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I created this circuit as general purpose MOSFET driver circuit for PWM frequency up to 10KHz (usually from microcontrollers):

enter image description here

But R3 and R5 are getting hot soon as I connect 5V supply and I did not connect PWM signal line (means that is floating). Could somebody please explain why? and what would be good values for resistors R3, R4 and R5?

They are 0805 resistors and SOT23 BJTs. Components are closer and hand soldered by me. Earlier I had experimented same circuit with normal through hole 1/4 watt resistors and TO-92 BJTs and I did not notice the heating problem that time.

I took this circuit from enter link description here, figure 4.

Thanks in Advance.

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    \$\begingroup\$ are you sure it is R3,4,5 ? they are like 1K and fed from a 5V circuit. Could it be Q2 and Q3 that are actually getting hot but their proximity to R3,4,5 is making them warm up \$\endgroup\$ – JonRB Sep 16 '18 at 14:50
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    \$\begingroup\$ Wouldn’t Q2 and Q3 be causing a short circuit? \$\endgroup\$ – KingDuken Sep 16 '18 at 15:10
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    \$\begingroup\$ @KingDuken depends if there is enough base current. but yes Q2 and Q3 where they are is questionable. it is typically the other way round. Q3 probaby is on via R3-R5 (2.5mA). Q2 the same amount but via R4-R5 \$\endgroup\$ – JonRB Sep 16 '18 at 15:38
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    \$\begingroup\$ @KingDuken It will if Q1 is not doing its job correctly, and during switching. Not a good design imo. \$\endgroup\$ – Harry Svensson Sep 16 '18 at 16:28
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    \$\begingroup\$ The OP never did mention if the PWM was ON or OFF when the 5 volts was applied. If the PWM input was at logic '0' then small smd resistors like R3 and R5 would heat up, using the BE junction of Q3 as a ground path. \$\endgroup\$ – Sparky256 Sep 16 '18 at 20:17
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Here is your circuit being discussed:

R3 and R5 should not be getting "hot".

Do the math. Even if Q1 were a perfect switch, there would only be 5 V across R3. (5 V)2(1 kΩ) = 25 mW. Unless this is a very tiny resistor, you wouldn't normally notice it getting warm. A 0805, for example, can usually dissipate about 150 mW safely in open air on a typical PCB. This is ⅙ of that.

It should be obvious that R5 dissipates less power. Even ignoring the B-E drops of Q3 and Q2, there is only 3.3 V across R5 when Q1 is off. That results in 11 mW. You're not going to feel that with a finger.

Therefore, something is not as your schematic shows. Perhaps the 24 V is applied somewhere unintended. Perhaps the resistors aren't the values you think they are. 1 kΩ resistors would be labeled "102" or "1001", depending on tolerance.

As an aside, this circuit has some questionable tradeoffs. I can see how it may appear to work, but you can do better with about the same parts and topology. However, that's not what you asked about. For example, I wouldn't have been surprised if you said Q2 and Q3 were getting hot. That would be expected due to the bad design. But, any of the 1 kΩ resistors getting hot makes no sense for the circuit as you show it.

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    \$\begingroup\$ you have found a good way to ensure that you will not start to look out insane after the questioner has changed his not so clever schematic. \$\endgroup\$ – user287001 Sep 16 '18 at 16:25
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    \$\begingroup\$ Did you print the schematic on a paper and then scan it? \$\endgroup\$ – Harry Svensson Sep 16 '18 at 16:29
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    \$\begingroup\$ @Harry: I started with the OP's full size (what you get when you click on it) schematic, then did a little image processing to make it look more presentable within the size limitation for pictures in answers here. \$\endgroup\$ – Olin Lathrop Sep 16 '18 at 21:31
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enter image description here

Figure 1. When Q1 is off both Q2 and Q3 are forward biased (green) and will turn on. The result will be shoot-through the two transistors (red).

If the supply voltage doesn't collapse (1) will be at about 4.3 V, (2) at 2.5 V and (3) at 0.7 V.

Ignoring R3 the current through Q2, R4, R5 and Q3 will be about \$ \frac {5 - 0.7=0.7}{2k} = 1.8 \ \text {mA} \$. I didn't look up the \$ \beta \$ for those transistors but if you do you can get an estimate of the shoot through current.

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    \$\begingroup\$ Last I remember, current flows from emitter to collector and base to collector in Q2, not OUT the base. The red arrow is right when both transistors are forward biased but I believe the green arrows are incorrect for Q2 and R4. \$\endgroup\$ – kronenpj Sep 16 '18 at 20:37
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    \$\begingroup\$ No, transistor current flows in the direction of the arrow. To turn on Q2, a PNP, current is drawn from the base. To turn on Q3, an NPN, current is driven into the base. Both transistors are shown forward biased and that is the problem. \$\endgroup\$ – Transistor Sep 16 '18 at 20:47
  • \$\begingroup\$ That's probably why I didn't go into EE as a job after my degree (embarrassing). \$\endgroup\$ – kronenpj Sep 16 '18 at 20:49
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    \$\begingroup\$ Stick at it. With very basic understanding of transistor operation you can work with a huge variety of circuits and understand the operation of many circuits just by looking at the schematics. If you carefully copy my schematic into a new question CircuitLab allows you to run a DC Solver simulator and monitor the voltages and currents at each point in the circuit. You can play all day without burning any components. See if my voltages (1), (2) and (3) work out correctly. I didn't check! \$\endgroup\$ – Transistor Sep 16 '18 at 20:54
  • \$\begingroup\$ After running it through Circuitlab, I see the current flows as Transistor describes. With "PWM" low, R4's (.8ma) current is small, compared to R3 (1.6ma). R5 is handling about 2.5ma. All these are far less than a 0.25 watt resistor can handle without significant heating. @Junaid - Is this built as a surface mount circuit? \$\endgroup\$ – kronenpj Sep 17 '18 at 1:49
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There's a continuous base current path for both Q2 and Q3 when Q1 does not conduct. You can easily expect continuous 200mA current through them from your +5V. That's about 1W dissipation which well can make also other parts behind short wires hot than Q2, Q3.

Thermal runaway can short your +5V source.

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