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Is the transistor calculation is right?

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medhatko

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TIP160.jpg
please tell me if my calculations with the attach picture True or False؟
 

from data sheet
Isn't it the saturation base emitter voltage?
12.jpg
 

The datasheet for this power Darlington transistor was 46 years old!
No transistor is biased with a single resistor like that because a transistor part number has a range of hFE and this one had a minimum hFE of 200 and most had a higher hFE, causing them to be saturated.
 
Hi,

I'd like to know: Is this a switching application, where the load is just switched fully ON and fully OFF, or is this a linear regulating application?

Klaus

And: is this just for learning or do you want to build a real circuit?
 
Something else doesn’t make sense here. You’ve got a 5A, 45W load. This implies Rload=1.8 ohms. 5Ax1.8=9. You have somehow gotten 9volts out of an 8 volt supply.
 

Something else doesn’t make sense here...

Yes. I see same. The OP needs a supply of at least (9+Vcesat) Volt. I would go for a higher voltage source though to allow the load resistance to limit the collector current.

From the look of things, it is a switching application but the diagram needs to give more info on biasing. That diagram makes it look like the OP just connected the base resistor directly to the supply rail.

The value of hFE in the calculation is way too high. I suggest 20 or so.
 

Hello,
Yes it is a switching application.
the load is a lamp with 6V , 5A
I just for learning
Is my calculations is right?
 

The data sheet DOES quote 2.2V maximum Vbe(sat) but at 10A pulsed collector current and at -30C temperature.
This is a rather unusual Darlington transistor for pulsed applications with 320V Vceo/Vbeo and a very wide operating temperature range. It also has an internal parallel diode.
hfe is around 250 at 5A collector current.

Brian.
 
practically ,
voltage drope between LED terminal is 4.66v even i exceed the Vcc to 9 or 10V??
Vbe=1.49V
Vbc=0.54V
 

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What's the intended circuit purpose, switch or current source?

Where did you get the assumed current gain of 30? Are you aware of the large type variations of actual current gain?
 

transistor 2n2222 is wrong picture
the main picture above with TIP160 work as switch
 
practically ,
voltage drope between LED terminal is 4.66v even i exceed the Vcc to 9 or 10V??
Vbe=1.49V
Vbc=0.54V

Hi,

Do not use Vcc. Use Vbb so that it would represent your IC output pin voltage, for instance.

Rb = (Vbb - Vbe)/Ib
Ib = Ic/hFEsat

hFEsat should be of low value, considerably lower than the hFEmin given in the datasheet. That way, the transistor will always stay in saturation and your load resistance will then limit Ic. Of course you should select a transistor that can comfortably handle the load current.

One thing you have to realize is that a diode will tend to fight the the voltage source if the source voltage is higher than the forward voltage of the diode. In this case, you have to connect a low value resistor in series with the diode.


What's the intended circuit purpose, switch or current source?

Where did you get the assumed current gain of 30? Are you aware of the large type variations of actual current gain?

It is a switching application.

When a low enough hFE is used in the calculation it ensures that the transistor actually remains in saturation.

Take for instance that needed Ic is 100mA and minimum hFE is 50. This means that at worst case, Ib should be Ic/hFE = 100mA/50 = 2mA. If you go ahead and use a hFE of 20, you will arrive at Ib = 100mA/20 = 5mA. However, at worst case, Ib of 5mA should drive an Ic of 5mA*50 = 250mA. Your load resistance can then be used to limit Ic and your transistor will always remain in saturation.

I recommend that the OP uses hFE = 20.

- - - Updated - - -

The resistor in series with the diode should actually be sized properly to drop the excess voltage at the required diode current.
 

transistor 2n2222 is wrong picture
the main picture above with TIP160 work as switch

You said in post #8 that your load is rated for 6V, 5A? Datasheet specifies Vcesat of 2.8V for the transistor. That means you need a source of (6 + 2.8)V = 8.8V. The 8V you provided may be adequate depending on the type of lamp it is. If it is an LED lamp, then the lamp may not come on at all until at about 8.8V of supply voltage.

The datasheet specifies minimum hFE as 200. Use 100 or lower for hFEsat.

Let's say you are using hFEsat = 80, then Ib = Ic/hFE = 5/80 = 62.5mA.

Rb = (Vbb - Vbe)/Ib

With this, your transistor will be in saturation all the time and your load will limit the current flowing through it.
 

The value of hFE in the calculation is way too high. I suggest 20 or so.
The old TIP160 transistor is a Darlington:
1) At a collector current of 4A its saturation Vce is 2.2V when its base current is 4A/200= 20mA.
2) At a collector current of 6.5A its saturation Vce is 2.8V when its base current is 100mA.
3) At a collector current of 10A its saturation Vce is 2.9V when its base current is 1A.

Now you show a 2N2222 that is not saturated as a switch. Also your LED has nothing to limit its current. Also the "3.7V" battery is Lithium rechargeable that is 3.7V when in storage but is 4.2V when fully charged. Also you do not have enough voltage from a 3.7V battery for the 4V LED plus a max saturation Vce of about 1V when its base current is supposed to be 250mA/10= 25mA. The "10" is the collector current/base current for saturation as shown in the datasheet for the 2N2222 and most other little transistors.
1)
 

An hFE of 10 gives about 3.3W power loss at the base resistor. That's not good. The goal is to get the transistor to always stay in saturation. Normally you need to use an hFE considerably lower than the datasheet spec for your maximum operating temperature. Since the datasheet for this transistor doesn't have such chart, then we can test with various hFE's ourselves and see how the behaviour goes around our maximum operating temp.

It is advisable to start with hFE = 100 here since the minimum spec for 25deg is 200. Our Rb power loss would be about 330mW which is better. If we find out that we need to use lower hFE, then we can then test with hFE = 50 and Rb loss of 660mW. We can go on like that till we determine a comfort zone.

In this case, the OP would have to change only one component which is Rb, and that shouldn't be a problem.
 

Hi,

For a transistor in saturation assume Ic=10*Ib
This is a rule if thumb for non darlingtons.

Let's see what the datasheet says:
For saturation mode there are a couple of values:
I_C 6.5A --> I_B = 0.1A --> ratio of 65 (I don't want to talk about h_FE here, because this usually is for linear mode)
I_C 10A --> I_B = 1A --> ratio of 10 ( here you have it ;-) )

Btw: In saturation mode you don't use h_FE (linear mode) of the datasheet. Neither table nor charts.

Increasing the base current further won't decrease V_CE much, I think. Thus there is no need to do so.
But you say "light bulb"..... and for a real circuit here arises a problem: a cold filament (at switching ON) easily draws 3 time the current when hot. With too low I_B you get too high V_CE and the bulb too less voltage to get hot in a short time.
--> Thus you get increased switch ON time with increased power dissipation. For a non heat sunk transistor .... in worst case ... his could cause overheat and destruction.

There are several solutions:
* increased base current (high power dissipation in base resistor)
* or increased base current just at switch ON - maybe with additional C and R acrosse the base resistor
* or - what I would do - use a Mosfet.

A Mosfet is the modern way to go. The voltage drop may be in the milliVolts, the gate current may be in the microamperes.
This means low power dissipation .. thus small cases without heatsink can be used. A SO-8 can handle this easily, maybe even a SOT23.

Klaus
 

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