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I have a very newbie and possibly stupid question, but I have been searching many sites and just can't find an answer.

Imagine you have a circuit, some device (electro-motor) that is powered with for example 20V dc power supply. Then you have another TTL circuit.

I am able to programatically change the output on TTL circuit to 0 (0,8V) or 1 (5V), but what I need to do is control the second circuit, using this digital output, that means, for example if I send 1 to output, the 20V device is switched on, when I send 0, it's turned off.

What is a simplest way to achieve this?

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2 Answers 2

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This is a very classic case to use a transistor as a switch.

Example:
enter image description here
Simulate circuit

  • The 300 Ohm resistor would be your load
  • The 1k resistor limits the base current
  • The 50Hz source simulates your 5V TTL output

Note that there are some things to consider when dimensioning the circuit. The transistor must be chosen to satisfy the collector current requirements. There are some other considerations, but this answer is not about transistor circuits in detail. There a plenty of resources in the WWW. Search for "transistor as a switch" and "emitter follower".

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  • \$\begingroup\$ I am still confused about 1 thing, I am testing this circuit on qucs and it tells me there is 20V on load even if there is a very low voltage (0.1V which represent logical 0 in TTL) on base. Is this some kind of bug in qucs or there is something more in this... \$\endgroup\$
    – Petr
    Commented Nov 23, 2013 at 21:58
  • \$\begingroup\$ @Petr: Given my example, the 20V will always be there. There is just no current flow "possible" as long as the transistor is "switched off" (base voltage at least about 1V). Or what do you mean? \$\endgroup\$
    – Rev
    Commented Nov 23, 2013 at 22:03
  • \$\begingroup\$ now I see I am getting 20V on 0V on base and nearly 0V with 5V on base, that explains this, so in this circuit logical 0 turns the secondary circuit on, that confused me \$\endgroup\$
    – Petr
    Commented Nov 23, 2013 at 22:04
  • \$\begingroup\$ What do you use to simulate the "load"? If there is no resistance, the emitter could indeed read 0V (and the transistor would blow up in real life). \$\endgroup\$
    – Rev
    Commented Nov 23, 2013 at 22:11
  • \$\begingroup\$ I created exactly the same schema as you sent here, there is 300 ohm resistor, when I put 0 - 0.8V on ttl output I see 19.5V on load, with 2 - 5V on ttl I see 0.1 V on load. This is basically what I needed. \$\endgroup\$
    – Petr
    Commented Nov 23, 2013 at 22:18
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A "high side" switch handles high current from low logic control to higher voltage output.

These are very low cost, smart , and tiny SMD which may be soldered with a tiny oven. There are many similar alternatives.

If this is too hard, look at how it is made in the schematic. http://www.fairchildsemi.com/ds/FP/FPF1203.pdf

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  • \$\begingroup\$ I have no idea if this answer is useful to me, but I would like to know why it got downvoted -2? Is it totally irrelevant or something? \$\endgroup\$
    – Petr
    Commented Nov 23, 2013 at 19:24
  • \$\begingroup\$ @Petr: The answer is somewhat confusing and bad choice of words. The second sentence is irrelevant to your question. Maybe the language barrier. \$\endgroup\$
    – Rev
    Commented Nov 23, 2013 at 21:58

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