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247 Threads found on edaboard.com: Darlington
The output darlington transistors conduct more when they heat which causes them to conduct more then heat more then conduct more then then heat more .... It is called "thermal runaway". An audio amplifier or opamp use diodes like you did in one circuit or a transistor to replace the diodes to also conduct when they are heated by the darlingtons to
Hi all, I have ORCAD Capture 16.6. In the software i implemented a darlington circuit taken from the example of an electronics book. The picture of the circuit is attached. The problem is that when i run the simulation i get double values of all the currents and voltages in the circuit. Can anyone help me in this regard. 134868[
TIP127 = PNP darlington. A typical use for pullup resistor is to shut off a P device. Whatever is the voltage at its emitter (more positive) terminal, that is the same voltage you should apply to the base. It may be okay to apply a slightly higher voltage, but there is a limit to this. Operating characteristics could be altered.
These are darlington transistors with a voltage drop hardly below 1V. A suitable MOSFET is clearly the better solution. Unfortunately you didn't mention the switched load current, but there are many "logic level" MOSFETs that fully switch on with 2 or 2.5 V Vgs.
It's a boost converter. The two transistors are arranged to increase gain (similar to a darlington or sziklai pair). The capacitor charge and discharges, causing voltage variations at the left-hand transistor. It creates hysteretic action turning the right hand transistor On and Off. The joule thief is a cousin of this circuit. The joule thief h
You find similar circuits in LM317 data sheets. Using a darlington doesn't make much sense, at least for output currents up to 10 A. Better use a regular PNP power transistor for lower voltage drop.
A TIP120 is not a single transistor, it nis two transistors connected as a darlington. Your circuit has nothing to limit the LED current so the LEDs will blow up. The maximum allowed current in a TIP120 is only 5A so you should use a powerful Mosfet for 25A.
Hello!!! i have a circuit whose output is 4.7 v and 143 micro Ampere now i want to drive simple led.. i want to increase current upto 10 mA , what should i do??? can i use darlington pair or else Thanks !!!
if output is collector, Zout is very high only due to leakage // Rc of the current sink. (NPN) so Zout on CB and CE is the same but ends being just the load resistor only. Base Input Zin=hFE*(Rbe + Re) where the latter is chosen to be larger than Rbe. So it can be high with a darlington or low with Re= zero Ohm Emitter impedance, Ze is the
The power darlington transistors need heatsinks. If the output voltage swing is as high as 8V RMS then the output power is 8W into 8 ohms and each darlington heats with almost 2W. The 741 opamp was designed 48 years ago! It is noisy and its slew rate cuts its high level frequencies above only 9kHz. A modern low noise audio opamp works perfectly to
You forgot to tell us how much current you need for the LEDs. The Texas Instruments datasheet for the CD4017 shows that it can typically supply an output current of 15mA into a 2V red LED with a 9V or more supply if you limit the power dissipation of each output transistor to below 100mW. darlington drivers are inverting and have outputs that go to
I have some ic which called sn754410 datasheet: This ic have two h-bridges with darlington transistors. My question is why there is a need in darlington transistors to create these h-bridges? there is no option to use regular transistors? maybe the high-current is the reason? if yes why darling
There is a difference between darlington transistor and darlington pair? What is darlington transistor in fact?
Good grief! You did not notice that many IC amplifiers are boosted like that? Guess what? The BD908 is an ordinary single PNP transistor but the TIP142 is an NPN darlington dual transistor that is completely different. The shorted schematic is from Unisonic in Taiwan (China). The TDA2030 was invented by ST Micro in Italy and their original
That is a darlington circuit, you could use a darlington-output opto and get rid of that external transistor. Not really sure what that diode is for. Sometimes a resistor put there to speed things up, as a darlington is relatively slow to turn off. Maybe it's there to sink leakage current from the input transistor. It's not clear, but (...)
Thanx for the schematic but there are not the same supply i the A and B variant. I did some fault finding yesterday and found a broken power darlington transistor "3-351" and it should be a 2N6053, so now I need to find a supplyer of that or a replacement.
ac coupled darlington inverter amp?
I presume you are referring to the circuit linked in post #1 126146 I would suggest some modifications. If no FET is available, you should at least use a darlington (e.g. BC517 for the NPN configuration) to keep the base current error negligible. Secondly, the output transistor involves a voltage gain for Rsens < 5
125970 Can i use this configuration to drive 12 VDC SSR Relay or 12 VDC Coil Relay ? Please suggest do i need to keep 220R Resistor in between 2 Transistor ? If i use as darlington pair directly will it damage PC817 ? Please suggest.
What does he mean by saying common emitter? Is he mean all emitters of the 7 darlington pairs are all connected together and with the common ground (pin #8)? Exactly. How to use pin #9? For inductive loads, such as the coil of a relay, could act as a protective device that prov
I think you should show a schematic to clarify about your darlington and level conversion problem. Generally, IR2110 can be driven by 3.3V logic signals without additional level conversion. VDD should be connected to 3.3V logic supply in this case. 3.3V logic with 5V VDD will "typically work", but not guaranteed by Vih datasheet specification. T
It is a really old obsolete optosiolator possibly with a darlington output. Closest functional match MAY be a TIL111M... TI, Fairchild Appears to be Signetics part.
The LM195 was basically a NPN common-emitter voltage amplifier with a threshold of 0.6V and an NPN darlington Emitter Follower. It's performance was like a FET with a threshold of Vth=0.6V witha n RdsON of ~1Ω but with an extra diode drop on the output. The advantage of the LM195 was that it was rugged with current limit protection but not
Circuit 1 is unstable without additional feedback loop compensation. Circuit 2 has errors, as already mentioned. Circuit 3 needs darlington transistors or additional current gain for high output current. Consider OP output voltage range and required supply.
What specs? They usually dont have specs on Ebay Maybe they meant surge current for 1ms. ...arg. Yes the Load regulation is essential the inverse of the output ESR where 50% load regulation uses a load equal to ESR of source. Active Load 1) Use a power darlington transistor and bias that easily with any 10k pot with gain from 1k to
A lock-in amplifier or synchronous detector is a simple AC-coupled opamp with a key operated from your refetence source (555). There are many good versions available, check google. Analog Devices offers one IC designed as the sync detector but is expensive. I used a LM324 quad opamp with CD4016 or CD4066 quad key in my circuits. After your dar
121142 i can't able to measure collector current using ammeter is there anyway or any circuit to measure?
I would recommend, if you can stand a fairly poor frequency response and have the headroom (you do not say about the input voltage range, which is a key design parameter) a single supply op amp driving the gate of a PMOSFET (if VIN > 8V) or maybe a NPN darlington (if you have an AUX supply but VIN is lower) or NMOSFET (if AUX > 8V and VINmin can be
My past experience of multiple transistor 'darlington' configurations in big PSUs is not good. Under short circuit, there is a risk (especially with slow op-amps) of excessive current causing failure of one of the drivers, BC337 & 2SC1173 in this design, which results in unregulated voltage and current flowing to the load. I would have some concer
To make some sense, assuming that R1 acts as the probe sensor, you should replace it by a smaller value in the range of some houndred kohms, as well tie Q1 emmiter to GND. The voltage measured at Q1 collector could be understood like the water detection. Obviously, Q1 should have a high gain, prefereably the darlington type. In my oppinion, you
Using the transistor as a saturated switch means knowing the Ic:Ib ratio. In this case it is limited by the R ratios. 100k/330R Reverse the LDR and R and use a darlington and adjust R for threshold, so that LDR conduicts current to drop 1.4V across R at required threshold.
Most opamps have emitter-follower darlington output transistors. Then they do not saturate, instead their emitter output follows the base with a 0.7V to 0.8V voltage drop. A darlington is two emitter-followers in series so of course each one produces a 1.5V voltage drop. Then with a plus and minus 5V supply the maximum output swing is plus and minu
First of all, a power transistor may be a darlington, or not. And a darlington may be a power transistor, or not. Secondly, I think you've misread the data sheet: Vbe for the transistor is not 5V. Maybe the MAXIMUM REVERSE Vbe is 5V. Third, what do you mean "told to go to 0V"???? The minimum output of that regulator is 1.2V. You need to show
Yes it can be used. All you need is a microphone, couple of transistors, a capacitor and a pot for adjusting dB level. Connect two transistors, say BC547, as a darlington pair, connect a pot of 100k or 500k with vcc to mic, then mic to gnd. Feed the pot output to a ceramic cap of 10nF, then to the darlington pair gate, and
The logic high output voltage from a very old 7408 TTL logic IC is much too low for your circuit where the darlington emitter-followers reduce the output voltage even more. It also has much too low current to saturate your darlingtons. The datasheet for the 7408 shows an output high of only 2.4V to 3.6V at a maximum of only 0.8mA.
Hello. I'm having difficulties deducing which addition of what type of transistor is the best option for me while enabling a variable linear regulator to use its max 50mA output to be used as a max 5A output. Here are two examples I know of, the NPN darlington is the one I went with first and have it on a prototype in front of me but I want t
EDITS: We cannot see what the emitter of the BC557 is connected to. Maybe you think the collector is called the emitter? darlington Q4 has its emitter and collector pins connected backwards. Please post the entire circuit showing what the emitter of the BC557 connects to
50W Class A darlington will be ~ 60% efficient with transformer or L with C coupling, tuned to f. Choose Q for 50R. ... but this is distorted.... due to variation in hFE with I. So depends on needs. thus use gain 50x and complementary x1 buffer. Does it need to be 50R output impedance? if so then V+=100V For low distortion use 2 stage compleme
The open loop gain of TL431 is quite small (it's the gain of the output darlington BJT configuration) so there's no need for pure resistive negative feedback to prevent oscillations. There's also an internal compensation (a 20 pF capacitor). How the gain has been set in the above error amplifier(TL431)? If you want a controlled
This is the basic idea I would replace the relay with the load This is for longer delay use a higher power transistor or power darlington there are many commercial modules - use google or bing;
What you need is a current amplifier as you micro can only deliver 10 mA and you need 5A (or more). So you need a FET or a darlington power transistor. Make sure to put a diode across the motor to suppress back EMF. I think you need a PI controller ( ). A good test would be to see wha
We think your old BDX53 darlington transistor might be destroyed because its voltage rating of only 45V is too low for your supply of 24V where its collector can easily exceed 48V. Its datasheet also shows that for it to saturate with a collector current of 3A then its base current must be at least 12mA. If the supply voltage of your PIC is 5V then
hello specsheet data The collector-current rating of each darlington pair is 500 mA. and , for global comsumption Total substrate-terminal current . . . .. . ?2.5 A you have 8x200 mA => 1,6A ! Could be OK .. Take care about over voltage.. (Max 50V) Add anti-retour diodes accross Solenoides valves .. or use "COM
You are designing a driver for an existing ignition coil. If I understand right, the motivation to replace the darlington with a MOSFET is mainly reduction of on-state losses. The darlington circuit already limits the output voltage by the Vce breakdown voltage of both transistors. In this relation, the MOSFET circuit can easily achieve better resu
The 'big' black thing is a darlington power transistor, the little black thing is a microcontroller. What it does is anyones guess. With it removed, what doesn't work on the bike? The transistor is rated at over 300V so I would guess it's part of a capacitor discharge ignition unit. Kam1787, you didn't mention the kafuffle clamp diode or the cros
Didn't you read the datasheet for the TPS7A4701? You might have destroyed it with an input voltage higher than the maximum allowed input of 36V and a maximum input of 35V is recommended. The datasheet does not show a slow darlington buffer transistor like you have. I think it will oscillate. Your voltage setting resistors have extremely low values.
darlington's have the disadvantage of almost 2 diode drops initially ( actually 1 diode drop & 1 saturation drop) but when conducting heavy currents with fast pulses in PWM due to low input capacitance compared to MOSFETS will have better current gain( ~100) . When the transistor is used as a switch and saturates, the hFE typically reduces to o
Does using the 4049 or 4069 hex inverters avoid the darlington problem? Why not look at the datasheet for the CD4069 hex inverters IC? When it has a 5V supply its minimum output current is only about 1.5mA. It can drive a ULN2803A. The input resistance of the CD4069 is infinite so you would use a high value resistor
Although the values doesn't exceed the absolute maximum ratings, there's no guaranteed maximum voltage drop for PIC IO-pins with 20 mA load current. An external driver is recommended. The standard solution is a BJT transistor (no darlington because it has too high voltage drop for a 5V relay) with base series resistor or a MOSFET, together with
I assume you also have current limiting resistors in each chain of LEDS. Try connecting a resistor (~ 15K) between the base and emitter. The TIP122 is a darlington with quite high leakage current which could be enough to light the LEDS. Brian.