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I have a Fairchild BC547 (NPN) that I'm going to use as a switch.

What parameter(s) should I look at, if I want to know how long time the transistor will take to go from cut-off to fully saturated?

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    \$\begingroup\$ At least draw few lines of schematics. \$\endgroup\$ – Marko Buršič Apr 5 '17 at 11:23
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    \$\begingroup\$ @OlinLathrop, please stop writing forms of "Pah, it's obvious!". They've read the datasheet and it is a fair question, it's not stated or obvious. Don't know if you just feel good kicking others but, regardless, try and help and don't downvote them. \$\endgroup\$ – TonyM Apr 5 '17 at 11:49
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    \$\begingroup\$ @Tony: There is no evidence at all that the OP has read the datasheet. If they had, there is good chance they would recognize one of the specs as answering the question. Either way, there should have been a link to the datasheet in the question. I was going to look in the datasheet and quote the relevant spec, but since there was no link, I thought to myself screw this. Linking to the datasheet really should have been obvious. \$\endgroup\$ – Olin Lathrop Apr 5 '17 at 11:56
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    \$\begingroup\$ @OlinLathrop the question suggests they have looked and can't find it. Don't need the schematic, I found the datasheet in seconds. Why the rage of 'screw this' - this site isn't obligatory or paid. And these people aren't punchbags, they're young or confused or uncertain. Don't use this site as primal scream therapy, you're angry so so easily and so often. Chill. Please, for goodness sake :-) \$\endgroup\$ – TonyM Apr 5 '17 at 12:02
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    \$\begingroup\$ The only thing related to time that I've found is "transition frequency", which says "150MHz typical". I don't know what to do with that value, hence my question. Does it mean one transition from off to high = 1/150e6? \$\endgroup\$ – bos Apr 5 '17 at 12:05
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It's the current gain bandwidth product, which your datasheet lists as 300 MHz typical.

You have your base current Ib and collector current Ic, so Ic/Ib is your required gain. If you then divide 300,000,000 by this, you will get a typical max. switching frequency. You could then take your maximum switch-on time to be half of the period of this frequency. Personally, I would derate this value considerably as it is a 'typical' and take a hundredth of it's value. If you don't get a switching time that you need, consider another transistor. (That's assuming you're designing anew and not fixing existing.)

So I would use:

$$t_{on} = \frac{I_c}{1,200,000 \times I_b}$$

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    \$\begingroup\$ Sorry but GBW only applies to linear region and not large signal Vce(sat) slew rate. Although GBW is related for the early linear region but not towards saturation <2V to 0.7 to 0.1, it depends on Ic/Ib ratio i.e=10 to achieve fast slew rate. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 5 '17 at 13:11
  • \$\begingroup\$ @TonyStewart.EEsince'75, you're quite right and saturated behaviour is considerably slower than linear mode behaviour. I thought that derating by 100 would remove the effects of saturation and get the OP a guide figure which I assumed would let them go ahead or not in their application, not get an accurate calculation. I am confessing my sins here...at least it's to another Tony :-) \$\endgroup\$ – TonyM Apr 5 '17 at 13:17
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    \$\begingroup\$ You are forgiven buddy \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 5 '17 at 13:26
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Normallly Vce(sat) is rated for Ic/Ib=10 This is because as Vce approaches 3 to 2V (depends on Ic) as it goes to Vce(sat) all transistor hFE's drops to about 10% of the linear region hFE. (Rule of thumb, expected worst case, not typ or best case) So taking min hFe and 10%hFE you get Ic/Ib=10, which has become a de facto transistor Vce(sat) standard. Special types, much more expensive, have Ic/Ib specs =50 and =10 for Vce(sat) made by Diodes Inc and others, which also have very high hFE >500 to 2k.

enter image description here Since C loads demand more current Ic=C dv/dt then your slew rate from 90 to 10% must overdrive the base current so depending on load capacitance, thus make Ib=3% to 10% and define your load reactance.

Small signal gain BW and transition frequency is almost irrelevant when used with large signal saturated switch application as this only applies to the design test circuit in spec.

In linear mode, Ib can be much smaller ( eg. Ib=1%Ic) as long as it is always < Ic/hfe or used with R ratios. Spec has tables of hFE and curves to show how it typically varies with Ic. In this case it drops near max Ic. But for fast saturation, output current must be "overdriven" by base current due to above reasons.

The table indicates hFE DC Current Gain V CE =5V, I C =2mA 110min 800max with different sort grades A,B,C,D. Higher hFE is better but again use Ib= 5 to 10% Ic for reliable fast switching. This is normal use. Variations depend on details of design.

In general we design for worst case hFE and temp effects, so it works for all production and not just 1 typical part. Load capacitance (pF) can be critical for fast Slew rates as it demands more slew current.

side note

As a matter of trivial interest, Relays can switch very high contact current and this ratio of Ic/Icoil can be as high as 2000 for <2A but typically <500 for faster operation but does not depend on contact current as in semi's.

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Good answers already given here. But I would add.

Arguably, most of the electrical specification numbers affect switching time in some way or another.

However, a transistor on it's own has an infinite switching time.

It's the components around it, the voltages or currents applied, in conjunction with the spec numbers that define the expected switching time. Which parameters are most crucial depend on how you drive it and what the loads look like.

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