Fairchild Data 625mW
Micro Commercial 600mW

absf thanks.