In this post I have explained the method of correctly replacing a BJT with a MOSFET, without affecting the final outcome of the circuit.
Introduction
Until MOSFETs arrived in the field of electronics, transistors or BJTs to be precise ruled the power switching circuits and applications.
Though even Bipolar Junction Transistors (BJTs) can not be ignored due to there immense flexibility and low cost, MOSFETs also have certainly become hugely popular as far as switching heavy loads is concerned and due to the high efficiency associated with these components.
However even though these two counterparts may look similar with their functions and style, these two components are completely different with their characteristics and configurations.
Difference Between BJT and MOSFET
The main difference between a BJT and a MOSFET is that, a BJT operation depends on current and needs to be proportionately increased with the load, whereas a mosfet depends on voltage.
But here the MOSFET gets an edge over a BJT, because voltage can be easily manipulated and achieved to the required degrees without much trouble, in contrast increasing current means greater power that's to be delivered, which results in bad efficiency, bulkier configurations etc.
Another big advantage of a MOSFET against the BJT is it's high input resistance, which makes it possible to be integrated with any logic IC directly, no matter how big the load may be that's being switched by the device. This advantage also allows us to connect many MOSFETs in parallel even with very low current inputs (in mA).
MOSFETs are basically of two types, viz. enhancement mode type and depletion mode type. Enhancement type is more frequently used and is the prevalent one.
The N-type MOSFETs can be turned ON or activated by applying a specified positive voltage at their gates while P-type MOSFETs will require just the opposite that is a negative voltage to get turn ON.
BJT Base Resistor vs MOSFET Gate Resistor
As explained above, a the base switching of a BJT is current dependent. Meaning its base current needs to be increased proportionately with increase in its collector load current.
This implies that the base resistor in a BJT plays an important role and must be correctly calculated to ensure that the load is optimally switched ON.
However, the base voltage for a BJT does not matter much, as it can be as low as 0.6 to 1 volts for a satisfactory switching of the connected load.
With MOSFETs it's just the opposite, you can switch them ON with any voltage between 3 V and 15 V, with current as low as 1 to 5 mA.
Hence, a base resistor may be crucial for a BJT but a resistor for the gate of the MOSFET may be immaterial. That said, a low value gate resistor must be included, just to safeguard the device from sudden voltage spikes and transients.
Since voltages above 5 V or up to 12 V are easily available from most digital and analogue ICs, a MOSFET gate can be quickly interfaced with any such signal source, irrespective of the load current.
How to Replace a Transistor (BJT) with a MOSFET
In general we can easily replace a BJT with a MOSFET, provided we take care of the relevant polarities.
For an NPN BJT, we may replace the BJT with a correctly specified MOSFET in the following manner:
- Remove the base resistor from the circuit because we don't typically need it anymore with a MOSFET.
- Connect the gate of the N-MOSFET directly to the activation voltage source.
- Keep the positive supply connected to one of the load terminals, and connect the other terminal of the load to the drain of the MOSFET.
- Lastly, connect the source of the MOSFET to ground.......DONE, you have replaced the BJT with a mosfet within minutes.
The procedure will remain as above even for a PNP BJT to be replaced with a P-channel MOSFET, you will need to just reverse the relevant supply polarities.
Compatible Pinout Replacement Diagram for PNP BJT with P-Channel MOSFET
Questions & Answers
Dear Swagatam
Thank you for writing this page.
Please take a look at the circuit “Figure 21. Series Pass Regulator” as seen on the page 8 of the Onsemi datasheet of TL431
https://www.homemade-circuits.com/wp-content/uploads/2026/09/series-pass-regulator-using-TL431.jpg
How can the NPN transistor in that circuit be replaced with a MOSFET?
Dear SA,
According to me you can directly replace the E, B, C, of the BJT with the S, G, D of a MOSFET, however a BJT drops around 0.6V at its emitter, but a MOSFET may drop a lot higher than that at its source.
I had connected Series Pass Regulator with a TIP41C and It has been working: Vin 18.7V, Vout variable between 3.5V to 18V. The components are 220Ω as base resistor, 43.77K as R1, 100K potentiometer as R2, and without capacitor.
Yesterday I replaced TIP41C with an IRF9540N MOSFET. Turning the potentiometer varies the Vout only from 18.7 to 18.2. It seems that this Series Pass Regulator did not work with only replacing the BJT with MOSFET.
Why my circuit can vary the Vout by only 0.5V? What is wrong with it?
IRF9540N is a P-channel MOSFET, please replace it with IRF540 or any other N-channel MOSFET…and then check the results…
Hello Swagatam
I used an AP40N03GP N-channel MOSFET. The output is now variable between 3.7 to 17.2 volts.
(As I in the previous comments said, the circuit is the one in the “Figure 21. Series Pass Regulator” on the page 8
https://www.homemade-circuits.com/wp-content/uploads/2026/09/series-pass-regulator-using-TL431.jpg
The input voltage is 18.7V. The components are 220Ω as Gate resistor, 43.77K as R1, 100K potentiometer as R2. I did not use any capacitor.)
The problem now is “shortage of output current” that causes voltage drop. The Input is supplied by a SMPS power supply with 18.7 volts and 3.4 A output. 18.7×3.4=63.58 watts. Let’s say 60 watts. When I adjust the output of the circuit on 12.09 volts, and connect a 42-watt 12-volt lamp to it, the output voltage drops to 11.74V. When I connect a 60-watt 12-volt lamp to it, the output voltage first drops to 0.1 to 0.2 volts and of course cannot turn on the lamp, but then rises to a fluctuating voltage between 2 to 3 volts and the lamp begins to flash (turns on and off).
Why cannot this circuit output even 3.5 A (42-watt 12-volt lamp) without dropping the voltage about 0.35 volts, let alone output 5 A (60-watt 12-volt lamp), whereas the input is at least 60 watts (18.7×3.4=63.58 watts)?
Hi SA,
I think it is happening due to the internal resistance of the pass transistor, as we know BJTs can be hugely inefficient in terms of current transfer, as these are current dependent devices.
Please try replacing the pass transistor with a Darlington BJT such as TIP122, and then check the results, or simply upgrade your existing pass power BJT with another NPN such as 2N2222 to form a Darlington pair….and then check the results. Let me know how it goes.
…Just wanted to add, TIP122 cannot handle 3.5 amps unless mounted on a very big air cooled heatsink, so better go with TIP142 (with heatsink) or a combination of 2N3055 and 2N2222 Darlington pair.
My last comment, #212460 (in reply to your #212033) was about using a N-channel MOSFET (AP40N03GP) not about using BJTs. But your reply in #212465 and #2124656 is about using BJTs. Do you imply not to use a MOSFET?
Hi SA, I am really sorry, seeing the BJT in the diagram, I just forgot that your question was based on a MOSFET, OK, here’s the correct answer now….
Since this is a linear regulator, so there is no power conversion taking place. The current through the MOSFET and load will be approximately the same as the input current.
With 18.7V input and 12V output, at 3.4A the MOSFET has to drop about:
18.7 – 12 = 6.7V
So the MOSFET itself will dissipate around:
6.7 * 3.4 = 22.8 watts, and the load will receive around:
12 * 3.4 = 40.8 watts
So yes, in theory your 3.4A SMPS can give around 40W at 12V through this linear regulator, provided the SMPS can actually maintain 18.7V at 3.4A.
But your 42W lamp needs:
42 / 12 = 3.5A
This is already slightly above the 3.4A rating of your SMPS. And most important aspect is that because it is an incandescent lamp, the filament is cold at the start, its resistance is much lower, so the initial current can be several times higher than the normal 3.5A.
Therefore the SMPS could be going into current limiting or hiccup protection mode. That could be why you see the output falling from 12.09V to 11.74V.
For the 60W lamp, the normal current is about 5A, which is far above your 3.4A SMPS rating. During startup it can demand even more. So the SMPS protection will most probably shut down, recover, shut down again, etc. This is exactly the kind of behaviour that can make the lamp flash.
There is also one important test I would suggest.
Please don’t measure only the regulator output. Connect the 42W lamp and measure the voltage directly at the SMPS output terminals at the same time.
If the SMPS voltage itself falls significantly below 18.7V when the lamp is connected, then the main limitation is the SMPS.
If the SMPS remains close to 18.7V but the regulator output falls badly, then we need to investigate the regulator itself — MOSFET heating, gate drive voltage, source/drain connection, wiring resistance, PCB tracks, or the TL431 control circuit.
Also check the voltage between MOSFET gate and source while the 42W lamp is connected. This is important because the AP40N03GP may require sufficient VGS (+15V) to conduct the required current.
And also remember about the MOSFET heating. At around 3.4A and 12V output, it can already be dissipating roughly 23 watts. So it needs a proper heatsink. If the MOSFET gets very hot, its behavior can also change and the output may start dropping.
So I would first check these three voltages under the 42W load:
SMPS output voltage = should remain close to 18.7V
Regulator output voltage
MOSFET VGS voltage
These three measurements will tell us where the actual problem is.
Hello Swagatam
Sorry for my long delay in reply; I was terribly busy.
Thank you very much indeed for your insights, including “linearity of this regulator, there being no power conversion, the current through the MOSFET and load being approximately the same as the input current”.
When I repeated the test, the output voltage of the regulator only dropped at the very moment of connecting the load only from 12.05 to 12.00, and quickly returned to 12.05. The SMPS output voltage, upon connecting the load, dropped from 18.7 to 17.96.
The voltage drop at the moment of connecting the load, led me to measure the current-drawing of the lamp. Somewhere at the outset, its current drawing spikes between 4.0 and 4.2 amps. So it seems that the voltage at the beginning is caused by these spikes.
You asked about “MOSFET VGS voltage”. When I measured it, it was 2.4V.
I have 2 questions:
1. What is the explanation of there being no power conversion in linear regulators?
2. in what types of regulator does power conversion happen? in SMPSs?
Hi SA,
VGS of 2.4V is quite low for the conventional MOSFETs, it should be at least 10V, otherwise the MOSFET can overheat and drop even more voltage at the output.
Power conversion happens in linear regulators but with bad efficiency of around maybe 20%, which depends on the difference between the input voltage and the regulated output voltage, where the output power reduces a lot compared to the input power.
In SMPS or all inductor based regulators, the power conversion happens with higher efficiency, about 80 to 90%
You said “VGS of 2.4V is quite low…”.
In the circuit (imagine a MOSFET instead of the BJT) that I introduced in the comment #211662, given that also the Cathode of TL431 is connected to the resistor that supplies the Gate of the MOSFET, can the VGS be increased? If yes how can it be increased without damaging the TL431?
Please disconnect the TL431 cathode fom the MOSFET gate and check the VGS, or disconnect the gate of the TL431 from resistive divider and check the VGS, it should be same as the V+ voltage. If it is, then the circuit is good and working.
Further you can try adjusting the R2 value until the TL431 gate voltage drops below 2.5V and the TL431 stops conducting, in this situation also the MOSFET gate voltage must be same as V+. This will prove all the components in your circuit are working good.
Are these only tests? After doing these tests and ascertaining that the MOSFET gate voltageis the same as V+, will it be OK to connect the TL431 as before?
Yes, that would mean the VGS 2.4V appears only while TL431 is regulating the output voltage.
…and yes, adding to my previous comment, alternatively you could definitely try a Darlington BJT such as TIP142 (with heatsink) and test the results….
The VGS(th) parameter of IRF540 is “minimum: 2.0V; maximum: 4.0V (conditions: VDS = VGS, ID = 250 μA)”.
Does this relates to what you said “a MOSFET may drop a lot higher than that at its source”, and means that IRF540 will cause 2.0 to 4.0 volts drop at the output of the Series Pass Regulator?
Yes, that’s correct!!
Most BJTs have low Beta and this limits the current that they can deliver (when saturated) in the output of that circuit.
Do you know if a MOSFET is affected by Beta or another such limitation that will limit the output current in that circuit?
MOSFETs efficiency is basically determined by their RDSon values, which is the minimum resistance value across drain and source of the MOSFET. Other factor is the voltage Vgs which must be above the minimum threshold, as specified in its datasheet.
“IRF9540N” MOSFET has the following specs:
RDS(on): Max. 0.117Ω (Condition: VGS= -10V, ID= -11A)
VGS(th): Min. -2.0V; Max. -4.0V (Condition: VDS = VGS, ID= -250µA)
Do you know how efficient is “IRF9540N” compared to other N-channel mosfets?
Sure, just compare its RDSon with other similar MOSFETs…you will be able to check which one has the lowest RDSon value, that one will be the most efficient one…
How would you manage to replace the bjt transistor with another type of Device?
What type of device are you referring to?
any replacement for D1 292 transistor. that might be similer to it
You can Google 100V 1 amp NPN transistor, you may find many good options.
Looooongest time sir.
I have been searching for your contacts or your site. I am happy I got you today. You helped me to understand a lot in the past
Thank you Ik, I am always happy to help….. All the best to you.
Hi SwagatamI sent you a msg in facebook asking for help with replacing mrf486’s and mrf422’s i sent it there because i attached a schematic if you could please help thank you
Thank you John,
According to me you cannot replace MRF486 with a MOSFET. You can try searching for 30 MHz 20 Amp Rf NPN power transistor, and find a suitable equivalent for the replacement.
maybe what i’m looking for is LDMOS not MOSFET’S ?
I think the basic replacement principle will be the same, just as any other mosfet, since LDMOS is also a type of MOSFET.
thanks for the reply , yes i been searching and found nothing even close to mrf486’s
i’m thinking they were only made for this one radio only
the next version of this radio that came out had mosfets in it
i compared both schematics to these radios and the pa unit circuit look pretty close to the same design
the thing is i don’t know that much as to how to make the proper modifications to make it work , i never tried it before
Yes, I agree, I too tried searching for “30 MHz power transistor” but could not find anything useful. Unfortunately in RF circuits a BJT cannot be easily replaced with a mosfet.
What ordinary transistor do I replace for Darlington 2SD1298 or D1298?
You can use BU406D and MJE13003 in Darlington pair.
Or you can search for any 400V 10 amp transistor and 400V 1 amp transistor and combine them to make a Darlington pair.
Hi sir how can i connect a universal power module to my smps 30amp cause i cannot get 13009 transistors on my local market
AS far as I know universal power module has a power mosfet or transistors with 3 wires corresponding to the 3 leads of the power device. You can carefully check these wire according to the given user manual and then connect these wires across the respective tracks of the SMPS PCB.
However I would strongly recommend replacing with a discrete transistor instead of a power module.
Hi,
Can you please review the MOSFET symbols given and the type of MOS labeled?
Irfp448n
Good day, i would like to know if its possible to change the 13009 transistors in a smps (cheap 12v 30amp) to N-Channel mosfets, the transistors are driven by a gate transformer. If its possible can you give me a example schematic of how to change it if possible, here is a schematic of the smps.
I would greatly appreciate any help you can give me.
Regards
Darius
Good day eant to add, primary has 33turns then a center tap with another 33turns after.
Then the secondary has two separate 9turns without a enter tap and then a feed back with 2 turns
The primary and secondary windings has a copper thickness of 0.4mm and the feedback is 0.65mm.
Regards
Darius
Hi, How is this transformer wired, which IC is used? I will need more information regarding your question to figure out the issue?
Hello, the IC is a TL494 and its a half-bridge topology, i see on my first post that i forgot to add the link to the schematic, i’ll add the link in the end.
(imajeenyus.com/electronics/20151028_smps_variable_voltage/index.shtml) or type in on google s-400-12 supply schematic and check on the images.
I hope this help, it has the full schematic and how the gate drive transformer is winded and the main transformer aswell.
Regards and thanks again
Darius
Thanks for the schematic, This is an SMPS circuit where the transformers are the most critical elements of the circuit. You can rewind it but the winding details must be exactly similar to the original data….you cannot change the data with your own specifications.
Hi, according to me a an 13009 BJT cannot be replaced with a MOSFET in an SMPS circuit.
Thank you for the fast reply, is there no way to rewind the gate drive transformer if I’m correct the transformer is wind in a 4:1 ratio and has a feedback winding aswell, the primary winding is 10v peak to peak and secondary is 2.2v peak to peak, so if you rewind it 1:1 will it work?
Thanks
Regards
Darius
Without seeing the full schematic it can be difficult for me to provide a useful suggestion.