# Transistor saturation question

Discussion in 'General Electronics Chat' started by J.Green, Jul 5, 2010.

1. ### J.Green Thread Starter Member

Jul 15, 2008
26
0
I am using a 2N3904 transistor to switch a 12v relay. The relay has a 25mA coil. The transistor is connected to a PIC.

The data sheet for the 2N3904 says the current gain at Ic=50mA is 60. To be conservative, I assumed 30 as the current gain. I calculated Ib as 25mA / 30 = 0.83mA. I calculated the resistor needed as (5.0v - 0.7v)/0.83mA = 5.16K.

Later I learned about collector-emitter saturation. I want to make sure that I am using the correct resistor at the base of the transistor to saturate the transistor. On a breadboard, with a 5.6K resistor, Vce=0.111v. The collector-emitter saturation graph in the data sheet looks like Vce(sat) = 0.06v for Ic=25mA @25 celsius.

Again on the breadboard, I replaced the 5.6K resistor with a 1K resistor and measured Vce=0.061v.

The table of the data sheet says Vce(sat) max for Ic=10mA and Ib=1.0mA is 0.2V. For Ic=50mA and Ib=5.0mA, Vce(sat) has a max value of 0.3V.

With a 5.6K resistor Vce is 0.111V, which is higher than 0.06V point (25mA at 25 celsius) for saturation according to the graph in the data sheet (fairchild). Switching to a 1K resistor, gives Vce of 0.061v. According to the table in the data sheet, as long as Vce is under 0.2V the transistor should be saturated. I am confused as to what value to expect for Vce when saturation occurs.

The table in the data sheet says the max Vce for saturation is 0.2V. The graph indicates the Vce = 0.06V for saturation. Will saturation occur at any value of Vce under 0.2V? In the graph, is the Vce shown the absolute minium that will be achieved or is a target that should be strived for? Practically, will there be any noticeable decline in transistor performance/life with a Vce of 0.111V versus Vce of 0.061V? Would Vc=0.111V still be considered saturated? During saturation, is there a range of Vce values that are expected when the the transistor is saturated or should the goal always be to minimize Vce?

A bunch of questions, but ultimately just trying to get a clearer understanding of when saturation is achieved and the Vce values I should expect when saturated

Thank you.
JG

2. ### Ron H AAC Fanatic!

Apr 14, 2005
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I'm not going to attempt to tackle each question individually, but the typical Vce(sat) for Fairchild's 2N3904 is summarized in this figure from the datasheet. Notice in the upper left corner it says "β=10". This means forced beta Ic/Ib=10. The saturation specs of almost all BJTs are specified with Ic/Ib=10.

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3. ### SgtWookie Expert

Jul 17, 2007
22,183
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The basic idea of "saturation" is that beyond a certain point, increasing the base current will not result in an appreciable increase of collector current. This is because Vce is as low as it is going to get.

As Ron_H mentioned, practically all transistors' datasheets have the saturation curves shown with Ib=Ic/10, or a forced beta/hFE/gain of 10.

The basic formula for determining a base resistor is:
Rb = (Vin-Vbe) / (Ic/10)
where:
Vin = the voltage on the end of Rb that is away from the base, with respect to Ve.
Vbe - typically around 0.63v to 0.7v, but this increases significantly as Ic approaches 50% of the maximum; for your transistor the maximum is 200mA, but the practical limit is 100mA.
Ic = desired collector current.

If you use the above formula for any given transistor, it'll be well saturated. If you use the minimum hFE specification, you risk having the transistor not fully saturated; thus having an excessive Vce, which results in higher power dissipation in the transistor.

4. ### J.Green Thread Starter Member

Jul 15, 2008
26
0
Using a gain of 10, the Ib would be 25mA / 10 = 2.5mA. The resistor would be (5-0.7)/2.5=1.72K.

If I used a 1K resistor, Ib would be (5-0.7v)/1K=4.3mA. The relay coil draws 25mA.

Would there be any downside in using a 1K resistor here?

Thank you.

5. ### timrobbins Active Member

Aug 29, 2009
318
16
JG, you've done a good job calculating and confirming the bjt performance. You may want to extend that to the relay wrt pull-in performance and current draw. If your PIC has no issue with the bjt base current requirement, then you final choice of base resistor is determined by a judgement on margins - margin of bjt operating point - and margin of coil operating voltage. Do you have any other issues, such as temperature range? If you're keen, you could determine the operating conditions for a few standard base resistor values - if they all give reasonable results then choose the middle value.

If you are trying to minimise supply power consumption then there are other things to consider.

Have you got a diode across the relay coil?

Ciao, Tim

6. ### SgtWookie Expert

Jul 17, 2007
22,183
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The downside is that you'd be using 1.8mA * 5v = 9mW more power than you really needed in order to saturate the transistor. That's not a lot, but it can add up rather quickly - particularly if your uC is going to be battery powered, or if you have a lot of I/O pins driving loads.

But then, think about that power usage being drawn from a mains-powered supply, where you're paying for every KWH flowing through the meter. How much will that 9mW cost over the lifetime of the circuit?

In lots of hobby projects that are simply put together just to see something work, then taken apart and used for another project, it really doesn't matter very much. However, if it's going to be in use for a long period of time, you should go to lengths to make it as efficient as possible, while staying well within recommended operating parameters.

1.8k is a standard value of resistance that is within 4.7% of 1.72k.
Decade table of standard resistance values: http://www.logwell.com/tech/components/resistor_values.html
Bookmark that page. Use the E24 values (green columns), as they are most commonly available with a low cost.

If you use E48 or higher values, the cost of the resistors will go up significantly - so reserve those values for special requirements, such as very accurate voltage dividers.

Don't forget that you should also calculate the power dissipation for the resistor, and double the requirement to determine the power rating requirement.

If you used a 1k resistor, then you'd need 4.3v*4.3mA= 18.5mW; x 2 = 37mW; you could use a 1/25 Watt or higher rated resistor. SMT/SMD resistors come in power ratings that small; for example size 0201. They are smaller than I'd care to handle; even 0805's are tiny - but a pick and place machine handles them just fine in a production environment.

I'm going to go on a tangent for a moment here on improving the efficiency of your circuit, specifically regarding the relay coil current.

Relays have a "pull-in" or "must operate" voltage specification, and a "drop out" or "must release" specification. For example, a 12v relay might actually operate at around 9v, and release at about 5v.

If you are controlling the relay using a uC that has PWM, you can use the PWM feature to reduce the average current flow through the relay coil to perhaps 1/2, saving that power, yet keeping the relay energized - after an initial application of full current for perhaps 50mS to 100mS. You would need a Schottky or fast recovery diode across the relay coil, of course - if you were not using PWM, you could simply use a 1N400x series diode, but those would not work well for PWM due to their slow recovery time.

Reducing total current use by half is a very significant power savings over time, and should be considered for projects that are going to be in continual use, or even intermittent yet regular use.

7. ### J.Green Thread Starter Member

Jul 15, 2008
26
0
Thanks for the replies.

Tim - no extreme temperature requirements. I have a 1N4004 across the relay coil. The PIC outputs can supply 25mA. I have run a few bench tests with various resistors to determine Vce.

Sgt - the power consumption is not really an issue. This is an automotive application. However, your PWM idea is very cool. I really like the detailed analysis/selection process for components you obviously are capable of. I need to build my knowledge for optimal component selection too. While throwing components at a project may work, my concern is longevity of the circuit.

BJT saturation: the saturation graph in the data sheet gives a single point at which Ic and Vce intersect. Using a gain of 10, the measured Vce is higher than than read from the graph. To get Vce of around 0.06V, I would need a 1K resistor. If I were to select a resistor based on the graph, as Sgt has pointed out, I would be wasting power. What is the indented purpose of the graph as it appears as to not help with selecting the optimal resistor?

Diode selection: I am using a 1N4004, which is rated for 400V, forward current of 1A, and reverse peak surge of 30A for 8mS. Interested in learning the correct way of selecting a diode for protection across a coil, as opposed to 'put it in and see if it works'. The coil is 12V, drawing 25mA.

Thank you.
JG

8. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
JG,
I'm afraid that you are incorrect in your "no extreme temperature requirements" assumption.

Automotive environments are among the most brutal on the planet. Temperature extremes, mechanical shock/vibration, electrical noise/power spikes, heinous chemicals, moisture/corrosion - it's all there, even if you're using a circuit in the passenger compartment.

Basically, your vehicle will beat your circuit to death very quickly unless ALL of your components are rated for the full automotive temperature range.

PIC uC's with the -E suffix are extended temp range, which covers the full automotive temp range of -40°C to 125°C. The -I suffix is for industrial temp range, which is not suitable for automotive use.

Your voltage regulator for your PIC's supply is a vulnerable spot. If it is not rated for at least 60v input, it is not suitable for automotive use. For example, the LM317 series and 78xx series are not suitable.

You are also mistaken in that power consumption is not an issue. In your vehicle, electricity is generated by spinning the alternator at 3x the speed of your engine, which can take a considerable amount of power. Automotive engines are in reality very inefficient; if you are getting 30% efficiency, you are doing extremely well.

Every milliwatt of power you use has to be generated by burning fuel, very inefficiently.

9. ### Jaguarjoe Active Member

Apr 7, 2010
770
90

As a rule of thumb, 1/3 of the energy consumed by an engine goes out the exhaust, 1/3 goes to the cooling system, and 1/3 gets to the crankshaft.

10. ### timrobbins Active Member

Aug 29, 2009
318
16
JG, datasheets that show max curves are meant as a bound. You have actual sample data. Sometimes you can overlay/translate an appropriate typ spec onto the max curve to get a feel for where the mean population will reside - but otherwise you would just use the max level as a worstcase calculation parameter, and as long as you were always on the good side of worst-case (for all worstcase parameter combinations) then your design is valid. But to do worstcase you need to fully appreaciate all the other parameter variations that can be an influence.

I wouldn't get hung up on 'saturation' per se - it is just a pointer for when gain has reduced to a certain level (eg. 10) - sort of similar to 'clipping' when gain =1. You need to focus on worstcase zones - eg. say to keep Vce<1V worstcase, as that may match the supply rail regulation and relay coil pull-in worstcase levels to give you a valid safety margin of say X volts. Your automative application may require you to operate successfully during cranking - where supply rail worstcase can get pretty low!

Ciao, Tim

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The curves I posted are typical. Your results may vary.
However, for 2N3904, the optimum base current to guarantee saturation is Ib=Ic/10. You don't need to get involved in trying to determine an optimum value.