I'm working on project that requires I switch a 12v power supply on the high side of a circuit, using a 3.3v micro controller. To make this work, I'm thinking a can connect a PNP transistor on the 12v high side, and connect the base to the collector of an NPN. Then connecting the base of the NPN to the 3.3v pin on the microcontroller and the emitter to ground. I found this example online: enter image description here

This leads to my question...I need to create 8 separate switches on this one circuit...Does this PNP to NPN high-side switch still work if I connect multiple in series?? Here is my current schematic... enter image description here

The LED represent the load for the 12v supply.
In this configuration, does the transistor still work like if there were only one NPN and one PNP...Or does the current get amplified with each NPN/PNP in the series?

  • \$\begingroup\$ Why R18? If that wasn't there (and assuming your 'LEDs' are proper 12V loads as you say) then this would work fine. The 8 circuits are in parallel, not in series and without R18 they shouldn't interact with each other. \$\endgroup\$ – brhans Jan 5 '16 at 16:08
  • \$\begingroup\$ Thanks for the quick reply! I added R18 to limit voltage from my power supply. Sorry, should have explained that part. 12v is a little high, so I'm using a resistor in this example to reduce to 10v. \$\endgroup\$ – Josh Jan 5 '16 at 16:14
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    \$\begingroup\$ Ahh, well using a resistor like that probably won't work the way you want it to (dropping 2V regardless of the load attached). You would need a resistor for each load. \$\endgroup\$ – W5VO Jan 5 '16 at 16:16
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    \$\begingroup\$ Also, this is not putting multiple switches in series. Does each 3.3V represent a digital output from a microcontroller? \$\endgroup\$ – W5VO Jan 5 '16 at 16:18
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    \$\begingroup\$ Why you actually a NPN transistor excess?.Just only by using PNP transistor itself,you can construct a simple switch. \$\endgroup\$ – Aadarsh Jan 5 '16 at 16:18

You should put a resistor in series with each LED and eliminate R18. If you are just switching LEDs it's perhaps wasteful to construct a high-side switch- if you can switch the low side it only requires one NPN transistor per LED.

The high-side switch you show (top diagram) will work, however you can only switch a fairly small current due to the 10K base resistor. At 12V you'll get about 1mA base current so most transistor will be well saturated for up to ~20mA load current, but if you want to switch 100mA reliably you should reduce that resistor in most cases.

Also, it's good practice to add a resistor from base to emitter on the PNP. The reason is that the leakage in the NPN can be amplified by the PNP and result in excessive current at the output (particularly at high temperatures). Something like 20K to 100K will work fine (it's not critical). That said, you can guess that the gain of the PNP will be low at low base current and the circuit will typically work fine at moderate temperatures.

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  • \$\begingroup\$ Could you loose a word about WHY the base-emitter resistor helps to help with the leakage issue? \$\endgroup\$ – Rev1.0 Jan 9 '17 at 8:33
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    \$\begingroup\$ @Rev1.0 Consider say a 50K resistor. To get 300mV (which will result in negligible base current in the PNP, you would need leakage of 6uA, which is enormous assuming a modern jellybean silicon transistor at reasonable temperatures. To put it another way, most of the leakage current will flow through resistor so there will be little base current to be amplified. Since the base current-voltage relationship is nonlinear, when the current gets up to 100's of uA most of it will flow through base and only about 15uA through the resistor (since Vbe will be about 0.7V). \$\endgroup\$ – Spehro Pefhany Jan 9 '17 at 13:05

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