NPN + PNP switch

friis

Senior Member
Hi,
I have on Dec 3. 2019 had Ingelwoodpete's commentary concerning a hardware switch using a NPN transistor and a PNP transistor (https://picaxeforum.co.uk/threads/timeout-again.31701/page-2#post-329521 shows the diagram), which has worked OK. I am now taking a closer look at the switch. I have looked at 2 situations:

1. when there is no load to be controled attached to the PNP transistor
2. when thare is a load to be controled attached to the PNP transistor. The load is a HC-12 Transceiver and a controller turning a motor either way

and I get:
Code:
No load:
Base of NPN     Base of PNP        P output of PNP
    0                  4.5                   0
    0.730           3.87                 4.64
    
Load:
Base of NPN     Base of PNP        P output of PNP
    0                  4.45                  0.7
    0.730           3.862                4.63
My problem is that "P output of PNP" not is 0 (but 0.7) when there is a load. I thought that "p output of PNP" only was dependent on the 3 inputs to PNP.

I am getting power from a wall-wart.
The use of R3 (10 k Ohm) makes no difference.
I am using the switch to save battery power.

Can anyone explain?

best regards
torben
 

AllyCat

Senior Member
Hi,

A PNP has two "P"s (Emitter and Collector) and you must use them the correct way around. ;) Also, specifying the PNP voltages is not very useful unless you also state the (emitter) Supply Voltage.

The value of R2 is generally unimportant, but the value of R3 (in post #43) can be very important, what value of R3 and transistor are you using? I believe that the peak current consumption of the HC-12 can be very high, even (or particularly) in its "low average drain" mode.

Cheers, Alan.
 

friis

Senior Member
Hi AllyCat
Supply voltage is 4.6 V
The Value of R3 is 10 kOhm. I have made all the runs without the R3. Adding the R3 made no difference.
I have tried to change R2 from 370 Ohm to 330 Ohm and to 1 kOhm. It made no difference.
I think the problem is the effect of adding the load. The switch is always depicted as an open or closed switch as if the load does not matter.
The only thing I can change is the base voltage - but to what?
The NPN switch works OK independently of what I do with the PNP switch.

best regards
torben
 

inglewoodpete

Senior Member
I'm away camping at the moment, with very poor phone reception (I have to stand on a cliff top). It appears that the PNP transistor is not working. When 'on', it's base voltage should be about 0.7v below it's emitter (emitter is connected to the positive rail).
 

friis

Senior Member
Hi Inglewoodpete
"On" is not a problem. I have Base/Collector voltage ratio at 3.87/4.64 and 3.862/4.63 in the case of "No load" and "Load".
For "Off" I get 4.5/appr. 0 and 4.45/0.7 in the case of "No load" and "Load".
In the latter case I am surprised at the difference between appr. 0 and 0.7 and the apparent fact that the load matters.
It is a bit difficult to get an overview of all the parameters.
I do not want you to fall off a cliff so I am looking forward to your being back from camping.

best regards
torben
 

AllyCat

Senior Member
Hi,
In the latter case I am surprised at the difference between appr. 0 and 0.7 and the apparent fact that the load matters.
That suggests that that the 0.7 volts is coming from the load. Is it connected to anything that might supply a small amount of voltage?

Try connecting a "real" resistive load (say 1k) across it. An "active load" (i.e. made with semiconductors) may draw almost no current at that voltage.

Cheers, Alan.
 

friis

Senior Member
Hi AllyCat,
You were right. If I replace the load I have with a resistor of 740 Ohm or 1 kOhm the collector voltage drops immediately from 4.63 V to 0 V.
If I use the load I have the collector voltage drops immediately to 0.7 V and then creeps down to 0.58 V just before it goes up to 4.63 V again.
There is no voltage applied to the load when the base voltage of PNP is 4.5 V, but it appears that the load (HC-12 and motor controller) is a capacitor that only slowly releases its charge to ground when the collector goes to 0 V.
When I measure the power consumption of the whole system I get 14 mA and 0.4 mA no matter whether the load is HC-12 and motor controller or it is a resistor of 470 Ohm or 1kOhm.
I do not know if the above makes sense, but the conclusion is that the load does not affect the PNP switch.
What do you think?
Best regards
torben
 

AllyCat

Senior Member
Hi,
the conclusion is that the load does not affect the PNP switch.
Why would you expect the load to "affect" a switch? A switch is a switch; it connects the load to the supply line when it's closed, and not when it's open.

Of course you must not exceed the "rating" or capability of the switch, but any switch will have some series resistance when closed and/or "leakage" when open, which may affect any measured voltages. Conversely, as you've discovered, if a "load" is constructed primarily from semiconductors, then it may draw very little current at low voltages, i.e. below their "bias" level of about 0.6v Vbe for a bipolar (PNP) or a few volts for the "gate-source threshold" of a FET.

But I'm not really sure that you understand that a bipolar transistor (PNP or NPN) amplifies currents and so any measured voltages are as much a consequence of the load that's connected to it.

Cheers, Alan.
 

friis

Senior Member
Hi sghioto,
It appears to make a small difference, but as you can see from my answer to AllyCat it does not matter because, as the measure of power consumption shows, the switch functions.
 

friis

Senior Member
Hi AllyCat,
I did not expect the load to affect the switch - indeed, I was surprised when it appeared to do so. As the measurement of power consumption shows, the switch functions and the load does not affect that.
I am operating the PNP transistor in the "cut-off region" .
best regards
torben
 

mortifyu

New Member
Hi,

Not sure if this thread is still current, but I am not really sure why in this day and age a P-channel FET would not be used in the place of Q2.

By doing this, R3 could happily be crazy high like 100K for all it matters. (R3 is just keeping the Q2 transistor in an OFF state when Q3 is not active) and R2 can be completely omitted from the circuit with Q3 collector and Q2 GATE directly connected together.

Power ON: MCU pin HIGH ---> 1K --->Q3 BASE.
Q3 Collector pulls Q2 GATE straight to GND which turns Q2 fully ON, thus providing the supply rail to the LOAD.

Power OFF: MCU pin LOW --->1K --->Q3 BASE.
Q3 Collector is now open circuit from GND and has no effect on Q2 GATE which now lets R3 do it's job, thus providing no rail voltage or current to the LOAD.

The immediate advantages of using a FET instead of a bi-polar transistor:
Negligible (uA) GATE current required. A bi-polar transistor does require some BASE current.
The junction closure of a FET is often milliohms and therefore almost zero voltage drop when used in this configuration.



Just my view of this application.



Regards,
Mort. ;)
 

AllyCat

Senior Member
Hi,

Would you care to give the manufacturers' part number of a (low-cost) Through-Hole, Logic-Level (preferably 3.3 volts), P-Channel FET, good for (at least) 200 mA?

Cheers, Alan.
 

inglewoodpete

Senior Member
Not an easy one. Element 14/Farnell have the Microchip VP2206N3-G, good for up to 640mA, RDSOn=1.1 ohms @VGS=-5v in a TO-92 package. Can carry 1 Amp pulse at VGS=-3v
 

WhiteSpace

Well-known member
If it's any help, I've been using the ON-SEMI NDP6020P for the top rails of my motor H-bridges. https://www.mouser.co.uk/ProductDetail/ON-Semiconductor-Fairchild/NDP6020P?qs=mdiO5HdF0Khl%2BuaDKGopOw==. Continuous drain current 24A, RDSOn 0.041 R, typical VGS -0.7 V. Does that fit the bill? I think I've understood the data sheet correctly. It's £1.52 for one from Mouser. It is no longer being manufactured. I bought several because there don't seem to be many other logic level P-Channel MOSFETs.
 

Goeytex

Senior Member
IRFD9120PBF(Vishay) Gate Threshold = 2.0V
VP3203N3-G (Microchip) Gate Threshold = 1.0V
 
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