Hello,
At first I try to find inverting buffer to work with 12VDC, and it is hard to find here. So I try to find idea how to work with transistor.
Attached is the circuit, I try to explain from left to right every component placed. Suggestion are very welcomed.
1. the input is open collector type, either Vcc or Gnd. The two of PNP will work either in saturation or cut off region.
2. the Rbase is a base resistor, it is there just to restrict the amount of current, the current is restricted to make the left PNP works in saturation (Ve>Vb<Vc).
3. The R-Pull-Up, since input (no1) is open collector I need a Pull up resistor. the placement of R-Pull-up is to the right of the Rbase to make the left PNP in saturation (Ve>Vb<Vc), so that the Vb is less than Vc.
4. the left PNP, is just like a switch. When the input start pull low, the PNP's base current will start to flow (from emitter to base). This base current (adjusted by Rbase) so that it works in saturation (act like a switch). thats make the collector current (pin 3) flow.
5. in the effect that the left PNP is turned on, that makes the right PNP base voltage to almost equal to Vcc (a little Vce drop).
6. I skip the R-Pull-down explanation as this is function for turning off the circuit.
7. the diode will start to conduct and makes the right PNP's emitter to be Vcc - VfDiode. The right PNP act as Cut off (Ve<Vb>Vc). that makes the output almost equal to Vcc (a little Vce and Vf drop)

This first part makes the output On.
The second part makes the output off as follow:

1. the turn off begin with floating input pin. this makes the Vbase of left PNP = Vcc.
2. Base current of left PNP = 0. then the collector current also 0. The charge in Collector bleed to the ground via R-Pull-Down. The left PNP goes to active (Ve>Vb>Vc) and eventually off, NOT cut off (Ve<Vb>Vc). If the fast off is needed, adjust the R-Pull-up to minimum so that the Vb increase faster.
3. The R-Pull-Down now play its part to bleed current to Gnd. then the right PNP's base current begin to flow. Base voltage start to drop. The right PNP goes to active (Ve>Vb>Vc). To make right PNP faster, the R-Pull-down's value is reduced.
4. the right PNP begin to turn on. The emitter voltage will be drop as the current flow from emitter to collector.
5. the diode Vf maintain the Vb>Ve so it will be either cut off or active. cannot be in Reverse active or saturation because the emitter is grounded.

Principally:
the left PNP will be either Saturation or Active (Ve tied to Vcc => always higher than anything)
the right PNP will be either cut off or active (Vc tied to Gnd => always lower than anything)

The question: Does put the diode is good idea?
If No:
1. if there is no diode, will the right PNP goes cut off since Ve=Vb.
2. if there is no diode, on the second part (output is going off), is the right PNP do any useful to bleed current from emitter to the collector? I mean does it ever goes to active?

waiting for your idea and suggestion
Thank you

 

About the easiest way to get a high-voltage (up to 18 volts out) inverting buffer is to use a MOSFET driver chip, for example the MCP1406. These chips are TTL logic compatible and can drive or sink up to 6 amps of current. Digi-Key has them for $1.03 apiece in single quantity, in DIP.

 

Thank you for the reply.
It is a practical solution from you. Appreciate it.
In my case i need only 200mA which is overkill for gate driver. Those above PNP benefit that it has no drain current when turn off, well except for the pull up since open collector in the input side.
Anyway, I thank for the response @OBW0549

 

In my case i need only 200mA which is overkill for gate driver.

Fair enough, but I usually consider "overkill" more as "design safety margin" unless it adds unacceptable extra cost, takes too much power, or requires more board space. In this case, I don't think it does.

 

Thank for your quick reply. It does require more board space and extra placement cost. having IC is neater and smaller.

 

You mentioned power, so also note that the MCP1406 draws a maximum of only 250μA from the supply.

 

Would you not be able to use a single NPN? If your logic drive will (1) support a 5mA output current (is this 5V?) and (2) will pull down to <0.6V then a 1k resistor to Vcc would provide the current needed for the base of the NPN. The NPN base is connected directly to the IC output pin.
If this current is too high use a Darlington connected NPN pair with higher resistor values.

 

The question: Does put the diode is good idea?
Yes, possibly. it may prevent log term degradation of low current gain by protecting against repeated reverse breakdown of the emitter-base junction. Whether there is any possible benefit depends a lot on the nature of your load.

If you are driving a low current load, you might be able to do this with a single common-base NPN switch.

V+ is teh positive logic supply for the circuit providing the input signal. The emitter is the input and the collector is the output. This circuit is non-inverting. The transistor and base resistor need to be selected for the collector load and required speed.

 

If you want an inverting buffer then you just using the first stage and give up the second stage, and adds a 10K resistor to the ground from c of bjt, the Rb will be decide the output current.

 

You mentioned power, so also note that the MCP1406 draws a maximum of only 250μA from the supply.

Thanks @crutschow . that is a point. it is high hfe..

 

Would you not be able to use a single NPN? If your logic drive will (1) support a 5mA output current (is this 5V?) and (2) will pull down to <0.6V then a 1k resistor to Vcc would provide the current needed for the base of the NPN. The NPN base is connected directly to the IC output pin.
If this current is too high use a Darlington connected NPN pair with higher resistor values.

Thanks @neonstrobe for the reply. Certainly the NPN is good to go. Since there is a pull up resistor, having PNP will be no drain while it is off, on the other hand the NPN will do the other way around. I am currently at 12V.
Simple NPN transistor can do the job, and darlington is good idea as well to gain current, but the frequency is a tradeoff. It can be done.
I really appreciate your reply, we here could discuss various way of doing things. Thank you for it.

 

The question: Does put the diode is good idea?
Yes, possibly. it may prevent log term degradation of low current gain by protecting against repeated reverse breakdown of the emitter-base junction. Whether there is any possible benefit depends a lot on the nature of your load.

If you are driving a low current load, you might be able to do this with a single common-base NPN switch.
View attachment 139982
V+ is teh positive logic supply for the circuit providing the input signal. The emitter is the input and the collector is the output. This circuit is non-inverting. The transistor and base resistor need to be selected for the collector load and required speed.

Thanks @DickCappels .
Since I work I like it very much. Pointing right to the heart.

Since the circuit from input to output all running in 12V, there is no need of voltage gain. Rather current gain is more useful to save current leak in my humble opinion.
A common base configuration have voltage gain but no current gain, on the other hand common collector have current gain but no voltage gain.
But thank you for the very good suggestion. I take diode as my thinking in a dream tonight.

 

If you want an inverting buffer then you just using the first stage and give up the second stage, and adds a 10K resistor to the ground from c of bjt, the Rb will be decide the output current.

Thanks @ScottWang for the reply.
Sometime I just overthinking simple things. You have reminded me to make things simple.
Having a 10K resistor to the ground from C of BJT in which the C also connected to the output pin will make the output current sink limited by 10K. In my case 12V/10K = 1.2mA. I might need to resize the resistor to what current sink I need (200mA). My bad is that the "Buffer" word in the topic is not a correct word while having current source 200mA and at the same time sink 200mA. It is a push pull configuration would be an ideal, which is a gate driver IC actually.
I just try to find a little bit of trade off in between the push pull and common collector configuration.

 

The question: Does put the diode is good idea?
Yes, possibly. it may prevent log term degradation of low current gain by protecting against repeated reverse breakdown of the emitter-base junction. Whether there is any possible benefit depends a lot on the nature of your load.

If you are driving a low current load, you might be able to do this with a single common-base NPN switch.
View attachment 139982
V+ is teh positive logic supply for the circuit providing the input signal. The emitter is the input and the collector is the output. This circuit is non-inverting. The transistor and base resistor need to be selected for the collector load and required speed.

Pardon me to ask @DickCappels . Is there any useful to put a PNP there I decided not to put any diode?
Thank you in advance.

 

Thanks @ScottWang for the reply.
Sometime I just overthinking simple things. You have reminded me to make things simple.
Having a 10K resistor to the ground from C of BJT in which the C also connected to the output pin will make the output current sink limited by 10K. In my case 12V/10K = 1.2mA. I might need to resize the resistor to what current sink I need (200mA). My bad is that the "Buffer" word in the topic is not a correct word while having current source 200mA and at the same time sink 200mA. It is a push pull configuration would be an ideal, which is a gate driver IC actually.
I just try to find a little bit of trade off in between the push pull and common collector configuration.

You didn't tell us what are the input and output and how much current the input can be sink, if only using one stage then it should be concerned about the sink ability of input, if it is only using a switch then it's ok, otherwise you can use a TIP32C and try Rb=510Ω, Rbe=1.5K.

Why you need the Buffer inverting and not just a Buffer?

 

You could refer to the circuit below.

 

You could refer to the circuit below.
View attachment 140012

Thank you @ScottWang, you have been very helpful.
Sorry at the first place I didn't exactly tell about the input and output characteristic. the input is 12v optocoupler with open drain. the optocoupler instruction comes from PWM. About having the inverting is because giving value in PWM the "off" value is just not a good user friendly (one should calculate the period - "off" value to get "on" value).
The output is a small mosfet, so i really try to design a gate driver actually, but with a small gate current just 200mA. Since it is a gate driver, the output need current sink and current source.
Now I just realize I made you confuse because didn't tell you the input sink ability. Pardon me about that.

Thank you also for the second reply.
I understood what you are trying to give me. the first NPN is just inverting and the second NPN with PNP build a high gain non inverting amplifier. I try to modify from the diagram you gave. I haven't tried on bread board nor having any resistor value decided.
Shall try to do a breadboard in the following weeks.

 

The output is a small mosfet, so i really try to design a gate driver actually, but with a small gate current just 200mA. Since it is a gate driver, the output need current sink and current source.

If the output side is a small MOSFET then it just needs the Ig<10 mA, and if you only need a NPN and PNP to make the driver then you will be need a resistor to turn off the Vbe of PNP in your circuit, otherwise the PNP will be in a floating status when the NPN is turn off and that is not allowed in a circuit, because it will causing the noise and oscillation, please refer to my circuit in #16.

When we design the digital driver, normally we will use the hFE=10 to calculate the Ic and Ib current, you can try to calculate the resistor use current as 10 mA of the output(Ic) and refer to my circuit is 200 mA(actually is 198.2 mA).

Maybe you can show us the input and output circuit then it is more practical when we trying to help you could provide the precisely solution.

 

If the output side is a small MOSFET then it just needs the Ig<10 mA, and if you only need a NPN and PNP to make the driver then you will be need a resistor to turn off the Vbe of PNP in your circuit, otherwise the PNP will be in a floating status when the NPN is turn off and that is not allowed in a circuit, because it will causing the noise and oscillation, please refer to my circuit in #16.

When we design the digital driver, normally we will use the hFE=10 to calculate the Ic and Ib current, you can try to calculate the resistor use current as 10 mA of the output(Ic) and refer to my circuit is 200 mA(actually is 198.2 mA).

Maybe you can show us the input and output circuit then it is more practical when we trying to help you could provide the precisely solution.

Thank you so much for your helping hand @ScottWang
Here is my circuit. Vcc is 12V and the gate current I just calculate using gate current calculator based on mosfet parameter and frequency applied, it is turned out to be 200mA.
Thank you in advance for the help.

 

Thank you so much for your helping hand @ScottWang
Here is my circuit. Vcc is 12V and the gate current I just calculate using gate current calculator based on mosfet parameter and frequency applied, it is turned out to be 200mA.
Thank you in advance for the help.
View attachment 140102

The Pin 6 of photocoupler should be disconnect from Vcc.

The BJT is drive by Ib current, but the MOSFET is drive by Vgs, so they are complete different, where did you get the idea that it needs the 200 mA for frequency and how high of the frequency?

I think maybe you just need one PNP stage to drive the MOSFET and the Ig of MOSFET is <10 mA.