Open Collector (or Drain) question

Thread Starter

ak2011

Joined Oct 24, 2011
12
OK, so here's the deal....

BJT example:
If I use an NPN, a high (say 3.3V) at the transistor's Base makes my output (Collector with pullup) go low (near 0V).

I want the opposite effect. Meaning I want my output (with the pullup) to go low (0V or near 0V) when a low (0V) is applied at the Base/Gate. Was going to use a PNP instead, but everything I've read recommends the Emitter be tied to the supply (Vcc) which defeats the purpose of what I'm trying to accomplish.

How do I achieve the above with only one transistor?
 

crutschow

Joined Mar 14, 2008
23,500
Here's the deal... You can't. You need two transistors.

Incidentally, you don't put 3.3V on the transistor base unless you want to blow it, you need to have a series current limit resistor in series with the base.
 

Thread Starter

ak2011

Joined Oct 24, 2011
12
:) ...I give in

Can you explain why the emitter of the PNP "should" be tied to Vcc? Recommendation say it should...with no clear explanation.

In theory (or practical) a PNP with the collector pulled up (and emitter tied to GND), will turn on with 0V at the base causing the transistor to conduct....thus making the node right at the collector low (close to 0V)....

Why won't this work?
 

praondevou

Joined Jul 9, 2011
2,942
Why does it need to be only one transistor? If the problem is space. there are transistor arrays in tiny smd packages that could be used for your purpose.
 

crutschow

Joined Mar 14, 2008
23,500
:) ...I give in

Can you explain why the emitter of the PNP "should" be tied to Vcc? Recommendation say it should...with no clear explanation.

In theory (or practical) a PNP with the collector pulled up (and emitter tied to GND), will turn on with 0V at the base causing the transistor to conduct....thus making the node right at the collector low (close to 0V)....

Why won't this work?
It's related to the polarity required for proper PNP transistor operation. To pull a PNP collector "up" requires a negative collector-to-emitter voltage. That is why, with a positive supply, the emitter is tied to V+ and the collector load goes to ground (give a negative collector-emitter voltage). Then if you connect the base to ground through a resistor, the base-emitter junction is forward biased and the transistor turns on, applying a plus voltage to the collector load.

Is that clear enough?
 

BJT_user

Joined Oct 9, 2011
35
Greetings ak2011. I'll explain a few basics of transistor circuitry. What you are describing, a NPN transistor with the signal applied to the base and the load on the collector, is your basic common emitter amplifier. There are three very basic amplifier styles... common base, common emitter and common collector. Of the three, the common base and common collector are known as non inverting amplifiers. IE, the same direction of voltage change at the input, shows up at the output. The common emitter amplifier, however, the one you are describing, is unique among the three in that it takes the signal direction applied to the input and inverts it at the output.

For what you are describing, you might want to use a different configuration... like the common collector. This is a non inverting configuration and In this configuration, you will get the same voltage direction change at the output as you have at your input. All you nee do is tie the collector to +Vcc and apply your pullup (or pulldown) resistor between the emitter and ground. The downside of this configuration is there is absolutely no voltage gain. So if you apply 3.3 volts at the transistors base, all you will get is 3.3 volts at the emitter ( minus the internal bias required to activate the emitter/base junction).

The plus of this configuration is the current amplification is maxed out. Very little current is required at the base to allow the circuit to conduct as much as the transistor can handle. It's not a big deviation from your original circuit, just switch the load from the collector to the emitter.
 

Thread Starter

ak2011

Joined Oct 24, 2011
12
Greetings ak2011. I'll explain a few basics of transistor circuitry. What you are describing, a NPN transistor with the signal applied to the base and the load on the collector, is your basic common emitter amplifier. There are three very basic amplifier styles... common base, common emitter and common collector. Of the three, the common base and common collector are known as non inverting amplifiers. IE, the same direction of voltage change at the input, shows up at the output. The common emitter amplifier, however, the one you are describing, is unique among the three in that it takes the signal direction applied to the input and inverts it at the output.

For what you are describing, you might want to use a different configuration... like the common collector. This is a non inverting configuration and In this configuration, you will get the same voltage direction change at the output as you have at your input. All you nee do is tie the collector to +Vcc and apply your pullup (or pulldown) resistor between the emitter and ground. The downside of this configuration is there is absolutely no voltage gain. So if you apply 3.3 volts at the transistors base, all you will get is 3.3 volts at the emitter ( minus the internal bias required to activate the emitter/base junction).

The plus of this configuration is the current amplification is maxed out. Very little current is required at the base to allow the circuit to conduct as much as the transistor can handle. It's not a big deviation from your original circuit, just switch the load from the collector to the emitter.

Ok, this sorta makes sense....with exception to the voltage@base = voltage@emitter part. So Vcc is irrelevant? I thought turning on the NPN means current flow (same current) from Collector to Emitter? Y won't the voltage at collector (Vcc) be the same as Emitter (Ve), but Ic = Ie?
 

Thread Starter

ak2011

Joined Oct 24, 2011
12
It's related to the polarity required for proper PNP transistor operation. To pull a PNP collector "up" requires a negative collector-to-emitter voltage. That is why, with a positive supply, the emitter is tied to V+ and the collector load goes to ground (give a negative collector-emitter voltage). Then if you connect the base to ground through a resistor, the base-emitter junction is forward biased and the transistor turns on, applying a plus voltage to the collector load.

Is that clear enough?

Thanks for the response. Do current flow from Emitter to Collector on a PNP? I always thought current on a Bipolar, regarless if NPN or PNP, flows from Collector to Emitter.

Your explanation makes a lot of sense of current flow on NPN is opposite of PNP
 

crutschow

Joined Mar 14, 2008
23,500
Thanks for the response. Do current flow from Emitter to Collector on a PNP? I always thought current on a Bipolar, regarless if NPN or PNP, flows from Collector to Emitter.

Your explanation makes a lot of sense of current flow on NPN is opposite of PNP
The current flow (plus to minus) is collector to emitter for an NPN and emitter to collector for a PNP (collector is biased negative with respect to the emitter). The arrow on the transistor icon shows the direction of the forward-biased base-emitter current flow (and thus also the direction of the emitter to collector current).
 

otisman37

Joined Dec 21, 2011
1
I use a circuit that sounds very much like what you want to achieve. Except it's not really one transistor, more like one PNP and the DMOS output of a shift register...

What we actually have is a bank of PNP (BD140) transistors (common collector to ground - load on the emitter) driving solenoids and stepper motors, DC pump motors, etc. These have either 12 or 24V supply... The PNP's are driven by a couple of TPIC6B595N shift registers. The output of the shift register goes directly to the base of the PNP - with no other external components.

Advantage is that there is no need for resistors or even snubber diodes. Snubbing is internal in the TPIC shift register. The TPIC has DMOS outputs, so when an output is off, it floats up to the 12 or 24V supply voltage - turning off the PNP. When the DMOS turns on, pulls the base of the PNP down very close to ground - emitter follows the base, turning on the device. Base current is only about 5mA with a 400mA load on the PNP.

I found it hard to find examples of this configuration, but it works very well. Good for a low component count, and allows various supply voltages and current loads. Some of the outputs we have protected with PTC thermistors for safety.

Disadvantage of this configuration I guess is that you can't drive the PNP directly from a PIC or microprocessor because the PNP base needs to be able to float up to the supply voltage to turn off the device.
 

jegues

Joined Sep 13, 2010
733
As BJT_user has already suggested an emitter follower should do the job. It will also work as an excellent buffer between any other sections of circuitry you have.

That is a clever implementation otis.
 
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