Questions on using PNP transistors to control rocket avionics

Thread Starter

TelluriumCrystal

Joined Nov 14, 2014
32
I am currently designing a circuit that will allow the remote startup of some avionics on-board a rocket. The general idea is that PNP transistors are used to keep the avionics in a default on state, and that an external control box can be hooked up during preflight assembly to keep them disabled until it is time to initialize them on the pad. There are two flight computers, and the design has an ATmega328 to grab diagnostics data from them during startup and drive a small OLED display. I'm pretty sure I got that bit set up properly, so you don't need to look at that part of the circuit unless you want to be extra helpful.

What I am asking for help with is setting up the PNP transistors properly. The way they are set up right now is as if they are controlled by simple digital logic: 0 V to be turned on and +7.4 V to be turned off. However, I am still very new to the concept of transistors, and I have just discovered the need for current-limiting resistors on the transistor base. This has proceeded to get me very confused as to what I need to do to not fry my transistors. So my questions are as follows:

1. If a PNP transistor is in a saturated state, does the current from the base matter? That is to say, once I've got a transistor saturated will it act like a true short circuit as long as the base is connected to ground or do I need to properly throttle the current from the base using the correct resistor? Right now I am using 10k pulldown resistors.

2. If a PNP transistor is in a cut-out state, I assume current does not flow through the transistor, but rather flows past the transistor, through the resistor, and into the battery ground? That is the idea I designed my circuit around, and I assumed that if there is no current through the transistor then I don't need a current limiting resistor on the line that gets toggled to turn off the flight computers.

I greatly appreciate any advice!

Information that might be needed to answer the above questions:

Expected maximum current draw by either flight computer: 1 A
Transistor datasheet: http://www.mouser.com/ds/2/308/D44H-D-111810.pdf

Images:

Rocket-side components:


External components:
 

#12

Joined Nov 30, 2010
18,217
The calculation for using a transistor as a switch is that the base current should be 1/10th of the collector current.
7.4V applied to the emitter, comes out the base as 6.8 volts.
6.8V/10K = 0.68 milliamps through the base.
0.68 milliamps through the base allows 6.8 milliamps through the collector (with the transistor at minimum interference with the power supply voltage).

A bipolar transistor used as a switch will usually collapse its emitter to collector voltage to about 0.3 volts and that's the amount of battery voltage used up by the transistor used as a switch. You can also do this with mosfets, and they will use up less than 0.3 of the battery volts, but they need several volts to turn on. 7.4V should be plenty to turn on many suitable mosfets.

If you want an amp through the collector, the base resistor should be 68 ohms. This is an unusual requirement so I recommend using a mosfet. Their gate current is nearly zero for a full amp of flow.
 

MrChips

Joined Oct 2, 2009
23,234
I believe you are misrepresenting how PNP transistors are employed.

Bipolar junction transistors come in two varieties, PNP and NPN.
They can be used as current switches, to turn currents on and off.
The current load can be placed in one of two locations, on the collector leg or on the emitter leg.

Hence we now have four possible configurations:

PNP with load on the collector
PNP with load on the emitter
NPN with load on the collector
NPN with load on the emitter

The only difference between the PNP and NPN circuits is the polarity of the applied voltage and the direction of the current.
The difference between a collector load or emitter load will be witnessed in the amount of voltage and current required at the base of the transistor in order to place it into saturation.
 

Thread Starter

TelluriumCrystal

Joined Nov 14, 2014
32
The calculation for using a transistor as a switch is that the base current should be 1/10th of the collector current.
7.4V applied to the emitter, comes out the base as 6.8 volts.
6.8V/10K = 0.68 milliamps through the base.
0.68 milliamps through the base allows 6.8 milliamps through the collector (with the transistor at minimum interference with the power supply voltage).
Ok. So I want to use a resistor that will give me the necessary amp flow through the base to ground. That makes sense. I am curious as to where you are getting these values though. I seem to recall reading somewhere that the base-collector amperage can be related to the emitter-collector amperage using a value. Is that what you are using to get the 1/10 ratio?

Also, I looked it up and apparently the Stratologger CF actually only draws 1.5 mA; so if I went with 3 mA just to be safe, then I would want a resistor of around 23k ohms between the base and ground. I don't actually know the current draw of the Marsa54, but I imagine it shouldn't be much different. I originally went with 1 amp because that is around what you need to fire an ematch, but the flight computers probably just use capacitors to deliver that much amperage briefly.

This brings me to one of the bits I am confused about with transistors and current flow: does the transistor try to "push" 3 mA through to the load or is that just the maximum current it will allow the load to "pull"? It is my understanding that transistors themselves only cause a voltage drop, but do not otherwise consume current themselves. So when the current pulled by the load is less than the maximum amp throughput, as determined by the amp flow through the base, does only that amount of current flow through the circuit? Or does the transistor pull all 3 mA and burn off the excess itself (which would not make much sense at all to me). And what happens if, for instance, the load tries to pull 5 mA but the transistor only wants to supply 3 mA?

I believe you are misrepresenting how PNP transistors are employed.
But I already understand all that. I'm not entirely sure what you are trying to tell me. Am I using the PNP transistor wrong?

As an aside, I am kinda curious as to why you never see BJTs with the load on the emitter in the example diagrams I've found online.
 

#12

Joined Nov 30, 2010
18,217
does the transistor try to "push" 3 mA through to the load or is that just the maximum current it will allow the load to "pull"?
Switch: A device that connects or doesn't connect.
When you turn on the ceiling light in your house, you flip the switch. Does the switch push current or pull current?
Neither. A switch is not a generator and it is not a battery. It's a switch.
The only difference between a mechanical switch and a transistor used as a switch is that the transistor is less efficient (and it only works with DC). It needs base current to allow collector current and it never has a zero voltage loss from the collector to the emitter.

The 10 to 1 ratio is derived from the fact that the DC gain of a transistor diminishes as the voltage from collector to emitter approaches zero. If you want the collector to emitter voltage to approach zero, like a switch does, you expect a current gain of about ten. You can try 5 or 20 and get pretty much the same results, or you can believe a thousand people that went before you and decided ten is a good enough number.
 

Thread Starter

TelluriumCrystal

Joined Nov 14, 2014
32
Alright, I'm starting to get it now. So the transistor is opened by allowing sufficient current to flow from the base to ground, and it is closed by raising the voltage at the base thus preventing current from flowing.

Now apparently one of the people I'm working on this project with just acquired a few D45H11Gs, set them up as shown in the circuit, and fried a breadboard - so something is obviously wrong with the circuit I designed. He recommended we switch to MOSFETs and considering you mentioned those as well I'm thinking that might be the way to go. The reading I've done on them so far seems to indicate they are easier to use anyways.

Edit: It turns out he didn't fry anything, he just assumed the circuit would because it will pull 6 amps when each E-match fires. Which, because that would require a 600 mA base current, means BJTs are out of the question. So I'm definitely going with MOSFETs now.

Here's the new design using Depletion mode N-channel MOSFETs:


How does that look?
 
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