# PNP Switching Circuit Question

Discussion in 'General Electronics Chat' started by N2HMM, May 18, 2015.

1. ### N2HMM Thread Starter New Member

May 18, 2015
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Hello,

Assume the following basic PNP switching circuit as shown below. If the input of the first NPN transistor goes HI then the output of that transistor goes LO. If I connect +Vcc directly to the collector of the PNP transistor at DC2 then per the table on the far right the emitter of the PNP transistor goes HI. How? Isn't the PN junction reversed bias? The table states the emitter of the PNP goes HI and indeed in the application it does go HI but i don't understand why?

2. ### dl324 Distinguished Member

Mar 30, 2015
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If you want the NPN to act as an inverter, connect it's collector resistor to VCC and take the output from the collector.

As drawn, the circuit won't do anything (useful).

3. ### N2HMM Thread Starter New Member

May 18, 2015
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Hi Dennis, thanks for the replay. This circuit is operational as drawn and i don't understand why. There is +28V on the collector of the PNP transistor and a relay on the emitter. When the base goes low, the relay energizes.

4. ### ian field AAC Fanatic!

Oct 27, 2012
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The base only goes low by about 0.7V, the rest is dropped by the resistor that limits the base current.

The VCE sat of a typical transistor is around 0.4V, so when the transistor is fully switched on by the base current, its collector will only be about 0.4V less than the Vcc rail.

Aug 23, 2012
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6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Normally for PNP transistor we connect emitter to VCC not the collector. But even if you connect the collector to Vcc transistor will start conduct current when PNP base goes low. Because now PNP transistor will work in reverse active mode or inverted. So the emitter will act just like a collector and emitter will act just like a collector in normal active region. Also in reverse active mode the current gain is very low compare to normal active region, but we have lower saturation voltage.

Bernard likes this.
7. ### dl324 Distinguished Member

Mar 30, 2015
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You're not using the PNP transistor correctly. You should be using the NPN to turn on the PNP and the PNP should have it's emitter connected to 28V and it's collector to the relay coil.

8. ### N2HMM Thread Starter New Member

May 18, 2015
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This is the best explanation I heard Jony130, thank you. This circuit is being used successfully but i didn't understand why. It's use in a transverter to switch an RF power relay. Thanks!

9. ### MikeML AAC Fanatic!

Oct 2, 2009
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Here you go:
Exercise for the student: Why does the collector of Q1 go slightly negative as it turns off?

10. ### N2HMM Thread Starter New Member

May 18, 2015
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Is it because the colapsing magentic field from the relay causes a back EMF?

11. ### MikeML AAC Fanatic!

Oct 2, 2009
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Yep. D1 is the snubber diode, which clamps the collector voltage one diode drop below ground (in this case). Bad things happen to Q1 if you leave it out.

(de wa7ark)

12. ### ian field AAC Fanatic!

Oct 27, 2012
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Tyco published an appnote about back emf clamps for their relays, they point out that the old faithfull clamp diode causes current to flow during the period of spike suppresion and that diode current slows down contact release - they say this increases contact damage from arcing.

Last time I searched for that appnote it was hard to find - there was a copy on the Element14 website last time I saw one.

13. ### MikeML AAC Fanatic!

Oct 2, 2009
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My take:

1. no diode at all, you blow up the transistor the first time you switch the relay off.
2. Simple diode snubber, you protect the transistor, relay contacts last 10 years due to slow release. (good enough)
3. Fancy Zener/resistor/capacitor, faster release on the relay, the relay contacts might last 20 years. (better is the enemy of good?)

14. ### ian field AAC Fanatic!

Oct 27, 2012
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The tyco appnote reccomends a zener to clamp the back emf to a safe value, and a series diode with the zener to prevent it forward conducting.