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Turning on a TIP35C Transistor from a CD4017 IC

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Beewild

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I have adapted a circuit for my own purpose. Currently the outputs of the CD4017 go to a 4k7 resistor and then to the base of a BC337 transistor. Can I swap out the BC337 for a TIP35C? Do I need to change the resistor value? Do I have to drive the TIP35c via a smaller transistor like the BC337? Your help would be much appreciated. Thanks.
 
What are you driving with the BC337/TIP35C?
What voltage for the CD4017?
The two transistors are very different. The BC337 has a gain more like 100 while the TIP35's gain is maybe 25. The little transistor is designed for 500mA of current while the big one 5A is more typical. Current gain is very dependent on how much collector current.
 
If you need less than 250mA per output, you could use a driver IC like a ULN2003a instead of discrete transistors.
 
I have attached the circuit. The supply voltage will be 12vdc. I am looking at a load of 8 amps max. According to the book TIP35c is 125w, 100v, Ic cont 15A, Hfe 15/75.
 

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The recommended current source on the cd4017B is about 1mA to 1.2mA. So, you need current gain ~of 6700.

also, when switching is done, we generally want more than the minimum gain. So let's boost the 6700 to about 10,000 to 15000.

To do that, you'll need a darlington transistor or a pair of transistors connected as a Darlington before your TIP35c. transistor (left circuit).

Alternatively, if your goal is to power the 8A load (instead of powering the 8-amp load specifically with the TIP35c transistor, you could use a MOSFET transistor. A mosfet only needs a voltage change (essentially no current flow) at the GATE pin (instead of a current flow at the base like the TIP35c will need) - (right circuit)

EB8ECA5C-8E2F-46A8-9831-72DCFA5E32F0.jpeg
 
I'll say again, as I did in the previous similar thread, it's very poor to use a darlington in that way, as there's very little collector voltage on the BC337 collectors, and thus little current to drive the TIP35. It's FAR better to drive the BC337 collectors from the higher supply rails.

Even better, as there's three transistors, is to make the middle one PNP, fed from the collector of the first one, both of course powered directly from the supply rail - maximising base current for the TIP35.

Or, as you say, use a suitable MOSFET.
 
Since your supply is already 12v, I would use a Mosfet.
The higher supply voltage would allow to properly enhance any power Mosfet, and would not require logic-level Mosfets.

Regardless of what you choose, the power devices must be adequately heat-sunk.
 
Since your supply is already 12v, I would use a Mosfet.
The higher supply voltage would allow to properly enhance any power Mosfet, and would not require logic-level Mosfets.

Regardless of what you choose, the power devices must be adequately heat-sunk.

The new MOSFETS have less than 5 milli-Ohms of on resistance so I^2*R = P = 8*8*.005 = 0.32W. No heat sink needed.

 
The datasheet of the CD4017 and most other CD4000 Cmos shows that with a 12V supply, an output of as high as 24mA is available which will fry it. If a current-limiting resistor is used then it can produce an output current of 12mA without getting hot.

The datasheet of the TIP35 and most other transistors shows that hFE is used when it is an amplifier with a collector voltage of at least 4V. But you want the load to get most of the voltage then the transistor must be saturated. The datasheet shows it saturating fairly well when its base current is 1/10th its collector current. Then 800mA is crazy, use a Mosfet instead.
 
Thanks for all your replies. I must admit I like the idea of using a mosfet, though I have never actually used one before. The buz71 is 50v 14amp?
 
It can handle that much voltage and current, it doesn't make that much current.

a Mosfet looks like a resistor when it is turned on whereas a bipolar transistor looks more like a diode.

the Mosfet I recommended has resistance of less than 5 mOhms, where power dissipated by the Mosfet is current squared * resistance = 8*8*0.005 = 0.320W. Under 0.5watts means no heat sink.

if you use the Mosfet you found, it has an on-resistance of 0.085 mOhms and that high resistance gives significantly higher heat to the device - I'll let you make the calculation.
 
Yes, that will work but put two in parallel to diffident the heat between them (commonly done with MOSFETs) or add a small heat sink.
 
The IRFP3206 does not seem readily available here. I can get a STP60NF06 which looks like 0.016 ohms ?
Probably.

At 8 amps, the power dissipated by the mosfet will be 1.024 watts.
The maximum junction temperature (Point of failure) is 175°C. My preference is to give that about a 50°C margin.
The Thermal resistance junction-ambient (Table 2) is 62.5 °C/W. In other words, the junction temperature will rise above ambient by 62.5 degrees C for every watt of dissipation, or about 64°C. If the ambient is 40°C, then the junction temperature will reach 104°C. That would be OK.

If the ambient is higher (closed box out in the sun, then you'll probably have a reliability problem. If it'll never see that hot of an environment, then you're probably OK. Another factor will be how long the mosfet is turned on and how frequently. The circuit in post #4 shows that each mosfet will not be on more than 1/6th of the time, so the average junction dissipation will be 1/6th of the 1.024 watt. Now, that doesn't mean that the junction is at the average all of the time. it means that the temperature ramps up and down. If the single event on time is long enough that the junction hits it's limit during that time, then the average no longer applies.
 
Use a better Mosfet then only one is needed and it will not need a heatsink.
 
So use 2 in parallel and use heat sinks?

or, if you are in a remote area Of the world without a good selection of mosfets available, use two or threeSTP60NF06 in parallel and you should be good.
Also, can you explain more about the cd4017 and the frequency you are feeding it. That is, does it turn on for one hour out of six hours or are you pulsing it 1000x/second and running at 1/6 duty cycle. These are much different demands on the Mosfet.
 
The outputs of the 4017 Q0, Q2, Q4 turn on and off sequentially at about 0.25 second intervals all controlling 1 transistor or mosfet and then Q5, Q7 & Q9 do the same for a second transistor or mosfet. Then repeat. The exact speed can be adjusted by the trimpot vr1 on the 4093. This produces 2 flashing channels for led emergency or safety vehicle lights. Each channel is 30 x 3w leds, set up as 10 arrays of 3 x 3w.
 
Last edited:
Ok, have decided to go back to the idea of transistors. Attached is my completed circuit. Your thoughts please!
 

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Ok, have decided to go back to the idea of transistors. Attached is my completed circuit. Your thoughts please!

I refer you back to post #6, it's a VERY poor way to do it - the TIP35 won't turn ON very hard, there will be a considerable voltage loss across it, and you will lose a lot of the power through it as heat.
 
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