PNP Switching Circuit Question

Thread Starter

N2HMM

Joined May 18, 2015
4
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?


upload_2015-5-18_14-0-52.png
 

dl324

Joined Mar 30, 2015
16,943
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).
 

Thread Starter

N2HMM

Joined May 18, 2015
4
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.
 

ian field

Joined Oct 27, 2012
6,536
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.
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.
 

ScottWang

Joined Aug 23, 2012
7,409
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.
Probably you missing the point on Vce is reversed in this thread, please reread again.
 

Jony130

Joined Feb 17, 2009
5,488
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.
 

dl324

Joined Mar 30, 2015
16,943
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.
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.
 

Thread Starter

N2HMM

Joined May 18, 2015
4
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.

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!
 

MikeML

Joined Oct 2, 2009
5,444
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.
Here you go:
Exercise for the student: Why does the collector of Q1 go slightly negative as it turns off?

217.gif
 

MikeML

Joined Oct 2, 2009
5,444
Is it because the colapsing magentic field from the relay causes a back EMF?
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)
 

ian field

Joined Oct 27, 2012
6,536
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)
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.
 

MikeML

Joined Oct 2, 2009
5,444
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?)
 

ian field

Joined Oct 27, 2012
6,536
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?)
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.
 
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