BJT Switch Design question.

[ghani256]

Ok so I am revising some transistors and I came across using BJT as a switch, so I decided to design a simple circuit to control an LED. I want someone to kindly check the steps that I followed and offer any suggestion, corrections or improvements that can be made.

The LED needs 20mA for full brightness and at that current has a forward voltage drop of 2.0V. The transistor I am using, 2N3904 has a saturation voltage of 0.2V. Also, I am using a 5V supply. So the collector resistor comes out to be Rc=(5-2-0.2)/(20*10^-3) = 140Ohm.

Now the minimum Hfe specified on the datasheet of the 2N3904 is 30, so I used that to find the base current as Ib = 20/30 = 666uA. Assuming Vbe of 0.7 volt, the base resistor can be found to be Rb = 6456Ohm. I can use the closest 6.2k.

Below is a schematic of the circuit I made in proteus, along with the simulated values of collector current and voltage.

[Zero999]

Now the minimum Hfe specified on the datasheet of the 2N3904 is 30, so I used that to find the base current as Ib = 20/30 = 666uA. Assuming Vbe of 0.7 volt, the base resistor can be found to be Rb = 6456Ohm. I can use the closest 6.2k.

You've done everything correctly except the base current may be a little low. Note that the Hfe is specified with a V_CE of 1V, so with a forced beta of 30, it might not quite saturate. However in practise, I think it will be fine. The transistor will probably be close enough to saturated for your application.

https://www.onsemi.com/pub/Collateral/2N3903-D.PDF

[danadak]

A rule of thumb I learned somewhere is to design for a forced beta of 10
to handle product and temp variation. In fact if my memory serves me
many datasheets show Vcesat at a forced beta of 10.


Regards, Dana.

[Ian.M]

That's a good rule of thumb, but if you need to be more conservative with the load on your logic output, calculate Ib from the minimum H_FE and double it.   Vce_sat will be slightly higher but that's often less of a problem than overloading a MCU I/O pin would be.

However if your operating Ic is far from the current the minimum H_FE was specced at, you'll have to search through the transistor datasheet for a H_FE vs Ic graph to see if H_FE may be lower than the spec at your operating Ic.

[danadak]

Excellent tip, I always thought the rule resulted in a lot of wasted power and
not exactly good LP design. And if one is driving lots of pins that way the
UP has internal power rail issues, as well as the output drivers.

I am curious, I might try your method out to see how much of a power savings
it portends.

There is one problem I see, the min Hfe, or Beta, not speced in sat, so I assume
thats why you double for safety....?



Regards, Dana.

[Ian.M]

Doubling the base current is just to ensure its actually saturated.  If you didn't double it, you'd be relying on the H_FE being far enough above the minimum specified.  When calculating the resistor, don't forget to allow for the drop in the logic '1' output voltage when sourcing the desired base current.

Anyway, its all obsolete nowadays with the easy availability of low Vgs threshold MOSFETs.

[BigBoss]

Don't put the transistor into hard saturation region.Arrange the resistors so how the transistor will work near saturation with a margin.

[Zero999]

In this case, I'd configure the transistor as an emitter follower and reduce the value of R1 to compensate for the extra voltage drop. Then there would be no base resistor to worry about.

[Benta]

Don't put the transistor into hard saturation region.Arrange the resistors so how the transistor will work near saturation with a margin.

Why not? This response interests me.

[BigBoss]

Don't put the transistor into hard saturation region.Arrange the resistors so how the transistor will work near saturation with a margin.

Why not? This response interests me.

Saturation region is inverse in BJTs compare to MOS.If you drive hardly the BJT, a Collector-Base current wants to flow through BC junction and this current returns to signal source.

[Zero999]

Don't put the transistor into hard saturation region.Arrange the resistors so how the transistor will work near saturation with a margin.

Why not? This response interests me.

Saturation region is inverse in BJTs compare to MOS.If you drive hardly the BJT, a Collector-Base current wants to flow through BC junction and this current returns to signal source.

We know that. The original topic of the discussion and the person you replied you mentioned saturating a BJT, not a MOSFET.

[danadak]

There is one additional consideration, using transistor in inverted modem, that is
collector and emitter swapped. That results for some devices in lower Vcesat, some
approaching 10 mV. Tradeoff is breakdown voltages and lower gain. This approach
was used in old designs for analog switching.



Regards, Dana.

[BigBoss]

We know that.

If you know that, you know what I'm talking about..

[Ian.M]

Here's a further scenario for young players:

You have a logic level output that's too weak to drive your NPN power transistor into saturation at your expected load current.   You have another NPN transistor with Ic_max greater than the power transistor Ib_sat you require.   You DO NOT have a heatsink for the power transistor, and your load cant tolerate a significant extra voltage drop. 

The solution is to use the smaller transistor as an emitter follower to boost the logic output's current capability, then use a suitable resistor between it and the power transistor base to provide Ib_sat at your expected Ic_sat.   Don't forget the Vbe drops of *BOTH* transistors when calculating the resistor.    The emitter follower reduces the load on the logic output by a factor of its H_FE, e.g making it easy to provide >100mA base drive to the power transistor from a logic output that is only rated for <4mA.

A pull down resistor may also be added at the driver emitter if you need faster turnoff  or lower off-state leakage current through the load at high temperatures.

[Zero999]

We know that.

If you know that, you know what I'm talking about..

You should have made it clear you were referring to a MOSFET.

Here's a further scenario for young players:

You have a logic level output that's too weak to drive your NPN power transistor into saturation at your expected load current.   You have another NPN trancistor with Ic_max greater than the Ib_sat you require.   You DO NOT have a heatsink for the power transistor, and your load cant tolerate a significant extra voltage drop. 

The solution is to use the smaller transistor as an emitter follower to boost the logic output's current capability, then use a suitable resistor between it and the power transistor base to provide Ib_sat at your expected Ic_sat.   Don't forget the Vbe drops of *BOTH* transistors when calculating the resistor.    The emitter follower reduces the load on the logic output by a factor of its H_FE, e.g making it easy to provide >100mA base drive to the power transistor from a logic output that is only rated for <4mA.

A pull down resistor may also be added at the driver emitter if you need faster turnoff  or lower off-state leakage current through the load at high temperatures.

That's a good idea. The downside is slower turn off. One way to improve that, is to connect a small capacitor (100pF to 10nF) between the logic output and the higher current transistor, bypassing the emitter follower and base resistor. Note though that there will be a high current spike, taken from the logic pin, when it goes high. If that's an issue (it shouldn't be, as the logic pin should be able to handle the surge) add a low value resistor 100R to 1k in series with the capacitor.

[danadak]

Actually his sim shows ~ 20 mA needed, so 2 mA into base easily
achieved by Arduino output to sat the transistor. The ATMEGA328
will drive up to 20 mA at 4.1 V (Vcc = 5V).

Rbase = (Vcc - Vbe) / 2 mA


Regards, Dana.

[Zero999]

Actually his sim shows ~ 20 mA needed, so 2 mA into base easily
achieved by Arduino output to sat the transistor. The ATMEGA328
will drive up to 20 mA at 4.1 V (Vcc = 5V).

Rbase = (Vcc - Vbe) / 2 mA


Regards, Dana.

I think he knows that and had no intention of suggesting two transistors to drive a 20mA LED off an ardunio.

Here's a further scenario for young players:

Often providing more information, even though it's not directly useful to the immediate problem is a good idea, because it helps people solve other problems in the future. What if the next question was going to be about driving a 2A motor? Then that would be useful.

[eetech00]

Hi

Add a resistor from transistor base to ground (about 10 * base resistor=62k) to ensure transistor turns off when switch is open. If you do this, then you don't need a SPDT switch. You can use a SPST switch.