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I'm a big fan of relays in this sort of application. Relays have very low resistance, so little power loss. Most arduino users drive the relay with a bjt. If you do this, use a diode across the coil to absorb any inductive spikes when turning off the relay. Simple and effective. Use a relay that is open when unpowered.


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Ltspice can simulate the noise contributions. But from looking at the schematic a few things become apparent: the BJT are off at low signal level, so the noise of the feedback block is dominated by the three feedback resistors. at high signal level, the feedback impedance drops (hence rolling back gain), so that means that noise drops too. and anyway the ...


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As you will most likely want to have a common ground potential between your sensors and controller, a low side switch is not appropriate. Therefore, you need a PNP or PMOS high side switch. PNP can only supply as much current as allowed by its base current times its gain. Which means that when the PNP is on you will by wasting the max current divided by gain ...


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Simple answer is - MOSFET are superior to BJT as a load switch for your application. Long answer comes down to the details. There are varieties of both BJTs and MOSFETs. They have a lot a parameters which should be taken into consideration. But for the Arduino sensor modules application you should really stick to FETs. Just make sure you pick the right parts ...


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The MOSFET will have less IR drop when 'on' (only Rds(on)) than a BJT (Vce(sat)). Also, the BJT has base current, while FET gate current is essentially zero. However, when off, the FET might have more leakage than the BJT - check the datasheets of the comparable FET vs. BJT devices. That said, there are ultra-low power load switches, like this one: https://...


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Yes, you need an additional set of transistors. Basically, for each PNP, connect a R (2 k ?) from base to emitter. Then connect an NPN: collector -> PNP base; base -> GPIO, and emitter to GND via 2 kΩ. This NPN will then turn on the PNP. You can't use the same GPIO for the upper PNPs and lower NPNs -- as you toggle the H-bridge, small delays will cause ...


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Your first problem is your choice of a Darlington follower for your output stage. The voltage across a Darlington cannot be less than two diode drops, which would amount to about 1.5 volts. So the output can never be greater than about 7.5 volts for a 9-volt supply - and this is an absolute limit. In your case your getting about 7 volts swing, which is a ...


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Usually the 2SC460 fails after 30 years in the TR2300 synthesizer (Q16). Replace with BF199 of BF 224. I had this issue in at least 3 radios.


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But still, the two currents derived so far for the Q1 collector current (4.3 mA and 14 μA) are too different. No, they are not. The base-emitter junction of Q2 is a diode. The collector load for Q1 is this diode and a resistor in parallel. There is nothing strange about this. As the current increases from 0 mA, eventually the current through the resistor ...


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The voltage between the bases of transistors is highly dependent on the value of the resistors, even if both resistors are of the same value. This is actually the case when the input voltage applied to the midpoint between the diodes approaches the supply rails (large input signal). The problem of static quantities Static resistance. The problem can be that ...


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Bipolar drivers must be biased at middle of entire supply range, which is usually bipolar supplies. Then the Class AB uses the input diode pair to control the output Vbe pair which controls the output DC current. Since the diode is smaller it also has slightly higher internal resistance such that it's current and thus PN voltages are controlled by the equal ...


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The "assuming a beta of 100 for Q1, we have a collector current around 4.3mA" part is wrong. A more correct stance would be "assuming a beta of 100 for Q1, we have a max collector current around 4.3mA", but the collector current is, anyway, limited by the other parts of the circuit. Answered by @dim


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Q1, being an NPN transistor, offers some BJT functionality when the collector is used as an emitter and, the emitter is used as a collector. It doesn't work so well but it does provide some transistor functionality. This is what is happening here. So, when the "new emitter" (previously the collector) is taken to a voltage lower than the "new ...


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The answer from SteveSh is one reason for a slow rise time. Another is that saturation in Q3 will cause it to be slow to turn-off. One improvement that can be done is to add a small capacitor across R7, maybe 100pF. This will cause a negative base current and more quickly turn-off Q3. Also a resistor from Q4 collector to ground, maybe 1k, will help speed up ...


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You need to add an active pull-up stage to the output, rather then just relying on the collector resistor to pull the collector voltage up. Look at what's used in a typical TTL output stage, what's called a totem-pole output structure. Look at this question: TTL, Totem Pole vs. Open Collector Output


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The circuit you have is essentially a common emitter amplifier feeding itself through an LC circuit. This circuit is slightly different due to the RFC in place of a collector resistor but in both cases the tank circuit works in the same way so to simplify matters lets look at this setup with a normal CE amplifier to see what's going on. Image source: https:/...


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simulate this circuit – Schematic created using CircuitLab Figure 1. For the DC/DC converter with the Ctrl input a simple inverter is all that's required. If the Ctrl input has a pull-up then R2 may be omitted. I see that you're using 2.8 V logic so you need to check if the Ctrl input will work at such a low 'logic high' voltage.


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Literally, your circuit is right. No diodes needed. I know what you mean, you thought to add a diode before EN of TMUX1208 to step up a little trigger threshold voltage. Don’t do that, makes no sense. According to the basic circuit theory, if your Arduino is powered by 5V, the the ripple must less than 5%. So here it is, 5V*5%=250mV. But you never get 250mV ...


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You can't easily set the bias point in this circuit. Basically because the gain of the transistor is so high, minor errors in the choice of base voltage (which is around 0.7 V, but not precisely that, the collector current (and therefore V in this circuit) will have wide variations. The appropriate DC value of base V is usually generated using negative ...


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Try a 100 k pull down on VF3 @Andyaka,Thank you. It is working fine now. May I know what is the issue. Missing the load? Basically, without a load on the output of a switch that goes open circuit, there is nothing to collapse the voltage to 0 volts. Hence, you need some form of light load to entice the simulator to do the proper thing. If you look at the ...


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The transistor does not have constant beta. The beta reduces as the collector current reduces which causes the severe distortion where the tops of the waveform are squashed and are not clipping. You need to add negative feedback to reduce the distortion.


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Another answer says: In other words its inconclusive to determine if the transistor is either in the active, saturation or cut-off region. Since there is no reference made with the base terminal. However, the base being open does not prevent us from calculating \$V_{BE}\$. If the base is disconnected, then \$I_{B}=0\$. One can then use the Ebers-Moll ...


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<<< 1 or 2 V between diodes >>> Here what happens ... Ok, it will work ... warmly.


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What you missed here is that this is a principal schematic and does not reflect the complete practical design considerations. In practice, we don't expect the output transistors' VBE voltages to be exactly equal to the diodes' VF voltages, because they aren't and they will never be. And we don't expect the transistors' VBE voltages to be exactly ±0.7V or ...


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It works 'well enough'. The diode V and transistor VBEs are similar enough so the transistors never turn fully off (which is the intent of this circuit). Mathematically, you can show that the product of the diode currents and the product of the transistor currents are equal. What this means (if the diode currents are constant) is the the product of the ...


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I will answer only the first part of your first question because it is very fundamental and deserves special attention: Can the amplified voltage exceed the power supply voltage...? The answer is simple: The amplified voltage cannot exceed the supply voltage because it is a part of the latter. Literally speaking, the "voltage amplification" is ...


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Since I have no technical information on the cameras, I will take a SWAG: Place a resistor in the 5K range in series with the base of the transistor. The emitter is connected to ground and the grounds of the two cameras need to be connected. If connecting grounds causes conflict substitute an opto coupler for the transistor. Then place a 1K resistor in ...


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Not sure what you mean be 'amplification voltage', but in this circuit, if the input V exceeds the supply, nothing more serious than distortion will occur. If the input is extremely large, you would have to worry about power dissipation in the resistors and damage to the input NPN (above 6 V input because of VBE breakdown). The circuit won't be able to ...


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Can the amplification voltage exceed the power supply voltages or might it cause any trouble? It won't exceed the power supply voltages. However, if your small-signal model predicts that the voltage exceeds the rails, then you're well outside the range where your small-signal model is applicable. In reality, your amplifier will have saturated, which is a ...


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