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I have some problems to understand the datasheet of the transistor KSP44 (and transistor in general).

I need to control a LED strip. I want to switch it up through a transistor NPN. My transistor will be triggered by an Arduino, so I can apply maximum 20mA on the BASE of the transistor.

I have looked for a transistor, and I have found the KSP44.

I see on its Maximum Rating values that it can sink max 300mA on the COLLECTOR pin, and to swith on my LED strip I need 150mA.

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Where I have a doubted is about the Electrical Characteristics table of the datasheet. The maximum current in a saturated state of the transistor is 50mA. So could you tell me ?

Can I use this transistor to sink 150mA on the COLLECTOR pin ?

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Thank you really much.

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  • \$\begingroup\$ Do you really need a 400V transistor? \$\endgroup\$ – Steve G Aug 1 '18 at 11:19
  • \$\begingroup\$ Not at all, I need a solution to sink a current of 150mA through the COLLECTOR triggered by a BASE current not exceding 20mA \$\endgroup\$ – FrancNovation Aug 2 '18 at 7:48
  • \$\begingroup\$ A 400V transistor is designed to withstand 400V and so compromises other parameters such as DC current gain hFE. The KSP44 is not a good part to use in this application. \$\endgroup\$ – Steve G Aug 2 '18 at 9:00
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Remember that a Bipolar Junction Transistor is an amplifier.

By feeding the Base a current, it allows 40-200x more current to flow from Collector to Emitter (see hFE, Current Gain Characteristic).

To enable this amplification, you must saturate the Collector-Emitter field (Vce) and the Base-Emitter field (Vbe). Looking at the datasheet, we can see a potential difference of 0.75V is needed across the Collector to the Emitter, and also 0.75V across the Base and Emitter. The column labelled "Test Conditions" isn't useful for you.

Now, looking at Figure 1 of the datasheet, 150mA Collector current sets the DC Current Gain (hFE) at around 70x.

150mA / 70 = 2.1428mA is needed at the Base (at 10v Vce).

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  • \$\begingroup\$ Thank you really much for your answer ! How have you calculated the of hfe at 70x for 150mA sinked ? It is great if you can explain me. \$\endgroup\$ – FrancNovation Aug 1 '18 at 9:25
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The maximum current in a saturated state of the transistor is 50mA.

It still has a maximum rating of 300 mA but there's another problem. What the VCE(sat) value is implying is that you have only got a DC gain of 10 (Ic = 50 mA, Ib = 5mA) if you want to saturate the transistor.

I suspect that you want to pretty-much saturate the transistor so to drive 150 mA through the collector you will need at least 15 mA at the base (by implication) and I'm unsure if an Arduino pin is going to like doing this for sustained periods.

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  • \$\begingroup\$ I don't think you can draw that implication because the test would current limit Ibe and Ice. vishay.com/docs/80061/80061.pdf Unless I have missed something? \$\endgroup\$ – MIL-SPEC Aug 1 '18 at 9:15
  • \$\begingroup\$ It's a fairly simple and easy conclusion to draw. If the circuit relies on the transistor saturating, the hFE of the device is going to be about 10. I'm unsure what the relevance of the pdf is that you linked @MIL-SPEC \$\endgroup\$ – Andy aka Aug 1 '18 at 9:26
  • \$\begingroup\$ It is true, I hope that MIL-SPEC is right, in this case I could sink 150mA on the COLLECTOR without damage on the ARDUINO ouput \$\endgroup\$ – FrancNovation Aug 1 '18 at 9:27
  • \$\begingroup\$ @KevHV look at figure 5 in the data sheet for the device. However, you may have sufficient voltage in your circuit to be able to drive the LEDs sufficiently without having too much transistor saturation - this then requires less base current but, if the transistor is then dropping (say) 3 volts when on, the power dissipated by it will be 450 mW - can you live with this power loss? Maybe show your intended schematic. \$\endgroup\$ – Andy aka Aug 1 '18 at 9:32

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